An example of a space $X$ which doesn't embed in $\mathbb{R}^n$ for any $n$?

Question

Apologies if this has been asked before, but couldn't find it.

The definition of embedding that I'm using is this:

Suppose $X$ and $Y$ are topological spaces. We call a function $f:X\rightarrow Y$ an embedding if $f$ is a homeomorphism from $X$ to $f(X)$, equipped with the subspace topology.

I think the idea is to look for a space $X$ where any function $f:X\rightarrow\mathbb{R}^n$ does not have a continuous inverse. I can't seem to get anywhere.

Answer 1

Since each $\mathbb{R}^n$ is Hausdorff, and as all subspaces of Hausdorff spaces are Hausdorff, any non-Hausdorff space would suffice. A simple example would be $X = \{ 0 , 1 \}$ with the trivial topology.

Answer 2

Note that for any $n \in \mathbf N$, we have $|\mathbf R^n| = |\mathbf R|$. If we equip $X:= \mathcal P(\mathbf R)$ with any, to be concrete say the discrete, topology, there isn't even any one-to-one map $X \to \mathbf R^n$, hence no embedding.

Answer 3

To add to the answers above, to get a Hausdorff space of continuum cardinality you can take a disjoint union of countably many simplexes of unbound degree (i.e. a $1$-simplex, a $2$ simplex etc.). Since an $n$ simplex can only be embedded in $\mathbb{R}^m$ for $n\le m$ you get that their union fits the goal.

Answer 4

consider $X = \mathbb{R}$ with discrete metric space...then $X$ cannot be embedded in $\mathbb{R^n}$ for all $n$...since $f(X)$ would be an discrete set of $\mathbb{R^n}$..but any discrete set can atmost be countable in $\mathbb{R^n}$.