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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.

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  • $\begingroup$ Try something very "big". $\endgroup$ – Zhen Lin Apr 9 '15 at 13:36
  • $\begingroup$ @ZhenLin Like a space that has infinite dimension? $\endgroup$ – Jango Apr 9 '15 at 13:38
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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.

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  • $\begingroup$ This is just because no function $f:X\rightarrow \mathbb{R}^n$ can be continuous ? $\endgroup$ – Jango Apr 9 '15 at 13:51
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    $\begingroup$ @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 '15 at 13:57
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    $\begingroup$ @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. $\endgroup$ – sie es er Apr 9 '15 at 14:00
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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.

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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 $\mathbb{R^n}$.

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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.

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