Metric and topological foundations

Moving-metric arguments need uniform estimates for positive forms, smooth cutoffs, finite-dimensional measure changes, and locally finite sums. This chapter proves the needed forms of those facts, including their endpoints. It leads to Localizing symbols with moving metrics. The elementary calculus is proved in Section 13, arbitrary-metric compactness and the exact support arguments in Section 14, and the measure-theory entries in the linked Banach chapter. Basic references for those foundations are [Lebl] and [Axler]; the arguments developed here are complete relative to those stated prerequisites.

1. Entry facts and maximality

Sections 12.1–12.3 construct the complete ordered real field, prove its Archimedean property and compare it exactly with the original real numbers. The underlying set operations, natural-number arithmetic, induction and choice axioms are explicit. We use the finite-dimensional bases, matrix algebra, determinant identities, and elementary elimination proved in Section 10 of Polynomial and contour interfaces for stable boundary models; and the construction of complex numbers as pairs of real numbers. Sections 13.1–13.8 prove limits, derivatives, the mean value theorem, the exponential series and its derivative, real-power differentiation, and the Riemann integral of a continuous function on a compact interval, with order, linearity and its uniform bound. Sections 13.5–13.6 give the full finite-coordinate derivative, ordered higher product and Taylor maps used in Section 3.

Sections 15.0 and 16.1 of Banach estimates and measure foundations prove one-dimensional Lebesgue measure built from interval outer measure, completion, measurable functions, nonnegative integration, monotone convergence and bounded convergence on finite-measure spaces. Limits of measurable functions are measurable. For product spaces, Sections 16.2–16.5 of that chapter construct the full sigma-finite product measure, prove both Tonelli orders and prove the completed-factor comparison used in Section 6. The affine change-of-variables result is proved there, not assumed. The real-number construction is proved in Section 12; the measure and integral entry uses only the independent measure construction and convergence proofs, rather than a larger theorem that assumes Tonelli.

The maximality principle used below is Zorn’s lemma in this form: a nonempty partially ordered set in which every chain has an upper bound has a maximal element. For disjoint ellipsoid families, the union of a chain is again disjoint, because any two members belong together to one of its comparable families. The empty family makes the partially ordered set nonempty. This selection does not itself establish countability; Section 4 proves that separately.

2. Positive forms and tensor norms

Let $E$ be real of dimension $d\geq1$, as in the moving-metric lesson, and let $Q$ be a positive-definite quadratic form. Its polarization is the symmetric bilinear form $$\langle x,y\rangle_Q=\tfrac14\bigl(Q(x+y)-Q(x-y)\bigr).$$ In coordinates $Q(x)=x^{\mathsf T}Gx$ with $G$ symmetric; expansion shows the displayed polarization is $x^{\mathsf T}Gy$, and $\langle x,x\rangle_Q=Q(x)$. Starting with a basis, subtract its projections on previously constructed orthonormal vectors and divide the nonzero residual by its positive $Q$-length. The residual cannot be zero, since that would put the original basis vector in the span of its predecessors. This finite Gram–Schmidt procedure gives a $Q$-orthonormal basis. Section 10.5 of the polynomial and contour lesson retains every original residual, its length, the full original Gram matrix and both coordinate maps for this comparison; its Section 10.10 proves the exact receiving metric and volume factors.

For $y\ne0$, nonnegativity of $Q(x-t y)$ at $t=\langle x,y\rangle_Q/Q(y)$ gives $|\langle x,y\rangle_Q|^2\leq Q(x)Q(y)$; the case $y=0$ is immediate. Expanding $Q(x+y)$ and applying this inequality proves the triangle inequality for $\|x\|_Q=\sqrt{Q(x)}$. Thus it is a norm.

Let $T:E^k\to\mathbb C$ be multilinear, and let $M$ be the largest absolute value of its coefficients in a $Q$-orthonormal basis. Its operator tensor norm obeys $$M\leq\|T\|_Q\leq d^{k/2}M.$$ The first inequality evaluates $T$ on basis vectors. For the second, expand each unit vector $u_j=\sum_a u_{ja}e_a$. The absolute value of the resulting sum is bounded by $M\prod_j\sum_a|u_{ja}|\leq M d^{k/2}$, by the scalar Cauchy–Schwarz inequality. For $k=0$, both sides are simply the absolute value of the scalar and $d^0=1$.

If $cQ\leq P\leq CQ$, where $0<c\leq C<\infty$, comparison of the two unit balls in every tensor argument gives $$C^{-k/2}\|T\|_Q\leq\|T\|_P\leq c^{-k/2}\|T\|_Q.$$ These comparison factors are exact. They require no uniform condition-number bound over a family of forms.

3. Differentiation and segment integrals

Sections 13.1–13.6 prove the finite-dimensional derivative definition, the first-order chain rule, the continuous-partials criterion, and interchange of partials for $C^2$ maps. For $C^k$ maps we use the definition by continuous iterated derivatives. The chain rule requires differentiability at a point and at its image; the interchange theorem requires $C^2$. All domains are open subsets of finite-dimensional real spaces. These local statements restrict to smaller neighborhoods. [Lebl] gives a named reference. We now derive the vector-target and higher-order forms used below.

For one real variable with a vector target, apply the scalar derivative limit to every component. There are finitely many components, so the Euclidean norm of their remainder divided by the absolute value of the increment tends to zero. Continuity of the resulting derivative is likewise componentwise. Complex scalar and finite matrix targets follow by identifying them with their finitely many real coordinates.

For clarity, the higher-order and product interfaces need no further theorem. A fixed bilinear map $B$ on finite coordinate spaces satisfies $$D B(f,g)[h]=B(Df[h],g)+B(f,Dg[h]).$$ Indeed, expand the increments bilinearly. The term containing both increments is $O(|h|^2)$, and the two differentiability remainders are $o(|h|)$. Coordinate multiplication and ordered matrix multiplication are such bilinear maps. Applying the first-order rule repeatedly proves the iterated product rule, without interchanging matrix factors. Applying the chain rule repeatedly proves the iterated composition rules. Applying the continuous-partials criterion to each derivative’s finite coordinate list identifies $D^j f$ with a continuous $j$-linear tensor for every $j\leq k$. Adjacent derivative orders may be exchanged by the $C^2$ interchange theorem applied to lower derivatives; every permutation is a finite product of adjacent exchanges. This gives symmetry at every allowed order, including complex and matrix-valued functions.

If a function is continuous on $[a,b]$, differentiable in its interior, and has a Riemann-integrable derivative there, the fundamental theorem of calculus identifies its increment with the integral of that derivative. The oriented integral handles a negative increment. [Lebl] is a reference for this elementary theorem. A $C^1$ coordinate segment meets those hypotheses. Apply it componentwise to obtain $$f(x+t e_j)-f(x)=\int_0^t\partial_j f(x+s e_j)\,ds$$ whenever the segment lies in the coordinate box; for $t<0$, use the oriented integral. If $f_n$ and $\partial_jf_n$ converge uniformly on that segment, the integral error is at most $|t|\sup|\partial_jf_n-g_j|$, so the identity passes to the limit. Repeat for each derivative order. No exchange of an uncontrolled pointwise limit with an integral is asserted.

4. Countability, compactness and supports

Rational points are countable because the integer numerator and positive integer denominator can be enumerated in finite diagonal batches. Finite tuples are likewise countable. By the Archimedean property, for $a<b$ choose an integer $N$ with $N(b-a)>1$; an integer strictly between $Na$ and $Nb$ then gives a rational in $(a,b)$. Every nonempty Euclidean open set contains an open coordinate box, hence a rational-coordinate point. Fix an enumeration and assign to each member of a pairwise disjoint family of nonempty open sets its first contained point. The assignment is injective; the family is countable. The countability argument uses no regularity of a varying metric.

Sections 12.4–12.8 prove both the open-cover and sequential forms of compactness, their equivalence with closed and bounded sets, all required extrema and the countability statements below. These apply to the Euclidean spaces here, including the empty set, whose empty open subcover is finite. Section 14.1 gives the full arbitrary-metric cover/sequential equivalence using explicit countable choice, with no maximality assumption; Section 14.2 gives the common compact-support neighborhood. [Lebl] is a reference for these foundational theorems.

For a direct finite-dimensional sequence argument, begin with a bounded sequence in $\mathbb R^d$. Extract a subsequence convergent in the first coordinate, a subsequence of that one convergent in the second, and continue for exactly $d$ steps. Each previously convergent coordinate remains convergent. The final subsequence converges in Euclidean norm, since the sum of its finitely many squared coordinate differences tends to zero. A closed set contains the limit. The scalar subsequence fact follows from real completeness: repeatedly bisect a bounded closed interval, retain a half containing infinitely many terms, and choose increasing indices in the nested halves. Their lengths tend to zero and their common point is the subsequential limit. This argument works in every finite dimension.

A fixed positive form is a continuous coordinate polynomial. Its unit sphere comparison, or its orthonormal-coordinate isomorphism, shows that a closed frozen ellipsoid is bounded; it is closed by continuity and hence compact. A finite union of compact sets is compact by taking a finite subcover for each member. Empty sets cause no exception.

For a continuous function $f$, define $\operatorname{supp}f$ as the closure of its nonzero set. Outside that support it vanishes on a neighborhood, so all existing derivatives vanish there. If a family of sets is locally finite, cover a compact $K$ by neighborhoods each meeting only finitely many members. A finite subcover meets only finitely many members in total, both on $K$ and on the union of those neighborhoods. For empty $K$, take the empty neighborhood. These observations give a neighborhood conclusion, rather than merely pointwise finiteness.

5. Cutoffs and locally finite sums

Define $\eta(s)=0$ for $s\leq0$ and $\eta(s)=e^{-1/s}$ for $s>0$. Each derivative on the positive half-line is $e^{-1/s}$ times a polynomial in $1/s$, by induction. For every fixed integer $N$, $u^N e^{-u}\to0$ as $u\to\infty$: use $e^u\geq u^{N+1}/(N+1)!$. Thus every one-sided derivative tends to zero at the origin. The derivative there is also zero at every order, since dividing such an expression by $s$ still has the same vanishing property. Induction proves $\eta\in C^\infty(\mathbb R)$.

The denominator in $$\chi(s)=\frac{\eta(3/2-s)}{\eta(3/2-s)+\eta(s-1)}$$ is strictly positive for every real $s$. The function is smooth, lies between zero and one, equals one for $s\leq1$, and is zero for $s\geq3/2$. On $[0,\infty)$ its support is contained in $[0,3/2]$, a compact subset of both $[0,2)$ and $[0,4)$. The function $x\mapsto\chi(|x|^2)$ is a smooth compact cutoff equal to one near the origin. Each fixed derivative is bounded by continuity on a compact support, and vanishes outside it. Smoothness of the quotient follows from the single-variable reciprocal rule and Section 3. The strict support endpoints are retained.

Suppose each point has a neighborhood on which all but finitely many smooth functions $f_i$ vanish identically. On that neighborhood their pointwise sum is that finite sum, so it is smooth and its derivatives equal the finite sums of derivatives. The local descriptions agree because they are derivatives of the same function. No convergence argument or measurability assumption on the varying metric or weight is needed.

6. Lebesgue measure under affine maps

We use Tonelli’s theorem in its full $\sigma$-finite form: for a nonnegative function measurable for a product $\sigma$-algebra, its product integral equals either iterated integral, with $+\infty$ allowed. Product sections of Borel subsets of Euclidean space are Borel. [Axler] is a reference for the product-measure theorem. Sections 16.1–16.4 of Banach estimates and measure foundations prove that full statement; Section 16.5 proves its completed-factor version and exact original Euclidean comparison. The proof below supplies the affine change-of-variables result from this stated measure-theory prerequisite; it does not assume that result.

For a finite family of measurable rectangles, partition each factor by the finitely many membership patterns determined by their sides. Products of these disjoint pieces refine every rectangle simultaneously. In Euclidean spaces, the product sigma-algebra agrees with the Borel sigma-algebra of the product: the rational-coordinate open rectangles form a countable basis and belong to the product sigma-algebra; conversely each Borel rectangle is the intersection of two continuous-projection preimages. For each fixed coordinate in one factor, the collection of product sets whose section is Borel is a sigma-algebra containing every Borel rectangle. Hence every Borel product set has Borel sections. These are the section facts used in the affine calculation.

We derive the full completed Lebesgue affine-change statement. Let $m_d$ be the completion of the Borel product of $d$ copies of one-dimensional Lebesgue measure. First, interval coverings show for every subset $S\subseteq\mathbb R$, every $a\ne0$ and every $b\in\mathbb R$, $$m_1^*(aS+b)=|a|m_1^*(S).$$ Transform a countable interval cover to prove one inequality; transform back to prove the other. The equality, applied also to intersections and complements, preserves the outer-measure measurability criterion. It therefore gives one-dimensional translation and scaling for measurable sets. The corresponding substitution for integrals of nonnegative functions follows first for indicators, then finite nonnegative simple functions, then increasing simple approximations by monotone convergence.

For a Borel subset $S\subseteq\mathbb R^d$, Tonelli applied to $1_S$ shows that a coordinate translation preserves its product measure, a nonzero scaling of one coordinate multiplies it by the absolute value of that scaling, and a permutation of coordinates preserves it. A shear $x_i\mapsto x_i+c x_j$, with $i\ne j$, also preserves the measure: hold the other coordinates fixed and integrate first in $x_i$; its section is simply translated by the fixed number $c x_j$. Every section used here is Borel, and the coordinate measures are $\sigma$-finite. These arguments remain valid for sets of infinite measure.

An invertible real matrix is a finite product of the elementary scaling, shear and coordinate-interchange matrices, by Gaussian elimination. Their absolute determinants are respectively $|a|$, $1$ and $1$. Applying the preceding identities successively and using determinant multiplicativity gives $$m_d(AS)=|\det A|m_d(S)$$ for every Borel $S$ and every invertible $A$. The same coordinate argument gives translation invariance. Invertible affine maps and their inverses are continuous, so they preserve Borel sets.

To include every Lebesgue-measurable $S$, use the completion description: there is a Borel $B$ and a Borel null set $N$ with $S\mathbin{\triangle}B\subseteq N$. The Borel formula makes $AN$ null and Borel, and $AS\mathbin{\triangle}AB\subseteq AN$. Completeness therefore makes $AS$ measurable with the same measure as $AB$, proving the formula for $S$. Translation is handled identically. Finite additivity and monotonicity are immediate measure properties; no varying family is integrated.

Finally choose a linear isomorphism $L:\mathbb R^d\to E$ carrying Euclidean coordinates to a $Q$-orthonormal basis. In fixed volume coordinates on $E$, $$\{x:Q(x)<r^2\}=L\{u:|u|<r\},\qquad m_d(\{x:Q(x)<r^2\})=r^d v_Q\quad(r>0),$$ where $v_Q=|\det L|m_d(\{|u|<1\})$. The unit ball contains a nonempty coordinate cube and is contained in a bounded cube, so the original coordinate-box volume and monotonicity give $0<v_Q<\infty$. This is the exact finite packing input. It imposes no smoothness or measurability on $X\mapsto g_X$ or on the weight.

7. Uniform limits

A Cauchy sequence in $\mathbb C$ has Cauchy real and imaginary parts, which converge by real completeness. The same argument in each of finitely many coordinates proves completeness of finite-dimensional coordinate spaces. If a sequence of functions is uniformly Cauchy on a set, its pointwise limits exist in such a space. Given $\varepsilon>0$, choose an index after which all pairwise differences are at most $\varepsilon$ uniformly, then pass one index to the pointwise limit. The resulting bound remains uniform, proving uniform convergence.

If the original functions are continuous, choose one with uniform error less than $\varepsilon/3$, and use its continuity near any chosen point to make its change less than $\varepsilon/3$. The triangle inequality bounds the change of the limit by $\varepsilon$. Thus the limit is continuous; in particular this holds on each compact set. The tensor coefficient estimate of Section 2 gives convergence in each fixed tensor norm from coefficient convergence, and its reverse inequality gives the converse. The integral passage for these uniform limits follows from the segment estimate in Section 3.

8. A metric for countably many seminorms

Let $V$ be a real or complex vector space with a countable separating family of seminorms $(p_k)_{k\geq0}$. Define $$d(x,y)=\sum_{k=0}^\infty 2^{-k-1}\min\{1,p_k(x-y)\}.$$ The series converges because it is bounded termwise by a summable geometric series. Symmetry and nonnegativity are immediate, and separation makes $d(x,y)=0$ imply $x=y$. Subadditivity of each summand follows from the seminorm triangle inequality and $\min(1,s+t)\leq\min(1,s)+\min(1,t)$. Summing proves the triangle inequality. Thus $d$ is a translation-invariant metric.

The topology specified by the seminorms has zero-neighborhoods given by finite intersections $p_{k_j}(x)<\varepsilon_j$. For a fixed $k$, if $d(x,0)<2^{-k-1}\min(1,\varepsilon)$, then $p_k(x)<\varepsilon$; this also works when $\varepsilon\geq1$. Taking a minimum over finitely many indices produces a metric ball within any such neighborhood. Conversely, given a metric radius $\varepsilon>0$, choose $N$ with $\sum_{k>N}2^{-k-1}<\varepsilon/2$, and require $p_k(x)<\varepsilon/2$ for $0\leq k\leq N$. The initial sum is less than $\varepsilon/2$, so this finite seminorm neighborhood lies in the metric ball. The topologies coincide.

Finite intersections of seminorm balls are convex and balanced. Addition is continuous by subadditivity. Scalar multiplication is continuous because, at $(a_0,x_0)$, $$p_k(ax-a_0x_0)\leq |a|p_k(x-x_0)+|a-a_0|p_k(x_0),$$ and $|a|$ is bounded on a sufficiently small neighborhood of $a_0$. Apply these estimates to each of finitely many seminorm conditions. Hence the topology is a Hausdorff locally convex vector-space topology.

The same head-and-tail estimates, applied to differences of sequence terms, show that a sequence is metric Cauchy exactly when it is Cauchy in every seminorm. Applying them to differences from a fixed vector gives the corresponding convergence assertion. Therefore the metric is complete precisely when every sequence Cauchy in all seminorms has a vector limit in all seminorms. No completeness of $V$ has been assumed in establishing this metric. Section 6 of Localizing symbols with moving metrics applies it to a symbol space.

9. Local inverses and matrix series

If $f=u+iv$ is a nonvanishing complex $C^k$ function on a real domain, then locally $$\frac1f=\frac{u-iv}{u^2+v^2}.$$ The denominator stays positive on a sufficiently small neighborhood. Repeated single-variable differentiation gives $(s^{-1})^{(j)}=(-1)^j j!s^{-j-1}$ for $s>0$. The rules of Section 3 therefore prove the asserted $C^k$ regularity, including $k=0$ by continuity. For a matrix function taking invertible values, its determinant is nonzero and its adjugate is a finite polynomial in its entries. The identity $A^{-1}=\operatorname{adj}(A)/\det A$ proves the corresponding local matrix assertion without any commuting-coefficient assumption.

For induced operator norms, $\|ABx\|\leq\|A\|\|B\|\|x\|$, so $\|AB\|\leq\|A\|\|B\|$. In orthonormal coordinates the largest absolute matrix coefficient is at most its operator norm, and the operator norm is at most $n$ times that coefficient maximum for an $n$-by-$n$ matrix, by two scalar Cauchy–Schwarz estimates. Thus coordinate completeness implies matrix norm completeness. The zero-dimensional matrix space has its single element and satisfies the same assertions separately. This proves matrix norm completeness.

Suppose $\|B\|\leq q<1$, with $q\geq0$, and put $S_N=\sum_{j=0}^N(-B)^j$. For $M>N$, $$\|S_M-S_N\|\leq\sum_{j=N+1}^M q^j\leq\frac{q^{N+1}}{1-q}.$$ If $q=0$, then $B=0$ and the conclusion is immediate; otherwise this bound tends to zero. Matrix completeness gives a limit $S$, and the norm bound is $\|S\|\leq(1-q)^{-1}$. Direct finite multiplication yields $$(I+B)S_N=S_N(I+B)=I-(-B)^{N+1}.$$ Continuity of multiplication and $\|B^{N+1}\|\leq q^{N+1}$ give both inverse identities for $S$. The conclusion holds for every contraction constant, including $q=1/2$. No derivative of an infinite series is taken.

10. Frequency brackets and real powers

For $\xi\in\mathbb R^n$, direct differentiation gives $$\nabla\langle\xi\rangle=\frac{\xi}{\sqrt{1+|\xi|^2}},\qquad |\nabla\langle\xi\rangle|\leq1.$$ Apply the segment fundamental theorem to $t\mapsto\langle\eta+t(\xi-\eta)\rangle$, use the chain rule and Cauchy–Schwarz, and integrate on $[0,1]$. The result is $$|\langle\xi\rangle-\langle\eta\rangle|\leq|\xi-\eta|.$$ For a fixed $s\in\mathbb R$ and $0<a\leq r\leq b<\infty$, monotonicity of the real power, reversed when $s<0$, gives $$\min(a^s,b^s)\leq r^s\leq\max(a^s,b^s).$$ Both endpoints are finite and positive, and $s=0$ gives one. This holds for arbitrary real exponents, without an integer or sign restriction. If a zero-dimensional coordinate factor occurs, its bracket is the constant one and the estimate is immediate.

11. Exercises and solutions

Exercise 11.1 (An incomplete seminorm space). Let $V=c_{00}$ be the vector space of complex sequences with finite support, and let $p_k(x)=|x_k|$ for $k\geq0$. Show that the metric in Section 8 is not complete.

Solution. Put $x^{(N)}=\sum_{k=0}^{N}2^{-k}e_k$, where $e_k$ is the $k$-th coordinate vector. For each fixed $k$, $p_k(x^{(M)}-x^{(N)})=0$ once both indices exceed $k$. The head-and-tail estimate in Section 8 makes the sequence metric Cauchy. If it converged to $x\in c_{00}$, convergence in every $p_k$ would force $x_k=2^{-k}$ for all $k$. That sequence has infinite support, a contradiction. Thus countably many separating seminorms need not give a complete space.

Exercise 11.2 (A shear). Let $S=[0,1]^2$ and $A(x_1,x_2)=(x_1+c x_2,x_2)$, with $c\in\mathbb R$. Compute the area of $AS$ directly by sections.

Solution. At a fixed $x_2=t\in[0,1]$, the horizontal section of $AS$ is $[ct,ct+1]$, of one-dimensional length one. Outside that range of $t$ the section is empty. Tonelli therefore gives $m_2(AS)=\int_0^1 1\,dt=1$, in agreement with $|\det A|=1$.

12. Real numbers and finite-dimensional topology

The real field is constructed here with its full arithmetic, order and completeness. Its exact comparison with the real numbers used in this course retains the original values. The logical entry is natural-number arithmetic with induction, ordinary set theory and countable choice. The last two sections apply the construction to the original norms, quadratic forms, polynomial minima and spectral contours. These are standard foundations; no novelty is claimed.

12.1. Integer and rational arithmetic with all original fractions retained

Start with natural numbers and their addition, multiplication, order, cancellation and induction. Construct the integers as classes of pairs $(m,n)\in\mathbb N^2$, with $$(m,n)\sim(m',n')\quad\Longleftrightarrow\quad m+n'=m'+n. \tag{RT1}$$ Reflexivity and symmetry follow from equality. For transitivity, add the two defining equalities for consecutive pairs and cancel the common middle sum in $\mathbb N$. Addition and multiplication are $$[(m,n)]+[(p,q)]=[(m+p,n+q)],\qquad [(m,n)][(p,q)]=[(mp+nq,mq+np)]. \tag{RT2}$$ Adding equalities in (RT1) proves addition independent of representatives. For multiplication, replacement of $(m,n)$ by $(m',n')$ follows by multiplying $m+n'=m'+n$ by $p$ and $q$: the equality of the cross sums in (RT1) is the sum of those two equations, with terms reordered. Replacement of $(p,q)$ follows by the same explicit operation with $m,n$; hence replacing both pairs is valid. The additive inverse is $[(n,m)]$; both inverse sums are $[(m+n,m+n)]=0$. The identities are $0=[(0,0)]$ and $1=[(1,0)]$.

Addition is associative and commutative by coordinate addition. Multiplication is commutative by its two coordinate formulas. For associativity, the coordinates of the left associated triple product are $$\begin{gathered} mpr+nqr+mqs+nps,\\ mps+nqs+mqr+npr. \end{gathered} \tag{RT3}$$ The right associated product has these same four terms in each coordinate, in the respective order $mpr,mqs,nps,nqr$ and $mps,mqr,npr,nqs$. Distributivity follows by expanding (RT2) against $(p+r,q+s)$: the first coordinate is $mp+mr+nq+ns$, and the second is $mq+ms+np+nr$, which are the coordinates of the sum of the two products. Thus every ring identity used here is proved.

Declare $[(m,n)]>0$ when $m>n$. Equation (RT1) and cancellation preserve this comparison. Exactly one of positive, zero and negative holds. Two positive pairs have $m=n+k$, $p=q+l$ for positive natural numbers $k,l$; their product’s first coordinate exceeds its second by $kl>0$, on expanding all terms in (RT2). Their sum is positive by addition of inequalities. It follows that this is an ordered ring. A nonzero integer and another nonzero integer have nonzero product: apply the positive-product assertion to their respective signs. In particular, multiplication by a nonzero integer permits cancellation.

Construct the rational numbers from pairs $(p,q)\in\mathbb Z^2$ with $q\ne0$, without restricting the denominator’s sign or reducing the fraction. Put $$\begin{gathered} (p,q)\sim(p',q')\ \Longleftrightarrow\ pq'=p'q,\\ [p,q]+[p',q']=[pq'+p'q,qq'],\qquad [p,q][p',q']=[pp',qq'],\\ -[p,q]=[-p,q],\quad 0=[0,1],\quad 1=[1,1],\quad [p,q]^{-1}=[q,p]\quad(p\ne0). \end{gathered} \tag{RT4}$$ Transitivity follows by multiplying $pq'=p'q$ by $q''$ and $p'q''=p''q'$ by $q$, then cancelling the nonzero $q'$. For sums and products, cross multiplication after changing either representative expands into the defining integer equality times the unchanged numerator or denominator. Hence both operations and negation descend. Associativity, distributivity and commutativity follow by cross multiplication into the already proved integer identities: denominators are the full original products, and the numerator of a sum of three terms is $pq'q''+p'qq''+p''qq'$. The product $([p,q]+[p',q'])[r,s]$ has the original numerator $(pq'+p'q)r$ and denominator $qq's$. Adding the two separate products by (RT4) instead gives the full pair $[prq's+p'rqs,qq's^2]$. These pairs have identical cross products, namely $(pq'+p'q)r\,qq's^2=(prq's+p'rqs)\,qq's$; each side is the same full integer sum. Every factor $s$ and $s^2$, including its sign, remains in this comparison. Both inverse products in (RT4) are $[pq,qp]=1$, with $p,q\ne0$. This proves the field laws.

Declare $[p,q]>0$ precisely when $pq>0$. If $pq'=p'q$, multiply by $qq'$ to obtain $pq(q')^2=p'q'q^2$. Both denominator squares are positive, so signs agree. The numerator-times-denominator of a sum of two positive fractions is $$(pq'+p'q)(qq')=(pq)(q')^2+(p'q')q^2>0. \tag{RT5}$$ The product has numerator-times-denominator $(pq)(p'q')>0$. Trichotomy follows from the integer signs and nonzero denominator. The embedding $z\mapsto[z,1]$ preserves all operations and order and is injective. Thus $\mathbb Q$ is an ordered field with its full original numerator and denominator data.

For $q\ne0$, $|q|\ge1$, so $|[p,q]|\le |p|$. The integer $|p|+1$ is a strict upper bound. This proves the rational Archimedean property. Between rationals $a<b$, the rational $(a+b)/2$ is strictly between them, since its differences from either endpoint are $(b-a)/2>0$. These facts use no real completeness.

12.2. Constructing the real field and its exact inverse operations

Let $\mathcal C$ be the set of rational Cauchy sequences $a=(a_n)_{n\ge1}$, where for every rational $\varepsilon>0$ some $N$ has $|a_n-a_m|<\varepsilon$ for all $m,n\ge N$. Let $\mathcal N$ be the sequences converging to zero by the same rational definition. Define $$\begin{gathered} a\sim b\ \Longleftrightarrow\ a-b\in\mathcal N,\qquad \mathbb R_{\mathcal C}=\mathcal C/\mathcal N,\\ [a]+[b]=[(a_n+b_n)_n],\qquad [a][b]=[(a_nb_n)_n],\qquad -[a]=[(-a_n)_n]. \end{gathered} \tag{RT6}$$ A Cauchy sequence is bounded: choose $N$ with $|a_n-a_N|<1$ for $n\ge N$, and take the actual rational bound $$M_a=1+\max_{1\le j\le N}|a_j|. \tag{RT7}$$ This bounds every coordinate, including the entire initial segment. Sums are Cauchy by adding the two actual errors. For products use $$|a_nb_n-a_mb_m| \le M_a|b_n-b_m|+M_b|a_n-a_m|. \tag{RT8}$$ Given $\varepsilon>0$, choose the errors below $\varepsilon/(2(M_a+M_b+1))$; the displayed sum is less than $\varepsilon$. The same bound proves a bounded sequence times a null sequence null. The full identity $$a_nb_n-c_nd_n=(a_n-c_n)b_n+c_n(b_n-d_n) \tag{RT9}$$ proves the product independent of both representatives. Null sums, negations and the triangle inequality prove the equivalence relation and all other operations independent of representatives. Pointwise rational field identities now prove all commutative ring identities on the quotient.

If $[a]\ne0$, the negation of null convergence provides a rational $\varepsilon>0$ such that arbitrarily late coordinates have $|a_n|\ge\varepsilon$. Choose $N$ making every tail difference less than $\varepsilon/2$, and choose one $k\ge N$ with $|a_k|\ge\varepsilon$. Then every $n\ge N$ has $$|a_n|>\varepsilon/2=:\delta>0. \tag{RT10}$$ Define $b_n=1/a_n$ for $n\ge N$, and $b_n=1$ for $n<N$. Its tail differences satisfy $$|b_n-b_m|=\frac{|a_m-a_n|}{|a_n||a_m|} \le\delta^{-2}|a_m-a_n|. \tag{RT11}$$ Thus $b$ is Cauchy. Both pointwise products equal $1$ on the tail; their discrepancies on the explicitly retained initial coordinates form null sequences. Consequently $[a][b]=[b][a]=1$. Uniqueness of a multiplicative inverse follows from the ring identities, so this construction gives the same inverse for any representative. Constant rational sequences give an injective field map $\iota:\mathbb Q\to\mathbb R_{\mathcal C}$.

Call $[a]$ positive if $a_n\ge\delta>0$ eventually for some rational $\delta$. Changing representatives preserves positivity by choosing their difference below $\delta/2$. A nonzero sequence in (RT10) has one fixed tail sign: if $a_k\ge\varepsilon$, then $a_n>\varepsilon/2$ for $n\ge N$; if $a_k\le-\varepsilon$, then $a_n<-\varepsilon/2$. Thus every nonzero class is either positive or negative, but cannot be both. Positive sums are bounded below by the sum of the two original positive gaps, and positive products by their product. This constructs an ordered field, with $\iota$ preserving order.

For any $x=[a]$, (RT7) gives $$-\iota(M_a+1)<x<\iota(M_a+1). \tag{RT12}$$ A rational upper bound therefore gives an integer upper bound. This proves the real Archimedean property directly in the constructed field. A positive $x$ has a rational $\delta/2$ with $0<\iota(\delta/2)<x$, by its positive tail gap. Absolute values satisfy the triangle and product rules by the field order: $-|x|\le x\le |x|$, addition gives $-(|x|+|y|)\le x+y\le |x|+|y|$, and applying the four possible signs gives $|xy|=|x||y|$.

For every rational $\varepsilon>0$, sufficiently late $n$ have $$|x-\iota(a_n)|<\iota(\varepsilon). \tag{RT13}$$ Indeed choose a rational Cauchy tail with differences less than $\varepsilon/2$. The sequences representing $\varepsilon-(a_m-a_n)$ and $\varepsilon+(a_m-a_n)$ then have positive tail gaps at least $\varepsilon/2$. This proves (RT13) by the exact order definition. In particular, between $x<y$, choose a positive rational $\varepsilon$ with $4\iota(\varepsilon)<y-x$, approximate $x$ by $\iota(a_n)$ within $\iota(\varepsilon)$, and take $q=a_n+2\varepsilon$. Then $x<\iota(q)<y$. All original values and all approximation factors remain in these inequalities.

12.3. Suprema, Cauchy completeness and comparison with the given real numbers

Let $S\subset\mathbb R_{\mathcal C}$ be nonempty and bounded above by $u$. Choose $x_0\in S$. By (RT12)–(RT13), choose rational endpoints $l_0,v_0$ with $\iota(l_0)<x_0$ and $u<\iota(v_0)$. In particular $l_0<v_0$. Inductively put $m_n=(l_n+v_n)/2$. If $\iota(m_n)$ is an upper bound of $S$, set $l_{n+1}=l_n,v_{n+1}=m_n$. Otherwise set $l_{n+1}=m_n,v_{n+1}=v_n$. At every step $\iota(v_n)$ is an upper bound, while $\iota(l_n)$ is not an upper bound. Moreover $$l_n\le l_{n+1}\le v_{n+1}\le v_n,\qquad v_n-l_n=(v_0-l_0)2^{-n}. \tag{RT14}$$ Induction gives $2^n\ge n+1$. The rational Archimedean property therefore shows that the right side of (RT14) tends to zero. Both endpoint sequences are rational Cauchy, since every later endpoint lies between the displayed endpoints; their difference is null. Define $s=[(l_n)_n]=[(v_n)_n]$.

Formula (RT13) implies $\iota(l_n)\to s$ and $\iota(v_n)\to s$ in the field order, first for rational errors and then for any positive field error using the positive rational below it. If some $x\in S$ had $x>s$, eventually $\iota(v_n)<x$, contradicting its upper-bound property. If an upper bound $w$ had $w<s$, eventually $w<\iota(l_n)$; because $\iota(l_n)$ is not an upper bound, there is $x\in S$ with $\iota(l_n)<x$, contradicting $w$. Thus $s$ is exactly the least upper bound. Its uniqueness follows by applying each least-upper-bound property to the other upper bound. Infima are $-\sup(-S)$, with the full negation map proving both defining inequalities.

Every Cauchy sequence $x_n$ in this ordered field converges. The argument also applies to any complete ordered field. A tail is bounded by $|x_N|+1$, and finitely many initial terms give a bound for the whole sequence. Define the original tail endpoints $$L_n=\inf_{m\ge n}x_m,\quad U_n=\sup_{m\ge n}x_m,\quad x=\sup_{n\ge1}L_n. \tag{RT15}$$ They exist by the proved least-upper-bound property; the $L_n$ are increasing and bounded above. For every $n,k$, $L_k\le U_n$: for $k<n$ use $L_k\le L_n$, and for $k\ge n$ every member of the smaller tail is at most $U_n$. Hence $L_n\le x\le U_n$. If all tail differences are less than $\varepsilon$, each tail member is an upper bound of that tail minus $\varepsilon$; taking supremum and then infimum gives $U_n-L_n\le\varepsilon$. It follows that $|x_n-x|\le\varepsilon$ on that tail. Taking initially $\varepsilon/2$ proves the strict-error convergence definition for every $\varepsilon>0$.

We now compare with the course’s given real field $F=\mathbb R$, without replacing it. In a complete ordered field the positive integers are unbounded: otherwise their supremum $s$ would have $s-1$ not an upper bound, giving an integer $n>s-1$ and $n+1>s$, a contradiction. For $a<b$ in $F$, choose an integer $N>1/(b-a)$. Among the integers bracketing $Na$, choose $k$ with $k\le Na<k+1$. Such a $k$ exists by restricting to a finite integer interval whose endpoints exceed $|Na|+1$, then choosing the largest integer at most $Na$. Then $(k+1)/N>a$ and $(k+1)/N\le a+1/N<b$. Thus the unique rational field embedding $\iota_F$, defined by the original integer multiples of $1_F$ and actual nonzero denominators, has dense image.

Apply (RT15) in $F$ to the embedded rational Cauchy sequence. Define $$\Phi:\mathbb R_{\mathcal C}\longrightarrow F,\qquad \Phi([a])=\lim_{n\to\infty}\iota_F(a_n). \tag{RT16}$$ Null differences have limit zero, so this map is well defined. Limits are unique because two distinct limits have a positive distance and their two errors can each be made smaller than one third of that distance. Sums commute with limits by the triangle inequality. Products commute by the full difference $$\iota_F(a_n)\iota_F(b_n)-xy =\iota_F(a_n)(\iota_F(b_n)-y)+y(\iota_F(a_n)-x), \tag{RT17}$$ bounded using the original $M_a$ and $|y|$. Both terms tend to zero by their actual finite bounds. Thus $\Phi$ is a field homomorphism fixing every rational. A positive class has positive gap $\delta$; its limit is at least $\iota_F(\delta)>0$, since a smaller limit would contradict its tail lower bound. Consequently $\Phi$ preserves positivity and is injective. For every $x\in F$, rational density supplies $a_n$ with $|x-\iota_F(a_n)|<1/n$. This rational sequence is Cauchy, because its tail difference is at most $1/n+1/m$, and $\Phi([a])=x$. The map is surjective.

For every original rational sequence representative all coordinate values remain in (RT16)–(RT17). Conversely the approximation just constructed is an explicit inverse value for each original $x$. Two choices differ by at most $2/n$ and hence give the same class; thus the inverse is well defined. Any ordered field isomorphism fixing the rationals must preserve the two error bounds (RT13), so it must have the value (RT16). This proves uniqueness, both compositions and every domain/codomain. Suprema and infima are transported by $\Phi$: an upper bound, and then the least such bound, correspond exactly under its order isomorphism. Hence all subsequent constructions may use the original $\mathbb R$, with their complete construction and comparison already proved. The course complex scalar set is the literal ordered-pair set $\mathbb R^2$. The coordinatewise comparison $([a],[b])\mapsto(\Phi([a]),\Phi([b]))$ has the coordinatewise inverse already proved in (RT16). It preserves the full pair addition and the two-coordinate product by (RT17) in every real factor. The separate complete complex field-law calculation is (FA0a)–(FA0d); none of those later laws is needed to construct the real field.

12.4. Square roots and the full Euclidean distance

For $a\ge0$ in the original $\mathbb R$, take $T=\{t\ge0:t^2\le a\}$. It contains $0$ and is bounded by $a+1$, because $(a+1)^2>a$. Let $r=\sup T$. If $r^2<a$, choose $$0<\eta<\min\left(1,\frac{a-r^2}{2(2r+1)}\right). \tag{RT18}$$ Then $(r+\eta)^2\le r^2+\eta(2r+1)<a$, contradicting the upper bound. If $r^2>a$, then $r>0$. Choose $0<\eta<\min(r/2,1,(r^2-a)/(2(2r+1)))$. Now $(r-\eta)^2\ge r^2-\eta(2r+1)>a$. Every nonnegative $t\ge r-\eta$ has $t^2>a$, since $t^2-(r-\eta)^2=(t-r+\eta)(t+r-\eta)\ge0$. Therefore $r-\eta$ is an upper bound of $T$, contradicting minimality of $r$. We obtain $r^2=a$. Nonnegative square roots are unique: their square difference factors as the difference times their sum, and a positive difference would make that product positive. The same factorization proves their order preservation.

For $d\ge0$, use all original coordinates and define $$|x|_2=\left(\sum_{j=1}^d x_j^2\right)^{1/2},\qquad \langle x,y\rangle_0=\sum_{j=1}^d x_jy_j,\qquad \rho(x,y)=|x-y|_2. \tag{RT19}$$ When $d=0$, the empty sum is zero and $\mathbb R^0$ is the singleton empty tuple. For $y\ne0$, expansion of the nonnegative square sum gives $$0\le\sum_{j=1}^d\left(x_j-\frac{\langle x,y\rangle_0}{|y|_2^2}y_j\right)^2 =|x|_2^2-\frac{\langle x,y\rangle_0^2}{|y|_2^2}. \tag{RT20}$$ Each of the two cross terms is retained in that expansion; their sum is minus twice the full original pairing, and the final square contribution is once its square divided by $|y|_2^2$. For $y=0$ the pairing is zero. Thus $|\langle x,y\rangle_0|\le |x|_2|y|_2$. Expanding $|x+y|_2^2$ and applying this bound proves the triangle inequality. The other metric axioms follow from the coordinate squares. For $d>0$, $$\max_{1\le j\le d}|x_j|\le |x|_2 \le \sqrt d\max_{1\le j\le d}|x_j|. \tag{RT21}$$ No coordinate or dimensional factor is suppressed. Convergence and the Cauchy property are therefore equivalent to their coordinate versions. Coordinatewise Cauchy completeness from (RT15) proves completeness of the entire original Euclidean metric. For $d=0$ every sequence is the constant empty tuple and the assertion follows directly.

Open sets are those containing a positive-radius metric ball about each of their points. Closed sets have open complements. Convergence in this metric has unique limits by the triangle inequality and the one-third-distance argument. Every limit of a sequence in a closed set stays in that set: an outside limit would have a ball in the complement, contradicting eventual membership in that ball and in the closed set.

12.5. Bounded subsequences in every original coordinate

First consider a bounded real sequence $t_n$, with an original interval $[a_0,b_0]$ containing it. If its endpoints coincide the whole sequence is constant. Otherwise bisect this actual interval. At least one of the two closed halves contains infinitely many sequence indices, since their union contains every index and a union of two finite sets is finite. Choose such a half, retaining its infinite set of indices. Repeat within that half. This gives nested original intervals $[a_k,b_k]$ and nested infinite index sets $I_k$ with $$b_k-a_k=(b_0-a_0)2^{-k},\qquad t_n\in[a_k,b_k]\quad(n\in I_k). \tag{RT22}$$ Choose $n_k\in I_k$ with $n_k>n_{k-1}$; an infinite subset of natural numbers cannot be bounded, because a bounded subset is finite. The supremum $t=\sup_k a_k$ exists. For any $j,k$, nestedness gives $a_j\le b_k$, so $a_k\le t\le b_k$. Thus $|t_{n_k}-t|\le(b_0-a_0)2^{-k}\to0$. Every original interval length and actual selected index is retained.

For a bounded sequence $x_n\in\mathbb R^d$, (RT21) bounds every coordinate. Apply the preceding construction to coordinate one. From the selected subsequence apply it to coordinate two, and continue through all $d$ coordinates. Earlier coordinate convergence is preserved under each later strictly increasing index map: any increasing natural-number map has its $k$-th value at least $k$. The final subsequence has the full original index $$n(k)=n_1\bigl(n_2(\cdots n_d(k)\cdots)\bigr) \tag{RT23}$$ and converges in every coordinate, hence by (RT21) in the original Euclidean distance. When $d=0$, take $n(k)=k$. The resulting limit belongs to any closed set containing the original sequence by Section 4. This proves bounded subsequence existence without using compactness as its own prerequisite.

12.6. Open-cover compactness and the exact closed-and-bounded criterion

A subset $K\subset\mathbb R^d$ is compact if every cover of $K$ by open subsets of $\mathbb R^d$ has a finite subcover. The empty set is compact because the empty subfamily covers it. A subset is sequentially compact if every sequence of its points has a subsequence converging to a point of that subset. Section 12.5 proves that every closed bounded set is sequentially compact, with all original coordinates retained.

We prove that sequential compactness supplies open-cover compactness in this metric. First it supplies finite ball covers at every radius $\varepsilon>0$. Otherwise, for any finite number of stages, choose $x_1\in K$ and, after $x_1,\ldots,x_n$, choose $$x_{n+1}\in K\setminus\bigcup_{j=1}^n B(x_j,\varepsilon). \tag{RT24}$$ Every two distinct selected points in one finite tuple have distance at least $\varepsilon$. If an infinite extension were already selected, a convergent subsequence would have two sufficiently late terms each within $\varepsilon/3$ of its limit, giving mutual distance less than $2\varepsilon/3$. Such an extension would contradict sequential compactness, but the finite extension rule alone does not select it under countable choice. The following independent countable construction proves that finitely many balls $B(x_j,\varepsilon)$ with centers in $K$ cover $K$.

Here is that construction with the original Euclidean distance unchanged. The empty set needs no balls; in dimension zero the nonempty coordinate space is a singleton and one ball covers it. Suppose $K\ne\varnothing$ and $d\geq1$. Enumerate all open coordinate boxes $Q_j=\prod_{i=1}^d(a_{ji},b_{ji})$ with rational endpoints $a_{ji}<b_{ji}$. This enumeration follows from the existing integer-pair enumeration of rational numbers and the diagonal enumeration of finite tuples: each rational keeps its actual numerator and nonzero denominator. For an original $x\in\mathbb R^d$ and $r>0$, rational density gives endpoints on either side of each $x_i$, within $r/(2\sqrt d)$. That box contains $x$, and every $y$ in it has $|y-x|<(\sum_{i=1}^d r^2/(4d))^{1/2}=r/2<r$. Thus these original boxes form a countable neighborhood basis, without assuming compactness.

Fix one $x_0\in K$. For each already fixed box define the independent nonempty set $$H_j= \begin{cases} K\cap Q_j,&K\cap Q_j\ne\varnothing,\\ \{x_0\},&K\cap Q_j=\varnothing . \end{cases} \qquad z_j\in H_j . \tag{RT24a}$$ Countable choice selects this sequence once. Its set of values $S$ is dense in $K$: for each $x\in K$ and $r>0$, the preceding basis calculation gives a box containing $x$ inside $B(x,r)$, and its selected $z_j$ lies in that ball. If finitely many balls of radius $\varepsilon/3$, with centres $c_1,\ldots,c_q\in K$, covered $S$, then for each original $x\in K$ density gives $z_j$ with $|x-z_j|<\varepsilon/3$. Some cover ball contains $z_j$, so $$|x-c_i|\leq |x-z_j|+|z_j-c_i| <2\varepsilon/3<\varepsilon . \tag{RT24b}$$ These same finitely many centres would cover $K$ by its original $\varepsilon$-balls. Under the assumed failure of such a cover, $S$ therefore has no finite $\varepsilon/3$-ball cover.

Starting with $w_1=z_1$, after $w_1,\ldots,w_n$ take the least enumerated index $j$ with $z_j$ outside their finitely many open $\varepsilon/3$-balls, and let $w_{n+1}=z_j$. The preceding failure of a finite cover proves that this least natural-number index exists. This is a deterministic recursion on the single chosen countable sequence; it needs no choices from history-dependent arbitrary sets. It gives $$|w_i-w_j|\geq\varepsilon/3\quad(i\ne j). \tag{RT24c}$$ Sequential compactness would give a convergent subsequence. Two sufficiently late terms have distances to its limit less than $\varepsilon/6$, hence mutual distance less than $\varepsilon/3$, contrary to (RT24c). This proves the required finite-ball covering conclusion using exactly the declared countable-choice foundation. The original finite extension formula (RT24), the original distance and its full coordinate norm, and all earlier comparison factors remain. No Zorn or dependent-choice axiom has been inserted.

For an open cover $\mathcal U$ of nonempty $K$, some $\delta>0$ has this property: for every $x\in K$, one member of $\mathcal U$ contains $B(x,\delta)\cap K$. If this failed, for every positive integer $n$ choose $x_n\in K$ such that no member contains $B(x_n,1/n)\cap K$. Take a convergent subsequence $x_{n_k}\to x\in K$. A member $U$ containing $x$ contains $B(x,r)$ for some $r>0$. For sufficiently large $k$, $$|x_{n_k}-x|_2<r/2,\qquad 1/n_k<r/2,\qquad B(x_{n_k},1/n_k)\subset B(x,r)\subset U. \tag{RT25}$$ The middle inclusion follows by adding the two original strict distance bounds. This contradicts the selection of $x_{n_k}$. With the resulting $\delta$, cover $K$ by finitely many balls of radius $\delta/2$ centered in $K$, using the preceding result. For each center choose a member containing its radius-$\delta$ ball intersected with $K$. These finitely many original cover members cover $K$. This proves open-cover compactness from sequential compactness.

Conversely a compact $K$ is bounded. The balls $B(0,n)$, $n=1,2,\ldots$, cover $\mathbb R^d$ by the Archimedean property. A finite subcover has maximal radius, giving an actual bound. The empty set is bounded by any nonnegative bound.

It is also closed. Fix $a\notin K$. If $K$ is empty its complement is the whole open space. Otherwise each $x\in K$ has the original positive radius $r_x=|x-a|_2/3$. Finitely many $B(x_i,r_{x_i})$ cover $K$. Put $\delta=\min_i r_{x_i}>0$. For $y\in K$, choose $i$ with $|y-x_i|_2<r_{x_i}$. Then $$|y-a|_2\ge |x_i-a|_2-|y-x_i|_2 >3r_{x_i}-r_{x_i}=2r_{x_i}\ge2\delta. \tag{RT26}$$ Thus $B(a,\delta)$ is disjoint from $K$. Every outside point has an open ball in the complement, proving closedness.

We have proved all three implications, so for every original finite dimension, including zero, $$K\text{ compact}\quad\Longleftrightarrow\quad K\text{ closed and bounded} \quad\Longleftrightarrow\quad K\text{ sequentially compact}. \tag{RT27}$$ For the last implication from sequential compactness to closed and bounded, the just-proved route through open-cover compactness supplies it. No definition of compactness was substituted for another without proving the maps between their assertions.

A closed subset $C$ of a compact set $K$ is compact. To an open cover of $C$ add the open complement of $C$; the resulting family covers $K$, and a finite subfamily still covers $C$ after discarding that complement. If closedness is relative to $K$, use an ambient open set whose intersection with $K$ is the relative complement; the same cover argument applies. A finite union of compact sets is compact by taking a finite subcover for each member and then their finite union. Empty unions are covered by the already proved empty case.

Finite Cartesian products preserve the original coordinates. For $K_j\subset\mathbb R^{d_j}$ compact, each factor is closed and bounded. The product is closed because an outside point has a coordinate outside one closed factor, and the inverse image of an open coordinate neighborhood is open by (RT21). If $|x^{(j)}|_2\le M_j$ in that factor, the complete product satisfies $$\left|\bigl(x^{(1)},\ldots,x^{(r)}\bigr)\right|_2^2 =\sum_{j=1}^r |x^{(j)}|_2^2\le\sum_{j=1}^r M_j^2. \tag{RT28}$$ Consequently it is compact by (RT27), with every original bound retained. An empty factor gives the empty compact product; a product of no factors is the one-point zero-dimensional space.

12.7. Continuous maps, full extrema and uniform continuity

A map $f:K\to\mathbb R^e$ is continuous at $x\in K$ when for every $\varepsilon>0$ some $\delta>0$ ensures $|f(y)-f(x)|_2<\varepsilon$ whenever $y\in K$ and $|y-x|_2<\delta$. This definition proves directly that inverse images of relatively open sets are relatively open: choose a ball about $f(x)$ in the target open set, then its continuity preimage ball. Conversely that inverse-image property applied to every target ball gives the same definition. It also proves preservation of sequence limits, by applying the defining $\delta$ to an eventually close sequence.

If $K$ is compact, $f(K)$ is compact: the inverse images of any open cover of $f(K)$ are a relatively open cover of $K$; the proved relative-cover formulation yields finitely many of them, whose original target members cover $f(K)$. Thus continuous real functions on a compact set have bounded image.