

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

Hello, We know that in Hilbert space it is, but what about these topology: Let $\mathcal{T}_{\infty}= \{ U \subset \mathbb{R}^{\infty}: \ U \cap \mathbb{R}^n \in \mathcal{T}_n, \ for \ n=1,2,... \} $ which isn't metric space?

Of course $\mathcal{T}_{\infty}$ is topology in $\mathbb{R}^{\infty}$. How to prove or disprove that $S^{\infty} = \{ v \in \mathbb{R}^{\infty} : \ ||v||=1 \}$ is contractible?


Can we find homeomorphism without fixed point from $D^{\infty} = \{ v \in \mathbb{R}^{\infty} : \ ||v|| \le 1 \}$ onto $D^{\infty}$? I was trying to find such homeomorphism, but I failed...

share|cite|improve this question
up vote 16 down vote accepted

The question doesn't seem to be very well expressed, but the intended question might be as follows. Take $\mathbb{R}^\infty$ to mean the vector space consisting of real tuples $(v_1, v_2, v_3, \ldots)$ such that all but finitely many $v_i$ are zero, equipped with the coherent topology (so that $U \subseteq \mathbb{R}^\infty$ is open iff $U \cap \mathbb{R}^n \subseteq \mathbb{R}^n$ is open in the standard Euclidean topology on $\mathbb{R}^n$, here identifying $\mathbb{R}^n$ with the subset of tuples $(v_1, v_2, v_3, \ldots)$ such that $v_i = 0$ for $i > n$). Let $S^\infty \subseteq \mathbb{R}^\infty$ be the subset consisting of tuples such that $\sum_i |v_i|^2 = 1$, equipped with the subspace topology. Then $S^\infty$ is contractible.

The idea is easy enough: define a map $f: S^\infty \to S^\infty$ by $f(v_1, v_2, \ldots) = (0, v_1, v_2, \ldots)$, and define a homotopy $H_1$ from the identity on $S^\infty$ to $f$ by

$$H_1(t, v) = N((1-t)v + tf(v))$$

where $N: \mathbb{R}^\infty \backslash \{0\} \to S^\infty$ is the map $v \mapsto \frac{v}{\|v\|}$. Then define a second homotopy $H_2$ from $f$ to the constant function on $S^\infty$ valued at $p = (1, 0, 0, 0, \ldots)$ by

$$H_2(t, v) = (1-t)^{1/2}f(v) + t^{1/2}p.$$

Gluing these two homotopies together, we have a homotopy which contracts $S^\infty$ to a point.

Edit: The question arose in a comment as to why $H_1$ is continuous. It might help to bear in mind two facts: (1) that the coherent topology on an increasing union of spaces $X = \bigcup_i X_i$ is the colimit topology, i.e., the topology of the colimit in $\mathrm{Top}$ of the diagram $X_1 \subset X_2 \subset \ldots$, and (2) the functor $[0, 1] \times -: \mathrm{Top} \to \mathrm{Top}$ preserves colimits because $[0, 1]$ is locally compact Hausdorff (i.e., if $I$ is locally compact Hausdorff, then $I \times -$ is left adjoint to exponentiation $(-)^I$, and left adjoints preserve colimits). For example, to check that the map

$$\mathrm{colim}_n I \times S^n \cong I \times \mathrm{colim}_n S^n \to \mathbb{R}^\infty \backslash \{0\}$$

defined by $v \mapsto (1-t)v + tf(v)$ is continuous, it suffices to check that its restriction to each $I \times S^n$ is continuous (by definition of colimit). But this restriction is a composite of manifestly continuous maps, $I \times S^n \to \mathbb{R}^{n+1} \backslash \{0\} \hookrightarrow \mathbb{R}^\infty \backslash \{0\}$, where the first map is defined again by $v \mapsto (1-t)v + tf(v)$. Similarly, to check that the normalization map $N: \mathbb{R}^\infty \backslash \{0\} \to S^\infty$ is continuous, it suffices to check that its restriction to each $\mathbb{R}^n \backslash \{0\}$ is continuous, but this restriction is a composite of continuous maps $\mathbb{R}^n \backslash \{0\} \to S^n \hookrightarrow S^\infty$.

share|cite|improve this answer
The same proof shows contractibility of the unit sphere of $\ell^2$... – Alain Valette Jan 20 '13 at 11:16
Thank you Mr. Trimble :) But why $N((1-t)v+tf(v))$ is homotopy? I meen why it's continuous? Thanks. – barnasik Jan 20 '13 at 15:08
Note that the colimit topology depends on which category you are working in. The colimit in LCTVS is different to that in Top (but the sphere is contractible in both via the same argument). – Loop Space Jan 21 '13 at 10:53

Don't want to be too pedantical but, to set the record straight, both topologies coincide in this situation---one of the consequences of the Banach-Dieudonne theorem.

share|cite|improve this answer
Sorry. This was meant to be a comment on the above comment. Tried to delete but couldn't. Maybe somebody more powerful could erase it. – jbc May 12 '13 at 12:22
I wouldn't worry about it -- let it stand. It seems like an interesting remark. Since some like me might not be too familiar with the Banach-Dieudonne theorem and its consequences, you might also consider fleshing this out a bit (I quickly googled and the statement I found applied to Banach spaces, which $\mathbb{R}^\infty$ is not, so it would help me personally to have a little more explanation). – Todd Trimble May 13 '13 at 0:38

Your Answer


