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

Question

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.

Answer 1

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 $\Bbb R^N$.