Locally Euclidean Space has Countable Local Basis Homeomorphic to Open Balls

From ProofWiki
Jump to navigation Jump to search



Theorem

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

Let $m \in M$.


Then:

there exists a countable local basis $\family{U_n}_{n \in \N}$ of $m$ where each $U_n$ is the homeomorphic image of an open ball of $\R^d$


Proof

By definition of a locally Euclidean space:

there exists an open neighbourhood $U$ of $m$ which is homeomorphic to an open subset $V$ of Euclidean space $\R^d$.

Let $\phi: U \to V$ be a homeomorphism.


By definition of the Euclidean space $\R^d$ the topology on $\R^d$ is the topology induced by the metric:

$\ds \map {d_2} {x, y} := \paren {\sum_{i \mathop = 1}^n \paren {x_i - y_i}^2}^{1 / 2}$

where $x = \tuple {x_1, x_2, \ldots, x_d}, y = \tuple {y_1, y_2, \ldots, y_d} \in \R^d$.


By definition of the induced topology:

$\exists \epsilon > 0 : \map {B_\epsilon} {\map \phi m} \subseteq V$

where $\map {B_\epsilon} {\map \phi m}$ is the open ball of radius $\epsilon$ and center $\map \phi m$


Consider the set of open balls:

$\BB_m = \set{\map {B_{\epsilon / n}} {\map \phi m} : n \in \N_{>0}}$

From Sequence of Reciprocals is Null Sequence:

the sequence $\sequence{\dfrac \epsilon n}_{n \in \N_{>0}}$ is a null sequence

From Null Sequence induces Local Basis in Metric Space:

$\BB_m$ is a countable local basis of $\map \phi m$ in $\R^d$

From Local Basis of Open Subspace iff Local Basis:

$\BB_m$ is a countable local basis of $\map \phi m$ in $V$


For each $n \in \N_{>0}$, let:

$U_n = \phi^{-1} \sqbrk {\map {B_{\epsilon / n}} {\map \phi m}}$

Consider the set:

$\BB'_m = \set{U_n : n \in \N_{>0}}$


From Inverse of Homeomorphism is Homeomorphism:

$\phi^{-1} : U \to V$ is a homeomorphism

From Homeomorphic Image of Local Basis is Local Basis:

$\BB'_m$ is a local basis of $m$ in $U$

From Local Basis of Open Subspace iff Local Basis:

$\BB'_m$ is a local basis of $m$ in $M$

From Restriction of Homeomorphism is Homeomorphism:

$\forall n \in \N_{>0}: U_n = \phi^{-1} \sqbrk {\map {B_{\epsilon / n}} {\map \phi m}}$ is homeomorphic to the open ball $\map {B_{\epsilon / n}} {\map \phi m}$

The result follows.

$\blacksquare$