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 $\mathbb{R}^n$ . This number is well defined (please, see parent object for more details) and we will denote it by $\mathrm{dim}_{y}X$ .

Proposition. Function $f:X\to\mathbb{N}$ defined by $f(y)=\mathrm{dim}_{y}X$ 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 $\mathbb{R}^n$ . Obviously for any $x\in U$ we have that $U$ is an open neighbourhood of $x$ homeomorphic to $\mathbb{R}^n$ . 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. $\square$

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 $$\mathrm{dim}_{y}X=n.$$ Proof. Consider the mapping $f:X\to\mathbb{N}$ such that $f(y)=\mathrm{dim}_{y}X$ . 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. $\square$

Remark. Generally, local dimension need not be constant. For example consider $X_1,X_2\subseteq\mathbb{R}^3$ such that $$X_1=\{(x,0,0)\ |\ x\in\mathbb{R}\}\ \ \ \ X_2=\{(x,y,1)\ |\ x,y\in\mathbb{R}\}.$$ One can easily show that $X=X_1\cup X_2$ (with topology inherited from $\mathbb{R}^3$ ) is locally Euclidean, but $\mathrm{dim}_{(0,0,0)}X=1$ and $\mathrm{dim}_{(1,1,1)}X=2$ .