# Metrizable compact space not homeomorphic to any compact in finiete dimensional space

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.

• 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? – Hagen von Eitzen Oct 10 '19 at 10:02
• You mean it’s a contradiction because of cardinality? – Anton Zagrivin Oct 10 '19 at 10:13
• Yes, the Hilbert Cube is literally too big to fit into any finite-dimensional space – Hagen von Eitzen Oct 10 '19 at 10:16
• It does not embed, but the proof requires some highly non-trivial results. – Paul Frost Oct 10 '19 at 14:31
• @HagenvonEitzen . Cardinality is not the issue. The cardinal of any compact metric space is at most the cardinal of $\Bbb R.$ – DanielWainfleet Oct 11 '19 at 1:23

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 $$\Bbb R^N$$.