Locally Euclidean Space is Locally Compact

Theorem

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

Proof 1

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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