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