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I’m wondering if the set of non-negative sequences which sum to 1 is compact under the product (or weak) topologies.

That is: $(a_1,a_2,...)$ such that $\sum_n a_n=1$ where $a_n \geq 0 \forall n$

I realize that this set of sequences is not closed under the $l^1$-topology because the following sequence of sequences has no convergent subsequence.

(1,0,0,0,0,0,0,...) (0,1,0,0,0,0,0,...) (0,0,1,0,0,0,0,...)

But, this kind of example does not apply to the product topology, which is not metrizable.

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  • $\begingroup$ Isn't $U_n=\{a:a_n\gt3^{-n}\},\ n=1,2,3,\dots$ an open cover with no finite subcover? $\endgroup$ – bof Jan 16 '18 at 2:54
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    $\begingroup$ Your sequence of sequences also doesn't converge in the product topology. Convergence in the product topology is pointwise convergence, and each component of the sequence converges to $0$. But the zero sequence isn't in your set. $\endgroup$ – Qiaochu Yuan Jan 16 '18 at 3:57
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    $\begingroup$ Crosspost: mathoverflow.net/questions/290833/… $\endgroup$ – Qiaochu Yuan Jan 16 '18 at 6:38
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    $\begingroup$ The set of members of your " sequence of sequences with no convergent sub-sequence" is an infinite closed discrete sub-space. A compact space cannot have an infinite closed discrete sub-space. $\endgroup$ – DanielWainfleet Jan 16 '18 at 12:37
  • $\begingroup$ Hi, thanks for all of the comments. As it turns out, the set of nonnegative sequences whose sum is less than or equal to 1 is compact under the product topology, but not the set of nonnegative sequences whose sum equals to 1. This points to, as has already been kinda mentioned, that the set of nonnegative sequences whose sum is equal is 1 is not even a closed set because 0 is in the closure of that set. To see it, take any open nbhd of 0 (in the product topology) and notice that there are sequences with sum=1 inside. $\endgroup$ – user521875 Jan 17 '18 at 11:24
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I realize that this set of sequences is not closed under the $l^1$-topology because the following sequence of sequences has no convergent subsequence.

(1,0,0,0,0,0,0,...)
(0,1,0,0,0,0,0,...)
(0,0,1,0,0,0,0,...)

This sequence is rather a good start. You might try to show that this countable set does not have a limit point, and thus this space is not limit point compact.

But, this kind of example does not apply to the product topology, which is not metrizable.

In fact, product of countably many metric space is metrizable. For example: Show that the countable product of metric spaces is metrizable or Show this metric generates the product topology on $X$. So you can safely apply results which you are known to be true for metrizable spaces.

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  • $\begingroup$ Thank you for pointing out the error on my part in the editing. First of all, apologies and promise to be more careful in the future. No defence, but what happened here was while attending to the details in symbols in the edit (it was a long one), the eyes did not catch that one swear word. Sorry again $\endgroup$ – Shailesh Jan 18 '18 at 7:20

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