Show every compact subspace of $m$-manifold has topological dimension at most $m$.

Question

Notation $\bar{U}$ is the closure of $U$.

Definition $U$ is a neighborhood of $x$ means that $U$ is an open set containing $x$.

Definition An $m$-manifold is a Hausdorff space $X$ with a countable basis such that each point $x$ of $X$ has a neighborhood that is homeomorphic with an open subset of $\mathbb{R}^m$.

Answer 1

Theorem 1 Every compact subspace of $\mathbb{R}^N$ has topological dimension at most $N$.

Theorem 2 Let $X$ be a space having finite dimension. If $Y$ is a closed subspace of $X$, then $Y$ has finite dimension and dim $Y\leq$ dim $X$.

Theorem 3 Let $X=Y\cup Z$, where $Y$ and $Z$ are closed subspaces of $X$ having finite topological dimension. Then dim $X$ = max{dim $Y$, dim $Z$}.

Lemma 1 Let $X$ be a space; suppose that $h:X\rightarrow Z$ is an imbedding of $X$ in the compact Hausdorff space $Z$. Then there exists a corresponding compactification $Y$ of $X$; it has the property that there is an imbedding $H:Y\rightarrow Z$ that equals $h$ on $X$. The compactification $Y$ is uniquely determined up to equivalence.

My efforts:

Let $C$ be a compact subspace of $X$. For each point $x\in C$, we have a neighborhood $U_x$ of $x$ homeomorphic with an open subset of $\mathbb{R}^m$. The collection of $U_x\cap C$ is an open covering of $C$. Since $C$ is compact, we have a finite open covering whose members are denoted as $U_i\cap C$.

Let $f_i$ be the homeomorphism mapping $U_i$ to $f_i(U_i)$ of $\mathbb{R}^m$. Then by Theorem 1, $\bar{f}_i(U_i)$ has topological dimension at most $m$.

We view $U_i$ as a space and view $\bar{f}_i(U_i)$ as compact Hausdorff space. The $f_i$ is an imbedding of $U_i$ in $\bar{f}_i(U_i)$. By lemma 1, there exists a corresponding compactification $V_i$ of $U_i$; it has the property that there is an imbedding $F_i:V_i\rightarrow \bar{f}_i(U_i)$ that equals $f_i$ on $U_i$. The compactification $V_i$ is uniquely determined up to equivalence.

Then $V_i$ has topological dimension at most $m$.

Since $C$ is closed, we can write it as $C=\bigcup(\bar{U}_i\cap C)$. Due to equivalence of $V_i$ and $\bar{U}_i$, $\bar{U}_i$ has topological dimension at most $m$. By theorem 2, $\bar{U}_i\cap C$ has topological dimension at most $m$. By theorem 3, $C$ has topological dimension at most $m$.

Answer 2

There is the countable closed sum theorem in dimension theory (see Engelking's books):

Let $X$ be a normal (i.e. $T_4$) space. If $X = \bigcup_{n \in \Bbb N} X_n$ writes $X$ as an at most countable union of closed subspaces with $\dim(X_n) \le k (\in \Bbb N)$ for all $n$, then $\dim(X) \le k$ as well.

Apply this to the countable cover (of the closed subset) of the manifold by closed subsets of $\Bbb R^N$, which exists by having a countable base. By the subspace theorem all of these sets also have dimension $\le N$ etc.