Equivalence of Definitions of Closed Set in Normed Vector Space

From ProofWiki
Jump to navigation Jump to search

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:

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

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 have 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:

$(1): \quad \exists x \in X \setminus F : \forall \epsilon \in \R_{> 0} : \map {B_\epsilon} x \cap F \ne \O$

Set $\epsilon = \dfrac 1 n$ 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:

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

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

By $(1)$:

$x \in X \setminus F$

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.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalence_of_Definitions_of_Closed_Set_in_Normed_Vector_Space&oldid=590619"