Rational Number Space is not Locally Compact Hausdorff Space
Theorem
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is not a locally compact Hausdorff Space.
Proof
For $\struct {\Q, \tau_d}$ to be a locally compact Hausdorff Space, it is required that every point of $\Q$ has a compact neighborhood.
Let $x \in \Q$.
Let $N \subseteq \Q$ be a neighborhood of $x$.
Then:
- $\exists U \in \tau: x \in U \subseteq N \subseteq \Q$.
Aiming for a contradiction, suppose $N$ is compact.
By Compact Set of Rational Numbers is Nowhere Dense, $N$ is nowhere dense.
Thus $N^-$ contains no open set of $\Q$ which is non-empty.
But $U$ is a non-empty open set of $\Q$.
By Set is Subset of its Topological Closure, $N \subseteq N^-$.
Because Subset Relation is Transitive:
- $U \subseteq N^-$
This is a contradiction.
Hence $\struct {\Q, \tau_d}$ is not a locally compact Hausdorff Space.
$\blacksquare$
Also see
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $8$