The quotient of $\mathbb{R}^{n}$ by a closed subset Let $A$ be a closed subset of $\mathbb{R}^{n}$. Can the quotient space $\mathbb{R}^{n}/A$ be embedded in some Euclidean space $\mathbb R^{m}$? In particular, assume that $A$ is an algebraic variety of degree $k$, can we control $m$ in term of $n$ and $k$?  
 A: For $A$ compact the answer is that given by Joseph Van Name. If $A$ is not compact then the answer is negative. For example take the standard $\Bbb R \subset \Bbb R^2$. Then the quotient is not II-countable (the same holds for any unbounded closed subset of $\Bbb R^n$, with unbounded complement).
In the compact setting there is an important special case: that of cellular subsets. $A\subset \Bbb R^n$ is cellular if it is the countable intersection of a sequence of nested $n$-balls: $$A = \bigcap_{i\in \Bbb N} B_i,$$ $B_i \cong B^n$ and $B_{i+1} \subset \text{Int}\, B_i$. For cellular $A$, the projection map $\pi : \Bbb R^n \to \Bbb R^n/A$ can be approximated by homeomorphisms which are the identity outside a neighborhood of $A$, with respect to a distance function on the quotient (whose existence follows from the Urysohn metrization theorem). As a consequence, $\Bbb R^n/A \cong \Bbb R^n$. This fact plays an important role in the proof of the topological Schoenflies theorem in dimension $n$. A reference is M. Brown, "A proof of the generalized Schoenflies theorem", Bull. Amer. Math. Soc. 66 (1960),
74-76.
A: I shall prove the case for $S^{n}$ rather than $\mathbb{R}^{n}$ since $S^{n}$ works better being compact (for closed non-compact sets of $\mathbb{R}^{n}$ it does not work; see Daniele Zuddas's answer). Suppose that $A$ is a closed subset of $S^{n}\subseteq\mathbb{R}^{n+1}$. Then define a mapping $f:S^{n}\rightarrow\mathbb{R}^{n+1}$ by letting $f(x)=d(x,A)\cdot x$ where $d(x,A)$ denotes the distance from $x$ to $A$. Then clearly $f$ is a continuous map. Since $S^{n}$ is compact, the mapping $f$ is a quotient map and $f(x)=f(y)$ if and only if $x=y$ or $x,y\in A$. Therefore $S^{n}/A\simeq f[S^{n}]$.
