Between an open set and its closed subset [closed]

Question

Suppose $F\subset V\subsetneq \mathbb {R}$ where $F$ is closed and $V$ is open. I want to show that $\exists$ an open set $U\subset \mathbb {R}$ satisfying $F\subset U\subset \overline {U}\subset V$. My proof is below, but I think it's problematic since I never use the condition that $F$ is closed.

For each $x\in F$, pick $r_{x}>0$ such that $(x-r_{x},x+r_{x})\subset V$. Then define $U=\bigcup\{(x-\frac {r_{x}}{2},x+\frac {r_{x}}{2})\}\subset V$. Then in fact, $F\subset U$. This is because countable union of open sets is open and $U$ is an open cover of $F$. It's trivial that $U\subset \overline{U}$. Now, what remains is to show that $\overline{U}\subset V$. First of all, $U\subset V$ since $U$ is just a collection of open intervals, and each one of them is in $V$. In fact, under this construction, $U\subsetneq V$. Therefore, for $y\in \overline{U}\setminus U$, $\exists u\in U$ such that $y\in B(u,r_{u})$. Take $r_{y}=r_{u}-d(y,u)$. Then $B(y,r_{y})$ is an open ball inside of $V$. Thus, $y\in V$ and $\overline{U}\subset V$.

Answer 1

$\newcommand{\R}{\mathbb{R}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\de}{\delta}\newcommand{\ol}{\overline}$The property of $\R$ that you want to prove is that the topological space $\R$ is normal. All metric spaces are perfectly normal and hence normal. This proves your desired result.

That all metric spaces are normal immediately follows e.g. from Theorems 5.1.3 and 5.1.5 of the book by Engelking: Theorem 5.1.3 (due to Stone; see also the short proof by M. E. Rudin) states that every metrizable space is paracompact, and Theorem 5.1.5 (due to Dieudonné) states that every paracompact space is normal.

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Here is an elementary, self-contained proof:

For each $n\in\Z$, the set \begin{equation*} F_n:=F\cap[n,n+1] \tag{1} \end{equation*} is closed and bounded, and hence a compact subset of the open set $V$, so that there is some $\de_n\in(0,1)$ such that \begin{equation*} \ol{U_n}\subseteq V, \tag{2} \end{equation*} where \begin{equation*} U_n:=(F_n)_{\de_n}, \tag{3} \end{equation*} the open $\de_n$-neighborhood of $F_n$.

Let finally \begin{equation*} U:=\bigcup_{n\in Z}U_n, \tag{4} \end{equation*} so that $U$ is an open set containing $F$.

It remains to show that $\ol U\subseteq V$. Take any $x\in\ol U$, so that $x=\lim_{k\to\infty}x_k$ for some sequence $(x_k)$ in $U$. Without loss of generality, for some natural $N$ and all natural $k$ we have $x_k\in U\cap[N-1,N+1]$. The conditions $F_n\subseteq[n,n+1]$, (3), and $\de_n\in(0,1)$ imply \begin{equation*} \ol{U_n}\subseteq[n-1,n+2]. \end{equation*} Hence, by (4), for all natural $k$ we have \begin{equation*} x_k\in\bigcup_{n\colon\,[n-1,n+2]\cap[N-1,N+1]\ne\emptyset}U_n\ \subseteq\bigcup_{n=N-3}^{N+2}U_n. \end{equation*} So, \begin{equation*} x\in\bigcup_{n=N-3}^{N+2}\ol{U_n}\subseteq V, \end{equation*} by (2). Thus, indeed $\ol U\subseteq V$. $\quad\Box$

(The part $V\ne\R$ of your condition $V\subsetneq\R$ is not needed here; looking at this condition, it seems that you are using $\subset$ for $\subseteq$.)