Locally Euclidean Space is Locally Compact

Theorem
Let $M$ be a locally Euclidean space of some dimension $d$.

Then $M$ is locally compact.

Proof
Let $m \in M$ be arbitrary.

By definition there is an open neighborhood $U$ of $m$, homeomorphic to an open subset of $\R^d$.

By the definition of an open set, there is some open ball:


 * $B = B_\delta \left({\phi \left({m}\right)}\right) = \left\{{x \in \R^d: \left|{x - \phi \left({m}\right)}\right| < \delta}\right\}$

of radius $\delta$ containing $\phi \left({m}\right)$, contained in $U$.

By Closure of Open Ball in Metric Space and Topological Closure is Closed, the set:


 * $C = \left\{{x \in \R^d: \left|{x - \phi \left({m}\right)}\right| \le \dfrac \delta 2}\right\}$

is closed, and $C \subseteq B \subseteq U$.

Moreover, $C$ is trivially bounded, hence compact by the Heine-Borel Theorem.

Now if $\phi$ is a homeomorphism $U \to \R^d$, then by definition $\phi^{-1}$ is continuous.

Therefore by Continuous Image of Compact Space is Compact, $\phi^{-1} \left({C}\right) \subseteq M$ is compact.

Furthermore $m \in \phi^{-1} \left({C}\right)$ because $\phi \left({m}\right) \in C$.

Thus every point of $M$ has a compact neighborhood.