Sequences and convergence · Selected prerequisite proofs

L7.3.9: Coordinatewise and Euclidean convergence agree for every finite n

Proposition 7.3.9.

Let

\{ x_m \}_{m=1}^\infty

be a sequence in

\R^n\text{,}

where

x_m = \bigl(x_{m,1},x_{m,2},\ldots,x_{m,n}\bigr) \in \R^n\text{.}

Then

\{ x_m \}_{m=1}^\infty

converges if and only if

\{ x_{m,k} \}_{m=1}^\infty

converges for every

k=1,2,\ldots,n\text{,}

in which case

\begin{equation*} \lim_{m\to\infty} x_m = \Bigl( \lim_{m\to\infty} x_{m,1}, \lim_{m\to\infty} x_{m,2}, \ldots, \lim_{m\to\infty} x_{m,n} \Bigr) . \end{equation*}

Proof.

Suppose

\{ x_m \}_{m=1}^\infty

converges to

y = (y_1,y_2,\ldots,y_n) \in \R^n\text{.}

Given

\epsilon > 0\text{,}

there exists an

M

such that for all

m \geq M\text{,}

we have

\begin{equation*} d(y,x_m) < \epsilon. \end{equation*}

Fix some

k=1,2,\ldots,n\text{.}

For all

m \geq M\text{,}

\begin{equation*} \bigl\lvert y_k - x_{m,k} \bigr\rvert = \sqrt{{\bigl(y_k - x_{m,k} \bigr)}^2} \leq \sqrt{\sum_{\ell=1}^n {\bigl(y_\ell-x_{m,\ell}\bigr)}^2} = d(y,x_m) < \epsilon . \end{equation*}

Hence, the sequence

\{ x_{m,k} \}_{m=1}^\infty

converges to

y_k\text{.}

For the other direction, suppose

\{ x_{m,k} \}_{m=1}^\infty

converges to

y_k

for every

k=1,2,\ldots,n\text{.}

Given

\epsilon > 0\text{,}

pick an

M

such that if

m \geq M\text{,}

then

\bigl\lvert y_k-x_{m,k} \bigr\rvert < \nicefrac{\epsilon}{\sqrt{n}}

for all

k=1,2,\ldots,n\text{.}

Then

\begin{equation*} d(y,x_m) = \sqrt{\sum_{k=1}^n {\bigl(y_k-x_{m,k}\bigr)}^2} < \sqrt{\sum_{k=1}^n {\left(\frac{\epsilon}{\sqrt{n}}\right)}^2} = \sqrt{\sum_{k=1}^n \frac{{\epsilon^2}}{n}} = \epsilon . \end{equation*}

That is, the sequence

\{ x_m \}_{m=1}^\infty

converges to

y = (y_1,y_2,\ldots,y_n) \in \R^n\text{.}

L7.3.11: Metric and neighbourhood definitions of convergence agree

Proposition 7.3.11.

Let

(X,d)

be a metric space and

\{x_n\}_{n=1}^\infty

a sequence in

X\text{.}

Then

\{ x_n \}_{n=1}^\infty

converges to

p \in X

if and only if for every open neighborhood

U

of

p\text{,}

there exists an

M \in \N

such that for all

n \geq M\text{,}

we have

x_n \in U\text{.}

Proof.

Suppose

\{ x_n \}_{n=1}^\infty

converges to

p\text{.}

Let

U

be an open neighborhood of

p\text{.}

There exists an

\epsilon > 0

such that

B(p,\epsilon) \subset U\text{.}

As the sequence converges, find an

M \in \N

such that for all

n \geq M\text{,}

we have

d(p,x_n) < \epsilon\text{,}

or in other words

x_n \in B(p,\epsilon) \subset U\text{.}

Let us prove the other direction. Given

\epsilon > 0\text{,}

let

U \coloneqq B(p,\epsilon)

be the neighborhood of

p\text{.}

Then there is an

M \in \N

such that for

n \geq M\text{,}

we have

x_n \in U = B(p,\epsilon)\text{,}

or in other words,

d(p,x_n) < \epsilon\text{.}

L7.3.12: Closed sets contain convergent sequence limits

Proposition 7.3.12.

Let

(X,d)

be a metric space,

E \subset X

a closed set, and

\{ x_n \}_{n=1}^\infty

a sequence in

E

that converges to some

p \in X\text{.}

Then

p \in E\text{.}

Proof.

Let us prove the contrapositive. Suppose

\{ x_n \}_{n=1}^\infty

is a sequence in

X

that converges to

p \in E^c\text{.}

As

E^c

is open, Proposition 7.3.11 says that there is an

M

such that for all

n \geq M\text{,}

x_n \in E^c\text{.}

So

\{ x_n \}_{n=1}^\infty

is not a sequence in

E\text{.}