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Let $E$ be a Banach space, $X_1, X_2, \ldots $ be a numerable collection of finite dimensional subspaces $X_1\subset X_2$ with dimension tending to infinity, denote by $S^n$ the unit sphere in $X_n$ . Denote by $X$ the union of the $S^n$ and let Y be the completion of the union $\cup X_n$ in the ambient norm in E.

Is there a (weak?) retraction, homotopy equivalence or domination
between X and the unitary sphere in $Y$? Notice that $Y$ is not a CW complex.

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1 Answer 1

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Well, it is well-known that both of $X$ and $Y$ are contractible (hence they are homotopy equivalent, this should answer your question). Notice that $Y$ is exactly the set of unit vectors of $E$.

To see that $X$ is contractible consider the map $H : X \times [0, 1] \to X$ defined by $$H((x_1, \dots, x_n, 0,\dots), t) = {(t, (1-t) x)\over \text{Norm(t, (1-t) x)}},$$ which is a homotopy between $\text{id}_X$ and a constant (a basis of $\bigcup_i X_i$ is understood).

To see that $Y$ is contractible, let's show that in fact $Y$ is homeomorphic to $E$. This can be addressed by considering the map $f: E \to E \oplus \Bbb R$ defined by $$f(x) = {(x, \|x\| - 1)\over \text{Norm}(x, \|x\| - 1)}.$$ This map gives a homeomorphism between $E$ and the unit sphere of $E \oplus \Bbb R$, which in turn should be homeomorphic to $Y$.

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