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Let $X$ be a metric space with a metric $d$.

Why does $x\in \overline A \iff \forall \epsilon>0, A\cap V_{\epsilon} (x) \neq \emptyset \tag{1}$

?

There are many posts that take it as no-need-to-be-proven result, however I haven't found a proof, which is probably quite basic.

I know the definition of a limit point, but it is difficult for me to derive $(1)$ from it.

My definition of the closure is $\overline A = A \cup A'$ where $A'$ are the limit points of $A$.

A limit point of $A$ is a point $p$ such as for every neighborhood $V$ of $p$ there exists a point $x\in A$ such as $x\neq p$ and $x\in V$.

A neighborhood of $x$ is the set of $y$ such as $d(x,y)<\epsilon$ where $\epsilon \in \mathbb R^{+*}$.

PS: $V_{\epsilon} (x) $ is the set of $y$ such as $d(x,y)< \epsilon$.

niobium
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    It matters how you define "closed". There are a few ways to define this in $\Bbb R$ and they're all equivalent, but what a proof looks like will change – FShrike Feb 22 '23 at 17:09
  • A set $E$ is closed if $E'\subseteq E$ – niobium Feb 22 '23 at 17:11
  • ... and what is $E'$? – FShrike Feb 22 '23 at 17:12
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    You tag this "general topology", but the definition you give only holds in a very limited type of metric spaces. It isn't even sensible in $\mathbb{R}^2$... – Arturo Magidin Feb 22 '23 at 17:12
  • $x\in E' \iff \forall \epsilon, V_{\epsilon} \cap E\backslash {x} \neq \emptyset$ where $V_{\epsilon}$ is an open ball centered around $x$ – niobium Feb 22 '23 at 17:15
  • @ArturoMagidin Yes, sorry. Post edited – niobium Feb 22 '23 at 17:21
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    $V_\epsilon(x)$ is not "a neighborhood of $x$", It is the set of all $y$ such that $d(x,y)\lt\epsilon$, where $d$ is the metric in question. You need to be precise, explicit, and accurate, especially when struggling with basic definitions. – Arturo Magidin Feb 22 '23 at 17:23
  • And how do you define the closure of a set? The smallest closed set that contains $A$, or some other definition? – Arturo Magidin Feb 22 '23 at 18:12
  • The statement $A\setminus{x}=A$ is true if and only if $x\notin A$. It has nothing to do with metrics, closures, or limit points. – Arturo Magidin Feb 22 '23 at 18:13
  • @ArturoMagidin the closure of a set in my book is defined as the union of the set with its derived set. – niobium Feb 22 '23 at 18:18
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    Please put all the relevant information in the post, not buried in the comments. Note that there are multiple ways of defining all of these concepts, and proving something about them depends precisely on which definitions you are using. – Arturo Magidin Feb 22 '23 at 18:24

2 Answers2

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I'm assuming that with the limit definition you mean that $\overline A$ is the collection of points $a$ such that a path $\{x_i\}_{i = 1}^\infty \subseteq A$ exists that converges to $a$.

The proof is actually quite easy.

  1. if $a \in \overline A$, then such a path exists for $a$, then by limit definition, for all $\epsilon > 0$ we can find a point $x_i$ such that $d(a, x_i) < \epsilon$, hence the $V_\epsilon(a)$ has overlap with $A$
  2. if every $V_\epsilon(a)$ has overlap with $A$, we can find points $x_i \in V_\frac{1}{i}(a) \cap A$. Now $d(x_i, a) < 1/i$, so $\{x_i\}_{i = 1}^\infty$ converges to $a$, and for all $x_i \in A$. Therefore $a \in \overline A$.

I hope this gives you an idea on why these definitions are equivalent.

student13
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Here are the definitions you are say you are using, once we dig through the comments (please put them in the post, this time and next!)

  1. For $\epsilon\gt0$ and a point $x$, $V_{\epsilon}(x) = \{y\mid d(x,y)\lt \epsilon\}$.

  2. For a set $A$, the derived set of $A$ is $$A'=\bigl\{x\;\bigm|\; \forall\epsilon\gt0\ \bigl( (V_{\epsilon}(x)\cap A)\setminus\{x\}\neq\varnothing\bigr)\bigr\}.$$

  3. For a set $A$, the closure of $A$ is defined to be $\overline{A}=A\cup A'$.

For simplicity, let $$F=\{x\mid \forall\epsilon\gt0 \ (V_{\epsilon}(x)\cap A\neq\varnothing)\}.$$ We want to prove that $F=\overline{A}$.

To prove that $\overline{A}\subseteq F$, note that if $x\in A'$, then by definition we have that for all $\epsilon\gt0$, $V_{\epsilon}(x)\cap A\neq\varnothing$, so $x\in F$. Thus, $A'\subseteq F$. And if $x\in A$, then for all $\epsilon\gt 0$, $x\in V_{\epsilon}(x)\cap A$, so $x\in F$. Thus, $A\subseteq F$ also holds. Thus, we have $$\overline{A} = A\cup A'\subseteq F.$$

Conversely, let $x\in F$; to prove that $x\in\overline{A}$, we need to show that $x\in A$ or that $x\in A'$. If $x\in A$, then we are done. So assume that $x\in F$ and $x\notin A$. Let $\epsilon\gt 0$. We want to show that $(V_{\epsilon}(x)\cap A)\setminus\{x\}\neq\varnothing$.

Because $x\in F$, we know that $V_{\epsilon}(x)\cap A\neq\varnothing$. And because we are assuming that $x\notin A$, then it follows that $x\notin V_{\epsilon}(x)\cap A$. Thus, $(V_{\epsilon}(x)\cap A)\setminus\{x\} = V_{\epsilon}(x)\cap A \neq\varnothing$. We have thus shown that the intersection of $V_{\epsilon}(x)$ with $A$ contains points other than $x$. As this holds for every $\epsilon\gt 0$, it follows that $x\in A'$.

Thus, we have shown that if $x\in F$, then either $x\in A$ or else $x\in A'$. Thus, $F\subseteq A\cup A'= \overline{A}$.

This proves that $F=\overline{A}$, as desired.

Arturo Magidin
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