local dimension of a locally Euclidean space

Let $X$ be a locally Euclidean space. Recall that the local dimension of $X$ in $y\in X$ is a natural number $n\in \mathbb{N}$ such that there is an open neighbourhood $U\subseteq X$ of $y$ homeomorphic to ${ℝ}^n$. This number is well defined (please, see parent object for more details) and we will denote it by ${\dim }_{y}X$.

Proposition. Function $f:X\to \mathbb{N}$ defined by $f(y)={\dim }_{y}X$ is continuous (where on $\mathbb{N}$ we have discrete topology).

Proof. It is enough to show that preimage of a point is open. Assume that $n\in \mathbb{N}$ and $y\in X$ is such that $f(y)=n$. Then there is an open neighbourhood $U\subseteq X$ of $y$ such that $U$ is homeomorphic to ${ℝ}^n$. Obviously for any $x\in U$ we have that $U$ is an open neighbourhood of $x$ homeomorphic to ${ℝ}^n$. Therefore $f(x)=n$, so $U\subseteq f^{-1}(n)$. Thus (since $y$ was arbitrary) we’ve shown that around every point in $f^{-1}(n)$ there is an open neighbourhood of that point contained in $f^{-1}(n)$. This shows that $f^{-1}(n)$ is open, which completes the proof. $\mathrm{\square }$

Corollary. Assume that $X$ is a connected, locally Euclidean space. Then local dimension is constant, i.e. there exists natural number $n\in \mathbb{N}$ such that for any $y\in X$ we have

$${\dim }_{y}X=n.$$

Proof. Consider the mapping $f:X\to \mathbb{N}$ such that $f(y)={\dim }_{y}X$. Proposition shows that $f$ is continuous. Therefore $f(X)$ is connected, because $X$ is. But $\mathbb{N}$ has discrete topology, so there are no other connected subsets then points. Thus there is $n\in \mathbb{N}$ such that $f(X)=\{n\}$, which completes the proof. $\mathrm{\square }$

Remark. Generally, local dimension need not be constant. For example consider $X_1,X_2\subseteq {ℝ}^3$ such that

$$X_1=\{(x,0,0)|x\in ℝ\}\mathit{\hspace{1em}\hspace{1em}}{X}_2=\{(x,y,1)|x,y\in ℝ\}.$$

One can easily show that $X=X_1\cup X_2$ (with topology inherited from ${ℝ}^3$) is locally Euclidean, but ${\dim }_{(0,0,0)}X=1$ and ${\dim }_{(1,1,1)}X=2$.