Locally Euclidean Space is Locally Compact/Proof 1
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 of locally Euclidean space, there exists 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:
- $B = \map {B_\delta} {\map \phi m} = \set {x \in \R^d: \size {x - \map \phi m} < \delta}$
of radius $\delta$ containing $\map \phi m$, contained in $U$.
By Closure of Open Ball in Metric Space and Topological Closure is Closed, the set:
- $C = \set {x \in \R^d: \size {x - \map \phi m} \le \dfrac \delta 2}$
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, $\phi^{-1} \sqbrk C \subseteq M$ is compact.
Furthermore $m \in \phi^{-1} \sqbrk C$ because $\map \phi m \in C$.
Thus every point of $M$ has a compact neighborhood.
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$\blacksquare$