2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Try something very "big". $\endgroup$
    – Zhen Lin
    Apr 9, 2015 at 13:36
  • $\begingroup$ @ZhenLin Like a space that has infinite dimension? $\endgroup$
    – Maylor
    Apr 9, 2015 at 13:38

4 Answers 4

6
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ This is just because no function $f:X\rightarrow \mathbb{R}^n$ can be continuous ? $\endgroup$
    – Maylor
    Apr 9, 2015 at 13:51
  • 1
    $\begingroup$ @Jango Technically there are some continuous functions, but they are not homeomorphisms. For instance constant functions are continuous here (as everywhere). In this case it works out because you can't have a bijection, but $\mathbb{R}^n$ with the trivial topology isn't Hausdorff either, so it can't embed into $\mathbb{R}^n$. $\endgroup$
    – Ian
    Apr 9, 2015 at 13:57
  • 2
    $\begingroup$ @Jango While there are continuous function from any indiscrete space into any $\mathbb{R}^n$, they are all constant functions. So as long as the indiscrete space has at least two points, no continuous function $X \to \mathbb{R}^n$ can be one-to-one. $\endgroup$ Apr 9, 2015 at 14:00
5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $\mathbb{R^n}$.

$\endgroup$
0
2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .