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Suppose $X$ is a metric space. Let $C$ denote the collection of all dense subsets of $X$. Show that $\bigcap C = \mathrm{iso}(X)$, where $\mathrm{iso}(X)$ refers to the set of all isolated points of $X$.

Attempt:

$C$ denotes the collection of all dense subsets of $X \implies \overline D = X ~~\forall~~D \in C ~~~\dots\dots (1)$.

Hence, $\mathrm{dist}(~x,D~)= 0~\forall~x \in X ~~~\dots\dots (2)$

Let $i \in X$ be an isolated point. Then : $\mathrm{dist} (~ i, X \backslash \{i\} ~)>0. ~~~\dots\dots (3).$

$D \backslash \{i\} \subseteq X \backslash \{i\} \implies \mathrm{dist} (~ i, D \backslash \{i\} ~) >0.~~~\dots\dots (4)$

Now, since $D$ is dense in $X$ and $i \in X \implies \mathrm{dist} (~ i, D ~)=0$

$\implies i \in \overline D$

$\implies i \in D~ \bigcup~ acc ~D.~~~\dots\dots (5). ~\mathrm{acc} $ refers to the accumulation point

From $(4),(5):~i \notin \mathrm{acc}~D. \implies i \in D~~\forall~~D \in C$.

$\implies \mathrm{iso}~ (X) \subseteq \bigcap C~~~\dots\dots (6)$

Now,to prove the other way around :

Suppose $z \in \bigcap C.$ We need to show that $\mathrm{dist} (~ z, X \backslash\{z\} ~)>0$.

Suppose $z \notin \mathrm{iso}~X. \implies \mathrm{dist} (~ z, X \backslash\{z\} ~)=0$.

$\implies z\in \mathrm{Cl}~(X \backslash \{z\}).~\mathrm{Cl}$ refers to the closure.

Could someone please tell me how to move ahead?

Thank you very much for the help!

Theo Bendit
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MathMan
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2 Answers2

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That $\bigcap C\subseteq\operatorname{iso}(X)$ easily follows from the following:

Claim: If $x\in X$ is not isolated, then $X\setminus\{x\}$ is dense in $X$.

Proof: Suppose that $x\in X$ is not isolated. Let $V\subseteq X$ be a neighborhood of $X$. By definition, $V$ contains a point other than $x$, so $V\cap (X\setminus\{x\})\neq\varnothing$. This means that $x$ is an accumulation point of $X\setminus \{x\}$, so $x\in\operatorname{cl}(X\setminus\{x\})$. Therefore, $$X=\{x\}\cup(X\setminus\{x\})\subseteq\operatorname{cl}(X\setminus\{x\}),$$ which readily entails that $X\setminus\{x\}$ is dense. $\blacksquare$

triple_sec
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  • Thank you very much for the answer. Though, My textbook hasn't introduced sequences as of yet. Only reached till the chapter - Closed and Open Sets. Wondering if you could please tell me something not involving sequences. – MathMan Jul 22 '15 at 08:34
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    You can still adapt triple_sec's answer without using sequences. To say $x$ is not isolated means, by definition, that every open neighbourhood of $x$ has some other point in it. That is, every open neighbourhood of $x$ has an element of $X \setminus \lbrace x \rbrace$. That implies $x$ is in the closure of $X \setminus \lbrace x \rbrace$, as is all of $X \setminus \lbrace x \rbrace$. Therefore the closure is $X$, so $X \setminus \lbrace x \rbrace$ is dense. – Theo Bendit Jul 22 '15 at 08:39
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    @Wanderer I reworked the proof so that denseness is now shown via accumulation with no reference to sequences. Please do let me know if you need further clarification. – triple_sec Jul 22 '15 at 08:40
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By definition an element $x$ is isolated if $\left\{ x\right\} $ is an open set.

Let $x$ be isolated. If $D$ is dense then $x\in\overline{D}$ so any open set containing $x$ has a nonempy intersection with $D$. That results in $\left\{ x\right\} \cap D\neq\varnothing$ or equivalently $x\in D$.

Conversely if $x$ is an element of each dense set then the set $E:=X-\left\{ x\right\} $ is not dense. Then $\overline{E}\neq X$ or equivalently $x\notin\overline{E}$. So an open set $U$ exists with $x\in U$ and $U\cap\left(X-\left\{ x\right\} \right)=\varnothing$. That tells us that $U=\left\{ x\right\} $, hence that $\left\{ x\right\} $ is an open set. Proved is now that $x$ is isolated.

drhab
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