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...



Unit sphere in R^\infty is contractible?
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 $\mathbb{R}^n \backslash \{0\} \to S^n \hookrightarrow S^\infty$. 


