MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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 $Y$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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