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Prove that if $F=\{x\}\subset\mathbb{R}^n$, there exist open sets $U_1,U_2,\ldots$ such that $F = U_1\cap U_2 \cap \cdots$.

Thoughts so far: from the definition of any open set we know for any $U_i$ there is a open ball of radius $r$ around every vector $a \in U$ (i.e.$B_r(a)\subseteq U$). Secondly from the nested interval theorem the intersection of nested intervals is non-empty. I am not sure how to combine these two theorems.

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  • $\begingroup$ I'm not sure if I understand the question fully. Is $F$ an arbitrary closed set or is $F=\{x\}$? If $F=\{x\}$ you can use $U_i=B_{1/i}(x)$ for instance. $\endgroup$ – Olivier Moschetta Oct 20 '16 at 19:53
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    $\begingroup$ You don't need any fancy theorems for this. Just consider balls with center $x$ with smaller and smaller radius. $\endgroup$ – D_S Oct 20 '16 at 19:54
  • $\begingroup$ @OlivierMoschetta F is an arbitrary closed set with one element $\vec{x}$. Such that $F=\{\vec{x}\}$. I will add this to the problem to make it clearer. $\endgroup$ – AzJ Oct 20 '16 at 19:58
  • $\begingroup$ @AzJ: D_S has already answered your question. Also, $F=x$ is a typo. $\endgroup$ – parsiad Oct 20 '16 at 20:00
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    $\begingroup$ Open ball centered in $x$ with radius $r=1/i$. $\endgroup$ – Olivier Moschetta Oct 20 '16 at 20:08
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Let $(X,d)$ be a metric space, $x\in X$, and $F=\{x\}$.

Remark: In your case, $X=\mathbb{R}^n$ and $d(x,y)=|x-y|$ is the absolute value distance between numbers. If you get confused by the use of $d$ below, simply replace instances of $d(x,y)$ with $|x-y|$.

Definition: $B(v;r)=\{w \in X\colon d(v,w) < r\}$ is the open ball of radius $r$ centred at $v$.

Claim: there exist open sets $U_1,U_2,\ldots$ such that $F=U_1 \cap U_2 \cap \cdots$.

Proof: Let $U_n = B(x;1/n)$ for each $n \geq 1$. By the definition of $B$, $x\in U_n$ for each $n \geq 1$. Moreover, for any $v \neq x$, since $d(v,x)>0$, we can find $N$ large enough such that $v\notin U_n$ for all $n\geq N$. Therefore, $v\notin U$.

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    $\begingroup$ Also relevant are the following facts: (i) a singleton set is always closed in a metric space (see math.stackexchange.com/questions/1459067/…) and (ii) $\mathbb{Q}$ is an example of a closed set (in $\mathbb{R}$) that cannot be written as a countable intersection of open sets. See also the definition of a $G_\delta$ set: en.wikipedia.org/wiki/G%CE%B4_set. $\endgroup$ – parsiad Oct 20 '16 at 20:17
  • $\begingroup$ That you for your answer but I believe you mean intersection instead of union ($\cap$ instead of $ \cup$). $\endgroup$ – AzJ Oct 20 '16 at 20:19
  • $\begingroup$ Just one more question. What if we instead have a large number of vectors that made up $F$ (i.e. let $F$ be an infinite countable closed subset). What changes would we need to make to the proof. I think that instead of each open ball being around a single point in $\mathbb{R}^n$ there is a open ball around each point in the set $F$. If you could comment on this change that would be great. $\endgroup$ – AzJ Oct 21 '16 at 0:52
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    $\begingroup$ If $F$ is finite, this works. However, I have already given you (in the comment above) an example of a closed countably infinite set $\mathbb{Q}\subset\mathbb{R}$ that cannot be written as an intersection of open sets (i.e., your claim is not true). $\endgroup$ – parsiad Oct 21 '16 at 4:25
  • $\begingroup$ That makes sense thank you for the clarification. $\endgroup$ – AzJ Oct 21 '16 at 4:34

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