Show that if $X$ is a compact metric space and an ANE,  such that $H_n (X) \neq 0$, then $X$ cannot be embedded in $\mathbb{R}^n$. An absolute neighborhood extensor (ANE) is a space $Y$ such that for every metric space $X$, $A$ - a closed subset of $X$, and a map $f:A \to Y$, there exists an open set $U$ containing $A$ such that $f$ can be extended to a map $U \to Y$.
$H_n(X)$ is the $n$th homology group of $X$.

Suppose $i:X \rightarrow \mathbb{R}^n$ is an embedding; since $X$ is compact so is its image $i(X)$. Let $U$ be an open set containing $i(X)$ to which we may extend the map $i^{-1}$. By compactness of $X$ there is a finite union $C$ of simplices containing $X$ and contained in $U$. Since $C$ is a finite union of simplices, it may be triangulated so that it obtains the structure of a finite simplicial complex. Since it is a finite simplicial complex embedded in $\mathbb{R}^n$, we must have $H_n(C)=0$ (in fact, the group of $n$-cycles is already $0$). But now the composition $X \rightarrow C \rightarrow X$ of $i$ and the extension of $i^{-1}$ is equal to the identity of $X$, and hence we get $H_n(C) \neq 0$, contradiction.
