1
$\begingroup$

I'm wondering if there's metrizable compact space $X$ such that $X$ is not homeomorphic to any compact $K \subset \mathbb{R}^n$

I know that there exists metric compact space $X$(e.g. Hilbert Cube) such that $X$ is not isometric to any compact in $\mathbb{R}^n$, but the condition of isometricity is essential in those case.

$\endgroup$
  • $\begingroup$ Why is the Hilber Cube no example? It is homeomorphic to $[0,1]^\omega$ which does not fit into any $\Bbb R^n$, does it? $\endgroup$ – Hagen von Eitzen Oct 10 '19 at 10:02
  • $\begingroup$ You mean it’s a contradiction because of cardinality? $\endgroup$ – Anton Zagrivin Oct 10 '19 at 10:13
  • $\begingroup$ Yes, the Hilbert Cube is literally too big to fit into any finite-dimensional space $\endgroup$ – Hagen von Eitzen Oct 10 '19 at 10:16
  • 1
    $\begingroup$ It does not embed, but the proof requires some highly non-trivial results. $\endgroup$ – Paul Frost Oct 10 '19 at 14:31
  • 1
    $\begingroup$ @HagenvonEitzen . Cardinality is not the issue. The cardinal of any compact metric space is at most the cardinal of $\Bbb R.$ $\endgroup$ – DanielWainfleet Oct 11 '19 at 1:23
1
$\begingroup$

The hilbert cube is strongly infinite-dimensional, so it's not a subspace (isometric or not) of any finite dimensional $\mathbb{R}^n$.

But any $n$-dimensional separable metric space can be embedded into $\mathbb{R}^{2n+1}$ (Menger's universal spaces, or Nöbeling's). So infinite-dimensionality is the obstruction against embeddings into spaces $\Bbb R^N$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.