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Given $K^w$ equipped with product topology is an infinite product of countably infinite discrete space $K$ . Show that $K^w$ is second countable.

My Progress: Since the product topology means there exists a basis $\cup U_i$ such that $U_i$ is open in $K_i$ for each $i$ and $U_i\neq K_i$ at only finitely many values of $i$. Thus the basis is countable since we only care about the difference part between the basis and the product topology, and there are only finitely many $U_i$, which means our basis is countable.

I don't feel quite satisfied with this naive proof, and I think there must be some mistakes in it. Can someone please help correct/modify it?

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  • $\begingroup$ Is that really true? What if $K$ itself is uncountable and dicrete? $K$ itself is not second countable. $\endgroup$ – user99914 Mar 4 '15 at 5:32
  • $\begingroup$ Oh, I added a word countably into the problem statement. Yes, $K$ is not second countable, but $K$ is discrete. $\endgroup$ – user177196 Mar 4 '15 at 5:39
  • $\begingroup$ @John can you elaborate on how to solve the problem if my argument can't be fixed? The problem statement is correct though. $\endgroup$ – user177196 Mar 4 '15 at 14:48
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Let $\mathscr{U}$ be the set of all products of the form $\prod_{n\in\omega}U_n$, where each $U_n$ is open in $K$, and $U_n=K$ for all but finitely many $n\in\omega$. As you say, $\mathscr{U}$ is a base for $K^\omega$, but it is not countable: $K$ has $2^\omega=\mathfrak{c}$ open sets, so there are already uncountably many possibilities just for the factor $U_0$.

However, we don’t have to allow all open subsets of $K$ as factors. Let $\mathscr{B}_K=\big\{\{x\}:x\in K\big\}$; $\mathscr{B}_K$ is a countable base for $K$. Now let $\mathscr{B}$ be the set of all products of the form $\prod_{n\in\omega}U_n$, where $U_n=K$ for all but finitely many $n\in\omega$, and $U_n\in\mathscr{B}_K$ whenever $U_n\ne K$. Show that $\mathscr{B}$ is a countable base for the product topology on $K^\omega$.

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  • $\begingroup$ Thank you for your great detail, Prof. Scott! I can see that the reason why $\mathscr{B}$ is a countable base is because $U_n\in \mathscr{B_K}$ whenever $U_n\neq K$, so product of $U_n$ is a subset of $\mathscr{B_K}$, and $B_K$ is a countable. But is $K$ countably infinite the fact that explains why $\mathscr{B_K}$ is countable? $\endgroup$ – user177196 Mar 5 '15 at 2:21
  • $\begingroup$ @user177196: There is a bijection $x\mapsto\{x\}$ between $K$ and $\mathscr{B}_K$, so the countability of $\mathscr{B}_K$ follows immediately from the countability of $K$. But be careful: $\mathscr{B}$ is not a subset of $\mathscr{B}_K$. $\mathscr{B}$ is countable because (a) if $F$ is a finite subset of $\omega$, there are only countably many members of $\mathscr{B}$ that $U_n\in\mathscr{B}_K$ iff $n\in F$, and (b) $\omega$ has only countably many finite subsets. $\endgroup$ – Brian M. Scott Mar 5 '15 at 2:26
  • $\begingroup$ I understand why (a) and (b) are true, but how do they imply $B$ is countable, since $U_n\in \mathscr{B_K}$ for every $U_n\neq K$, not just for those with $n\in F$? $\endgroup$ – user177196 Mar 5 '15 at 2:44
  • $\begingroup$ @user177196: For each $U=\prod_nU_n\in\mathscr{B}$ there is a finite subset $F_U$ of $\omega$ such that $U_n\ne K$ if and only if $n\in F_U$. For each finite $F\subseteq\omega$ let $\mathscr{B}_F=\{U\in\mathscr{B}:F_U=F\}$. Then $$\mathscr{B}=\bigcup\{\mathscr{B}_F:F\subseteq\omega\text{ is finite}\}\;.$$ The (a) of my previous comment says that each $\mathscr{B}_F$ is countable, and my (b) says that there are only countably many $\mathscr{B}_F$’s. $\endgroup$ – Brian M. Scott Mar 5 '15 at 2:54
  • $\begingroup$ Many thanks for your elaboration later on. That helps a lot, since I'm quite dumb:( Btw, $w$ only has countably many finite subsets comes from the fact that it's countably infinite? $\endgroup$ – user177196 Mar 5 '15 at 3:14

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