It is known that the Euclidean space $\mathbb{R}^n$ is locally compact, and that all open or closed subsets of a locally compact Hausdorff space are locally compact. What can we say about neither open nor closed subsets in the Euclidean space? Are they locally compact?
Are neither open nor closed sets in the Euclidean space locally compact
compactnessgeneral-topologyreal-analysis
Best Answer
Here a subset $A$ is locally closed if it is open in its closure $\overline A$. Equivalently if $A$ is the intersection $U\cap C$ of an open subset $U\subseteq X$ and a closed subset $C\subseteq X$.
Proof of claim: $\Rightarrow$ Suppose $A$ is locally compact. To show that it is locally closed in $X$ it will suffice to show that each point $x\in A$ has an open neighbourhood $U\subseteq X$ such that $U\cap\overline A\subseteq A$. Proceed as follows. Since $A$ is locally compact there is a compact $K\subseteq A$ and an open $U\subseteq X$ such that $x\in U\cap A\subset K\subset A$. Then $K$ is also compact in $X$, and hence closed there too, since $X$ is Hausdorf. Thus $\overline{U\cap A}\subset K\subset A$. Now notice that $U\cap\overline A\subseteq \overline{U\cap A}$. For if $y\in U\cap\overline A$ and $V$ is any neighbourhood of $y$, then $U\cap V$ is a a neighbourhood of $y$, which therefore meets $A$, so $V\cap(U\cap A)=(U\cap V)\cap A\neq\emptyset$ and hence $x\in \overline{U\cap A}$. Thus we have $x\in U\cap \overline A\subset \overline{U\cap A}\subseteq A$, as required.
$\Leftarrow$ The subset $\overline A$ is locally compact as it is closed in the locally compact space $X$. Thus if $A$ is an open subset of $\overline A$, then $A$ is also locally compact. $\square$
Remarks: The first part of the proof establishes that any locally compact subspace of a Hausdorff space $X$ is locally closed (regarless of whether $X$ itself is locally compact). The second part of the proof establishes that any locally closed subset of a locally compact space is locally compact (regardless of any separation assumptions. Here the correct definition of local compactness is that each point have a base of compact neighbourhoods).
Examples: 1. Any half-open interval $[a,b)$ or $(a,b]$ is a locally compact subspace of $\mathbb{R}$ which is neither open nor closed. Since finite products of locally compact spaces are locally compact, $\prod^{n}_{i=1}[a_i,b_i)$ and $\prod^{n}_{i=1}(c_i,d_i]$ and mixed products thereof are locally compact subspaces of $\mathbb{R}^n$, and in particular locally closed there. 2. Neither $\mathbb{Q}$ nor its complement $\mathbb{P}=\mathbb{R}\setminus\mathbb{Q}$ is locally compact, since neither is locally closed in $\mathbb{R}$. Products like $\mathbb{Q}^n$ and $\mathbb{P}^n$ are not locally closed in $\mathbb{R}^n$ and are not locally compact.