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After reading about the remarkable result of Tychonoff's theorem, I've been going through some exercises to better understand the product topology. At the end of the day, this one has eluded me. Perhaps someone could shed some light on why the following is so?

Suppose I have the set $X=\{0,1\}^\mathbb{N}$ equipped with the usual product topology, and I let $\{0,1\}$ have the discrete topology. I can metrize $X$ with the metric $$ \rho(\{x_n\},\{y_n\})= \begin{cases} 2^{-\inf\{n\in\mathbb{N}\mid x_n\neq y_n\}}, &\{x_n\}\neq\{y_n\}\\ 0, & \{x_n\}=\{y_n\} \end{cases} $$ But why exactly does this particular metric induce the product topology on $X$?

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up vote 3 down vote accepted

The topology induced by the metric is generated by the open balls $B(x,r)=\{y\in\{0,1\}^\mathbb{N}:d(x,y)<r\}$ where $x\in\{0,1\}^\mathbb{N}$ and $r>0$. The product topology is generated by the basic sets $U(x,N)=\prod_{n=1}^N\{x_n\}\times\prod_{n=N+1}^\infty\{0,1\}$ where $x=(x_n)\in\{0,1\}^\mathbb{N}$ and $N\in\mathbb{N}$. These two families coincide since $B(x,r)=U(x,N)$, where $N$ is the least integer for which $2^{-(N+1)}<r$, so the topologies coincide too.

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Thanks LostInMath, do you mind explaining how exactly the sets $U(x,N)$ generate the product topology? – Gotye Sep 29 '11 at 0:29
You can use the following useful proposition: a family $\mathcal{B}$ is a base for a topological space $X$ if and only if (i) each $B\in\mathcal{B}$ is open in $X$ and (ii) for each $x\in X$ and open $U\subseteq X$ for which $x\in U$ there is $B\in\mathcal{B}$ such that $x\in B\subseteq U$. So let $X=\{0,1\}^\mathbb{N}$ and $\mathcal{B}$ be the family of $U(x,N)$'s and try to show (i) and (ii). Then the proposition tells that the family of $U(x,N)$'s generates the product topology. Let me know if you don't make progress and I'll try to explain more. – LostInMath Sep 29 '11 at 3:03

Hint: $x_k = y_k$ for all $k \le m$ iff $\ldots$.

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