Equivalence of Definitions of Closed Set in Normed Vector Space

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Theorem

The following definitions of the concept of Closed Set in the context of Normed Vector Spaces are equivalent:

Let $V = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.

Let $F \subseteq X$.

Definition 1

$F$ is closed in $V$ if and only if its complement $X \setminus F$ is open in $V$.

Definition 2

$F$ is closed (in $V$) if and only if every limit point of $F$ is also a point of $F$.


That is: if and only if $F$ contains all its limit points.


Proof

Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $F$ with a limit point $x$ in $X$.


Definition 1 implies Definition 2

Let $F$ be closed in $V$ according to Definition 1:

$F$ is closed in $V$ if and only if its complement $X \setminus F$ is open in $V$.


Aiming for a contradiction, suppose $x \notin F$.

Since $F$ is closed, $X \setminus F$ is open.

Let $\map {B_\epsilon} x := \set {y \in X : \norm {y - x} < \epsilon}$ be an open ball.

Then there exists $\map {B_\epsilon} x$, which belongs to $X \setminus F$:

$\forall x \in X \setminus F: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subset X \setminus F$

$\sequence {x_n}_{n \mathop \in \N}$ converges to $x$.

Therefore:

$\displaystyle \forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in N : n > N \implies \norm {x_n - x} < \epsilon$

Hence:

$\norm {x_{N + 1} - x} < \epsilon$.

Then we have:

$x_{N + 1} \in \map {B_\epsilon} x \subset X \setminus F$

But at the same time:

$x_{N + 1} \in F$

and we reached a contradiction.

Hence there is no such $x$ which is not in $F$.

It follows that every limit point of $F$ is also a point of $F$.

That is, $F$ is closed in $V$ according to Definition 2.

$\Box$


Definition 2 implies Definition 1

Let $F$ be closed in $V$ according to Definition 2:

$F$ is closed (in $V$) if and only if every limit point of $F$ is also a point of $F$.


We are given that $x$ is a limit point of $F$.

Let $x \in F$.

Aiming for a contradiction, suppose $X \setminus F$ is not open.

Then:

$\neg \paren {\forall x \in X \setminus F : \exists \epsilon \in \R_{> 0} : \map {B_\epsilon} x \subseteq X \setminus F}$

By Assertion of Existence, this is equivalent to $\paren \star$:

$\exists x \in X \setminus F : \forall \epsilon \in \R_{> 0} : \map {B_\epsilon} x \cap F \ne \empty$

Set $\displaystyle \epsilon = \frac 1 n$ with $n \in \N$.

Then:

$\exists n \in \N : x_n \in \map {B_\epsilon} x \cap F$

So there is a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $F$ such that:

$\displaystyle \forall n \in \N : \norm {x - x_n} < \frac 1 n$

and $x_n$ converges to $x$ in $F$.

By $\paren \star$:

$x \in X \setminus F$

That is:

$x \notin F$

This contradicts the condition in the definition that $x \in F$.

Hence, $X \setminus F$ has to be open.

That is, $F$ is closed in $V$ according to Definition 1.

$\blacksquare$


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