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I am trying to understand the proof showing why infinite product of separable space is separable.

The link is: https://math.stackexchange.com/questions/244427/proving-separability-of-the-countable-product-of-separable-spaces-using-density#:~:text=Proving%20separability%20of%20the%20countable%20product%20of%20separable%20spaces%20using%20density.,-real%2Danalysis%20general&text=The%20goal%20is%20to%20prove,has%20a%20countably%20dense%20subset.

I am reading the check marked answer. In the proof, the author does product the dense subset of each factor finitely up to $1<n<m$ and then for each factor with index $n>m$, the author fixes a point from each factor. So why is this done? Why can't we use the same method we did with the proof of finite product and simply product all the dense subset of each factor?

willyx888
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2 Answers2

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We cannot just pick a dense set $D_n$ in every $X_n$: yes, the set $D=\prod_n D_n$ is dense, but $D$ is uncountable when we have infinitely many $D_n$ that are not singletons. A set $\{0,1\}^{\Bbb N}$ (Cantor cube) is also uncountable, e.g. because it's in bijection with the power set of $\Bbb N$ (Cantor's theorem). And we need a countable dense set to show separability. For a finite power, a finite product of countable sets is countable, so there the naive idea works. You have to be more clever for infinite products. We do essentially need that all basic open sets only depend on finitely many coordinates.

So we take (as in that proof) all points (i.e sequences ) in $D$ that are constant after a finite index. Of these there are only countably many. It's the difference between all subsets of $\Bbb N$ vs all finite subsets of $\Bbb N$.

Henno Brandsma
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  • Ok I thought for a second that countable infinity product of countable infinity is countable Thats true for union but not for cartesian product which was why it was ok to take the union of all $E_m$ – willyx888 Feb 09 '21 at 08:21
  • Hold on, how exactly is E as constructed by the author different than taking cartisian product of all dense subsets? E1= D1 x {x2} x {x3} x... and E2=D1 x D2 x {x3}x {x4}x{x5}.... and so on. If we take the infinite union of all $E_m$ aren't we pretty much just taking the infinite direct product of all dense subsets? – willyx888 Feb 09 '21 at 08:31
  • @willyx888 No, that's different. It's $D_1 \times {x_2} \times {x_2} \ldots, D_1 \times D_2 \times {x_3} \times {x_3} \times {x_3} \ldots$ etc. so all the same value after finitely many free choices. The full product is all choices. Like I said, infinite subsets vs essentially finite ones. – Henno Brandsma Feb 09 '21 at 08:35
  • wait so you're saying that the infinite union D1×{x2}×{x3}…$\cup$ D1×D2×{x3}×{x4}×{x5}... $\cup$ D1×D2×D3×{x4}×{x5} is not equal to D1 $\times$ D2 $\times$ D2 $\times$ D4.....? – willyx888 Feb 09 '21 at 08:41
  • @willyx888 Yes. Set theory is subtle isn't it? You pick a member of the sense set by first specifying a cut-off index $n$, and picking finitely many $d_i$ from the first $n$ (finitely many!) $D_i$. Then we fix the point to constant values, like the second answer for your quoted answer does. So finitely many choices from countable sets, so countable. You forget the $(x_n)_n$ is a fixed point, no longer variable. – Henno Brandsma Feb 09 '21 at 08:45
  • oh yes the countable infinite union of sets each countably infinite is still countable, but the countable infinite cartesian products of sets each countably infinite is uncountable. So i shoulve realize that these two are not the same thing. – willyx888 Feb 09 '21 at 08:48
  • @willyx888 Indeed they are not. – Henno Brandsma Feb 09 '21 at 08:49
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In that proof if you take all sequence $(x_n)$ where $x_n \in D_n$ for all $n$ then you do get an dense set in the product but this set is uncountable. Even an infinite product of two point sets is uncountable.

Henno Brandsma
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Kavi Rama Murthy
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