Local Compactness in Hausdorff Space

From ProofWiki
Jump to: navigation, search

Theorem

Let $T = \left({S, \tau}\right)$ be a $T_2$ (Hausdorff) space.


Then in $T$ the following statements are equivalent:



$(1): \quad$ For each $x \in S$, there is a neighborhood base of compact subsets of $S$.
$(2): \quad$ Each $x \in S$ is contained in some compact neighborhood of $x$.


Proof


From the definition of locally compact, the first statement directly implies the second one.


For the converse implication, suppose that:

$\forall x \in S : \exists E^x \subseteq S$

where $E^x$ is a compact neighborhood of $x$.



Regard $E^x$ as a subspace of $T$.

From Neighborhood in Topological Subspace iff Intersection of Neighborhood and Subspace:

every neighborhood of $x$ in $E^x$ is the intersection of $E^x$ and a neighborhood of $x$ in $T$.

From Intersection of Neighborhoods in Topological Space is Neighborhood:

every neighborhood of $x$ in $E^x$ is itself a neighborhood of $x$ in $T$.



Therefore:

every neighborhood basis of $x$ in $E^x$ is a neighborhood basis of $x$ in $T$.


As a consequence, it suffices to prove that every $E^x$ is locally compact.

From T2 Property is Hereditary, $E^x$ is a Hausdorff space.

By hypothesis, $E^x$ is compact.

Hence from Compact Hausdorff Space is Locally Compact, $E^x$ is locally compact.

$\blacksquare$