how can I show that any finite CW-space can embedded into an euclidean space of some dimension? Any help or reference would be greatly appreciated.
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1See http://epub.ub.uni-muenchen.de/4524/1/4524.pdf – Vidit Nanda Oct 26 '13 at 16:40
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2This is proved in Hatcher's "Algebraic Topology" corollary A.10. Voted to close. – Igor Belegradek Oct 26 '13 at 20:09
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If your finite CW-complex is of topological dimension $n$, then it is an $n$-dimensional compact metric space, thus, by the The Menger-Nöbeling theorem (1932), it can be embedded in ${\mathbb R}^{2n+1}$. In this theorem $2n+1$ is the lowest possible dimension, since there exist $n$-dimensional simplicial complexes that cannot be embedded in ${\mathbb R}^{2n}$.
Wlodek Kuperberg
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Well, any simplicial complex can be realized as a subset of the simplex in $\mathbb{R}^V$ (where $V$ is the number of vertices). But a CW complex can only be embedded up to homotopy, it seems (see the answer to your duplicate question on math.stackexchange)
Igor Rivin
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1As long as $X$ is a separable metrizable space of finite dimension $n$, space $X$ can be topologically embedded in $\mathbb R^{2\cdot n+1}$. (Thus Wlodek's pointer is correct, and the author of the QUESTION doesn't even need to restrict the CW-complex to be finite--separability would be enough to assure embeddability). – Włodzimierz Holsztyński Apr 15 '14 at 07:46