Let $X=[0,1]^{\mathbb{N}}$. View $X$ as the space of sequences $(x_n)_{n=1}^{\infty}$ with $x_n \in [0,1]$ for all $n \in \mathbb{N}$. We define two topologies on $X$. First let $\tau_1$ be the smallest topology for which the projection map $\pi_i: X \to [0,1]$ defined by $\pi_i((x_n))=x_i$ is continuous for every $i \in \mathbb{N}$. Second, let $\tau_2$ be the metric topology on $X$ defined by the metric $$d((x_n),(y_n))=\sqrt{\sum_{n=1}^{\infty}2^{-n}|x_n-y_n|^2}$$

Let $\text{Id}:X \to X$ be the identity map.

(a) The map $\text{Id}:(X,\tau_1) \to (X,\tau_2)$ is continuous

$B_d((x_n),\epsilon)$ is a basic open neighborhood in $(X,\tau_2)$. There exists a $n_0(\epsilon)$ such that for all $n \ge n_0(\epsilon)$ we have $$\sum_{i \ge n_0(\epsilon)}\frac{1}{2^i} \lt \frac{\epsilon^2}{2}$$

Then $$\Pi_{n \in \mathbb{N}}B_n \subset B_d((x_n),\epsilon)$$ where $$B_n= \left\{ \begin{array}{ll} (x_n-\frac{\epsilon}{\sqrt{2}},x_n+\frac{\epsilon}{\sqrt{2}}) & \mbox{$n=1,2,\ldots,n_0-1$};\\ [0,1] & \mbox{otherwise}.\end{array} \right. $$ Let $(y_n)_n \in \Pi_{n \in \mathbb{N}}B_n$. Then $$\left(d((x_n)_n,(y_n)_n)\right)^2=\sum_{i=1}^{n_0-1}\frac{1}{2^i}|x_i-y_i|^2+\sum_{i \ge n_0}\frac{1}{2^i}|x_i-y_i|^2 \le \sum_{i=1}^{n_0-1}\frac{1}{2^i}\frac{\epsilon^2}{2}+\frac{\epsilon^2}{2} \le \epsilon^2$$ Thus, $(y_n)_n \in B_d((x_n),\epsilon)$. So, the claim holds.

(b) The map $\text{Id}:(X,\tau_2) \to (X,\tau_1)$ is continuous

Let $\Pi_n U_n$ be a basic neighborhood in $(X,\tau_1)$. Let $k \in \mathbb{N}$ be the smallest natural number such that $U_n=[0,1]$ for all $n \ge k$. Without any loss of generality assume that $U_i=(x_i-\epsilon,x_i+\epsilon)$ for some $\epsilon \gt 0$ ,for $i=1,2,\ldots,k-1$.

Then $$B_d\left((x_n)_n,\frac{\epsilon}{2^{\frac{k}{2}}}\right) \subset \Pi_nU_n$$ Let $(y_n)_n \in B_d\left((x_n)_n,\frac{\epsilon}{2^{\frac{k}{2}}}\right)$. Obviously, for $n \ge k, y_n \in U_n$. Now for any $i=1,2,\ldots,k-1$, since $$\sum_i \frac{1}{2^i}|x_i-y_i|^2 \lt \frac{\epsilon^2}{2^k}$$ we have $$\frac{1}{2^i}|x_i-y_i|^2 \lt \frac{\epsilon^2}{2^k} \implies |x_i-y_i| \lt \epsilon$$

(c) The claim that the topological spaces $(X,\tau_1)$ and $(X,\tau_2)$ are homeomorphic follows from (a) and (b).

Is this alright? Thanks for the help!!

  • $\begingroup$ Why PI_(n in N) B_n? Don't you mean CAP_(n in N) B_n? With that change your proof looks good. $\endgroup$ – William Elliot Apr 30 '17 at 5:38

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