Continuous functions on $\{0,1\}^\mathbb{N}$ I am trying to do excercise 4.1. in Continuous Time Markov Processes: An Introduction by Thomas Ligett.
There I have to provide some counterexample of a function in $C(\{0,1\}^\mathbb{N})$.
However I don't know what these continuous functions are.
The setting is the following:
We have $\{0,1\}^\mathbb{N}$ equipped with the product topology. According to the author we can metrize this space with the metric
$$\rho(\eta, \zeta) = \sum_{x\in\mathbb{N}} \alpha(x)|\eta(x)-\zeta(x)|$$
with $\alpha(x)$ is strictly positive and summable.
So my question is: How do I prove that a function $$f:\{0,1\}^\mathbb{N} \to \mathbb {R}$$ is continuous with respect to the product topology.
Can I just fix a sequence $\alpha$ and then apply the $\epsilon - \delta$ definition or do I have to account for something else?
 A: $\{0,1\}^{\mathbb N}$ has the product topology. A sequence converges in this space iff each coordinate converges. Since the coordinates are just $0$ or $1$ this means each coordinate becomes constant after some stage. An example of  dis-continuous function on this space is given by $f((x_n)) =\inf \{n: x_n=1\}$ if $x_n =1$ for some $n$ and $f(0,0,,...)=0$. Note that $f(e_n)=n$ if $e_n$ is the sequence qwith $1$ in the $n-$th place and $0$ elsewhere. Also, $e_n \to (0,0,...)$. Clearly $f(e_n)$ does not tend to $f(0,0,...)$ so $f$ is not continuous.
A: Your original question is about the definition of continuity in your topological space. (You, yourself, stated a valid metrization. That is a good way of understanding a topology for a lot of mathematicians working in probability theory.) It sounds like what you might want is an example for the problem you mention later. Here is an example that might be somewhat canonical: What about $f(\eta) = \limsup_{n \to \infty} f_n(\eta)$ where we define $f_n(\eta)=n^{-1} \sum_{i=1}^{n} \eta(i)$? This is a stand-in for the asymptotic density of $\eta$. Then $f(\eta^x)=f(\eta)$ (where $\eta^x$ is the configuration obtained by changing $\eta$ exactly at the site $x$) because for any given $n>x$ we have $|f_n(\eta)-f_n(\eta^x)|=n^{-1}$ and $\lim_{n \to \infty} n^{-1}=0$. So we definitely have $\sum_{x \in \mathbb{N}} \sup_{\eta \in \{0,1\}^{\mathbb{N}}} |f(\eta^x)-f(\eta)|$ equals $0$ because it is the series of all $0$'s. But we can also see that $f$ is not continuous. For each $m$ consider a configuration $\eta_m$ where $\eta_m(x) = \mathbf{1}_{\{1,\dots,m\}}(x)$. So this is the configuration with $1$'s at the sites of $\{1,\dots,m\}$ and $0$'s elsewhere. (I am assuming $\mathbb{N} = \{1,2,\dots\}$.) Then $f(\eta_m)=0$ because $f_n(\eta_m)=m/n$ and $n$ converges to $\infty$ while $m$ stays fixed. In other words, for each fixed $m$
the asymptotic density of $1$'s in $\eta_m$ is 0. But according to your metric above, taking $\alpha(n)=2^{-n}$, for example, we have that $\lim_{m \to \infty} \eta_m$ exists and it is $\eta_{\infty}$, where $\eta_{\infty}(x)=1$ for all $x \in \mathbb{N}$. Therefore, $\eta_m \to \eta_{\infty}$ in your topology when $m \to \infty$. But $f(\eta_m)=0$ for each fixed $m$, while $f(\eta_{\infty})=1$ (because $f_n(\eta_{\infty})=1$ for each $n$). So $\lim_{m \to \infty} f(\eta_m) \neq f(\eta_{\infty})$. If this example is of interest, it probably comes up in studying the tail $\sigma$-algebra. It would also arise wherever exchangeability is applicable (which happens in some of Liggett's proofs in Interacting Particle Processes via de Finetti's theorem.
A: You can prove continuity using the metric and $\epsilon-\delta$ argument, but this seems to be unnecessary. Just use the definition of the product topology, show that if $x=(x_1, x_2, \cdots)\in f^{-1}(U)$ where $U$ is an open set of $\mathbb R$, then there exists $n$ such that $(x_1, x_2, \cdots, x_n, y_1, y_2, \cdots)\in f^{-1}(U)$ for all $(y_1, y_2, \cdots)\in \{0, 1\}^{\mathbb N}$.
