Reading guide · Proof index

Limit superior, limit inferior, and Bolzano–Weierstrass

Jiří Lebl, Basic Analysis I–II, version 6.3. Free author edition of this section. Selection and attribution · Notation.

L2.3.2: Existence and bounds of liminf/limsup, with explicit local omitted steps

Proof.

Let us see why {an}n=1∞\{ a_n \}_{n=1}^\infty is a decreasing sequence. As ana_n is the least upper bound for {xk:k≥n},\{ x_k : k \geq n \}\text{,} it is also an upper bound for the subset {xk:k≥n+1}.\{ x_k : k \geq n+1 \}\text{.} Therefore, an+1,a_{n+1}\text{,} the least upper bound for {xk:k≥n+1},\{ x_k : k \geq n+1 \}\text{,} has to be less than or equal to an,a_n\text{,} the least upper bound for {xk:k≥n}.\{ x_k : k \geq n \}\text{.} That is, an≥an+1a_n \geq a_{n+1} for all n.n\text{.} Similarly (an exercise), {bn}n=1∞\{ b_n \}_{n=1}^\infty is an increasing sequence. It is left as an exercise to show that if {xn}n=1∞\{ x_n \}_{n=1}^\infty is bounded, then {an}n=1∞\{ a_n \}_{n=1}^\infty and {bn}n=1∞\{ b_n \}_{n=1}^\infty must be bounded.
The second item follows as the sequences {an}n=1∞\{ a_n \}_{n=1}^\infty and {bn}n=1∞\{ b_n \}_{n=1}^\infty are monotone and bounded.
For the third item, note that bn≤an,b_n \leq a_n\text{,} as the inf⁡\inf of a nonempty set is less than or equal to its sup⁡.\sup\text{.} The sequences {an}n=1∞\{ a_n \}_{n=1}^\infty and {bn}n=1∞\{ b_n \}_{n=1}^\infty converge to the limsup and the liminf, respectively. Apply Lemma 2.2.3 to obtain
lim⁡n→∞bn≤lim⁡n→∞an.\begin{equation*} \lim_{n\to \infty} b_n \leq \lim_{n\to \infty} a_n. \qedhere \end{equation*}

L2.3.4-upper: Fully written limsup subsequence proof; no assumption of the omitted liminf proof

Proof.

Define an≔sup⁡{xk:k≥n}.a_n \coloneqq \sup \{ x_k : k \geq n \}\text{.} Write x≔lim sup⁡n→∞xn=lim⁡n→∞an.x \coloneqq \limsup_{n\to\infty} x_n = \lim_{n\to\infty} a_n\text{.} We define the subsequence inductively. Let n1≔1,n_1 \coloneqq 1\text{,} and suppose n1,n2,…,nk−1n_1,n_2,\ldots,n_{k-1} are already defined for some k≥2.k \geq 2\text{.} Pick an m≥nk−1+1m \geq n_{k-1} + 1 such that
a(nk−1+1)−xm<1k.\begin{equation*} a_{(n_{k-1}+1)} - x_m < \frac{1}{k} . \end{equation*}
Such an mm exists as a(nk−1+1)a_{(n_{k-1}+1)} is a supremum of the set {xℓ:ℓ≥nk−1+1}\{ x_\ell : \ell \geq n_{k-1} + 1 \} and hence there are elements of the sequence arbitrarily close (or even possibly equal) to the supremum. Set nk≔m.n_{k} \coloneqq m\text{.} The subsequence {xnk}k=1∞\{ x_{n_k} \}_{k=1}^\infty is defined. Next, we must prove that it converges to x.x\text{.}
For all k≥2,k \geq 2\text{,} we have a(nk−1+1)≥anka_{(n_{k-1}+1)} \geq a_{n_k} (why?) and ank≥xnk.a_{n_{k}} \geq x_{n_k}\text{.} Therefore, for every k≥2,k \geq 2\text{,}
∣ank−xnk∣=ank−xnk≤a(nk−1+1)−xnk<1k.\begin{equation*} \begin{split} \sabs{a_{n_k} - x_{n_k}} & = a_{n_k} - x_{n_k} \\ & \leq a_{(n_{k-1}+1)} - x_{n_k} \\ & < \frac{1}{k} . \end{split} \end{equation*}
Let us show that {xnk}k=1∞\{ x_{n_k} \}_{k=1}^\infty converges to x.x\text{.} Note that the subsequence need not be monotone. Let ϵ>0\epsilon > 0 be given. As {an}n=1∞\{ a_n \}_{n=1}^\infty converges to x,x\text{,} the subsequence {ank}k=1∞\{ a_{n_k} \}_{k=1}^\infty converges to x.x\text{.} Thus there exists an M1∈NM_1 \in \N such that for all k≥M1,k \geq M_1\text{,} we have
∣ank−x∣<ϵ2.\begin{equation*} \sabs{a_{n_k} - x} < \frac{\epsilon}{2} . \end{equation*}
Find an M2∈NM_2 \in \N such that
1M2≤ϵ2.\begin{equation*} \frac{1}{M_2} \leq \frac{\epsilon}{2}. \end{equation*}
Take M≔max⁡{M1,M2,2}.M \coloneqq \max \{M_1 , M_2 , 2 \}\text{.} For all k≥M,k \geq M\text{,}
∣x−xnk∣=∣ank−xnk+x−ank∣≤∣ank−xnk∣+∣x−ank∣<1k+ϵ2≤1M2+ϵ2≤ϵ2+ϵ2=ϵ.\begin{equation*} \begin{split} \sabs{x- x_{n_k}} & = \sabs{a_{n_k} - x_{n_k} + x - a_{n_k}} \\ & \leq \sabs{a_{n_k} - x_{n_k}} + \sabs{x - a_{n_k}} \\ & < \frac{1}{k} + \frac{\epsilon}{2} \\ & \leq \frac{1}{M_2} + \frac{\epsilon}{2} \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{split} \end{equation*}
We leave the statement for lim inf⁡\liminf as an exercise.

L2.3.5: Convergence iff the two tail limits agree

Proof.

Let ana_n and bnb_n be as in Definition 2.3.1. In particular, for all n∈N,n \in \N\text{,}
bn≤xn≤an.\begin{equation*} b_n \leq x_n \leq a_n . \end{equation*}
First suppose lim inf⁡n→∞xn=lim sup⁡n→∞xn.\liminf_{n\to\infty} x_n = \limsup_{n\to\infty} x_n\text{.} Then {an}n=1∞\{ a_n \}_{n=1}^\infty and {bn}n=1∞\{ b_n \}_{n=1}^\infty both converge to the same limit. By the squeeze lemma (Lemma 2.2.1), {xn}n=1∞\{ x_n \}_{n=1}^\infty converges and
lim⁡n→∞bn=lim⁡n→∞xn=lim⁡n→∞an.\begin{equation*} \lim_{n\to \infty} b_n = \lim_{n\to \infty} x_n = \lim_{n\to \infty} a_n . \end{equation*}
Now suppose {xn}n=1∞\{ x_n \}_{n=1}^\infty converges to x.x\text{.} By Theorem 2.3.4, there exists a subsequence {xnk}k=1∞\{ x_{n_k} \}_{k=1}^\infty converging to lim sup⁡n→∞xn.\limsup_{n\to\infty} x_n\text{.} As {xn}n=1∞\{ x_n \}_{n=1}^\infty converges to x,x\text{,} every subsequence converges to xx and so lim sup⁡n→∞xn=lim⁡k→∞xnk=x.\limsup_{n\to\infty} x_n = \lim_{k\to\infty} x_{n_k} = x\text{.} Similarly, lim inf⁡n→∞xn=x.\liminf_{n\to\infty} x_n = x\text{.}

L2.3.8: Bolzano–Weierstrass via the first complete proof. Alternate bisection proof also read, but not needed by this graph.

Proof.

Theorem 2.3.4 says that there exists a subsequence whose limit is lim sup⁡n→∞xn.\limsup_{n\to\infty} x_n\text{.}