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.

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    $\begingroup$ Your example is not correct. Your sequence converges to the zero function in the product topology. $\endgroup$ Feb 11, 2019 at 12:18
  • $\begingroup$ @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. $\endgroup$ Feb 11, 2019 at 12:21
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    $\begingroup$ It does, the zero function is the indicator function of the empty set. $\endgroup$
    – Christoph
    Feb 11, 2019 at 12:22
  • $\begingroup$ I think your set $B$ is not open - in infinitely many indices it is not all of $\{0,1\}$. $\endgroup$
    – Micapps
    Feb 11, 2019 at 12:24
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    $\begingroup$ The elements of $\{0,1\}^{\Bbb R}$ are by definition indicator functions of subsets of $\Bbb R$. $\endgroup$
    – Christoph
    Feb 11, 2019 at 12:25

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Feb 11, 2019 at 23:25

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