# Constructing a sequence in $\{0, 1\}^{\mathbb{R}}$ (in the product topology) with no convergent subsequence

I want to know whether my proof is correct:

I claim that the sequence of indicator functions $$x_n = 1_{A_n}$$ (where $$A_n = \{n\}$$) does not converge in $$\{0,1\}^{\mathbb{R}}$$ and has no convergent subsequence. Indeed, if $$A \subset \mathbb{R}$$ is non-empty then there exists $$x \in A$$, so the basic open set defined by $$B = \displaystyle{\prod_{\alpha \in \mathbb{R}} U_{\alpha}}$$ where $$U_\alpha = \{1\}$$ if $$\alpha = x$$ or $$\alpha = c(x+1)\doteq d$$ (where $$c(x)$$ stands for the ceiling function) and $$U_\alpha = \{0,1\}$$ everyhere else obviously contains no tail of the sequence $$(x_n)$$ (since $$1_{A_n}(d) = 0$$ for all $$n > d$$ and so $$1_{A_n} \notin B$$ for all $$n > d$$ -it doesn't contain one tail, therefore it doesn't contain any other-). An almost identical argument applies to show that it no subsequence converges.

Is everything okay? Could I improve anything here?

EDIT: it's all wrong. I'll try to fix it.

• Your example is not correct. Your sequence converges to the zero function in the product topology. Feb 11, 2019 at 12:18
• @KaviRamaMurthy I see... I was under the misconception that it had to converge to an indicator function, I see now that is indeed not the case. Thanks. Feb 11, 2019 at 12:21
• It does, the zero function is the indicator function of the empty set. Feb 11, 2019 at 12:22
• I think your set $B$ is not open - in infinitely many indices it is not all of $\{0,1\}$. Feb 11, 2019 at 12:24
• The elements of $\{0,1\}^{\Bbb R}$ are by definition indicator functions of subsets of $\Bbb R$. Feb 11, 2019 at 12:25

Your example is wrong. I'll use that $$C:=\{0,1\}^\mathbb{N}$$ has the same cardinality as $$\mathbb{R}$$, so that $$\{0,1\}^C$$ is homeomorphic to $$\{0,1\}^\mathbb{R}$$.

On the space $$C$$ we have the projections $$p_n$$ mapping each element (a sequence of "bits") to its $$n$$-th component. On $$\{0,1\}^C$$ we have the projections $$\pi_c$$ for each $$c \in C$$ mapping $$f \in \{0,1\}^C$$ (functions from $$C$$ to $$\{0,1\}$$ to its value $$f(c)$$, and the product topology has the property that it is the smallest topolgoy making all $$\pi_c, c \in C$$ continuous, and also that $$f_n \to f$$ in $$\{0,1\}^C$$ iff $$f_n(c)=\pi_c(f_n) \to \pi_c(f)=f(c)$$ for all $$c$$ (so convergence is pointwise convergence).

Now define the sequence $$f_n$$ by $$f_n(c)=\pi_c(f_n)=p_n(c)$$ for all $$c \in C$$.

This sequence does not have a convergent subsequence: suppose it had, say $$f_{n_k}, k = 0,1,2,$$ converging to $$f \in \{0,1\}^C$$.

Define $$c': \mathbb{N} \to \{0,1\}$$ by $$c'_{n_{2k}} = 1$$ for $$k=0,1,2,\ldots$$ and $$c'_m=0$$ for all other $$m$$ not of that form. Then $$c' \in C$$.

Consider $$\pi_{c'}(f_{n_k}) = p_{n_k}(c')$$ for all $$k$$: for even $$k$$ its value is $$1$$, for odd $$k$$ its value is $$0$$. So $$\pi_{c'}(f_{n_k})$$ does not converge to any point in $$\{0,1\}$$ so certainly not to $$\pi_{c')(f)=f(c')$$, as it ought to. This is a contradiction: no convergent subsequence can exist.

• I had to do so many other exercises today that I forgot to try and fix my original post. Thanks for the answer, I appreciate it. Feb 11, 2019 at 23:25