Space not homeomorphic to subset of Euclidean space

Question

Is there a topological space which is not homeomorphic to a subset of $\mathbb{R}^J$ for some index set $J$?

If the answer is yes, is there a topological space which is not homeomorphic to a quotient of a subset of $\mathbb{R}^J$?

Answer 1

Your first question is easy: $\mathbb{R}^J$ is Hausdorff for any $J$ and a subspace of a Hausdorff space is Hausdorff, so any non-Hausdorff space cannot be homeomorphic to a subspace of $\mathbb{R}^J$ for any $J$. More generally, a space that embed in $\mathbb{R}^J$ for some set $J$ is called completely regular. A more convenient and well-known characterization of completely regular spaces is that they are $T_1$ spaces $X$ such that if $A\subset X$ is closed and $x\in X\setminus A$, there exists a continuous function $f:X\to\mathbb{R}$ such that $f(x)=0$ and $f(a)=1$ for all $a\in A$.

Your second question is more interesting: it turns out that every topological space is a quotient of a subspace of $\mathbb{R}^J$ for some $J$. Here is a sketch of a (somewhat complicated) proof. First, note that by my answer here, any topological space is a quotient of a disjoint union of spaces of one of the following two forms:

1. $\mathbb{N}\cup\{\infty\}$, the one-point compactification of $\mathbb{N}$

2. The space $X_{F,x_0}$ for a set $X$, a nonprincipal ultrafilter $F$ on $X$, and a point $x_0\in X$. The underlying set of this space is $X$, and a set is open if either it is in $F$ or if it does not contain $x_0$.

Both of these types of spaces are completely regular (for type (2), if $x\neq x_0$ then $\{x\}$ is clopen, and if $x=x_0$ and $A$ is a closed set not containing $x$ then $A$ is clopen). A disjoint union of completely regular spaces is completely regular, and hence every space is a quotient of a completely regular space.