How to show the diagonal in the extended plane is closed?

  • #1
psie
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TL;DR Summary
Consider . I've managed to show this set is closed in by showing its complement is open. Now using what I've already showed, how can I argue that is closed in ?
Let be a sequence of measurable functions from into . I'm reading a proof of the fact that the set of all for which converges in as is measurable. The proof goes like this (I'm paraphrasing):

Define by . Then is measurable and if , then So the measurability of follows since is measurable subset of (note that is measurable as a closed subset of ).

Why is closed in ? As I said, I can show the diagonal with is closed in , but if and the space is , how does one proceed?
 
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  • #2
psie said:
TL;DR Summary: Consider . I've managed to show this set is closed in by showing its complement is open. Now using what I've already showed, how can I argue that is closed in ?

Let be a sequence of measurable functions from into . I'm reading a proof of the fact that the set of all for which converges in as is measurable. The proof goes like this (I'm paraphrasing):



Why is closed in ? As I said, I can show the diagonal with is closed in , but if and the space is , how does one proceed?
Is the 1-pt compactification? What metric do you use in ; the product metric? Edit: Is ?

Edit 2: Isn't the compactification map an embedding , and, if so, it sends closed sets to closed sets.
 
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  • #3
Why does you proof for doesn't work for ? In any case for a topological space , the diagonal is closed in (with the product topology) if and only if is Hausdorff. This is easy to check. And your , I suppose is Hausdorff.
 
  • #4
Good questions.
WWGD said:
Is the 1-pt compactification? What metric do you use in ; the product metric? Edit: Is ?
Yes, is . The ad hoc metric is I think , so it is homeomorphic to . Yes, .
WWGD said:
Edit 2: Isn't the compactification map an embedding , and, if so, it sends closed sets to closed sets.
I have only very rudimentary tools at my disposal (metric spaces and the fact that is metrizable), so I don't understand your Edit 2.
martinbn said:
Why does you proof for doesn't work for ?
It doesn't work because the proof uses the Euclidean norm on . We can't equip with the Euclidean norm. :frown:
 
  • #5
psie said:
It doesn't work because the proof uses the Euclidean norm on . We can't equip with the Euclidean norm. :frown:
You take something off the diagonal , this means . Then you take to open sets and that don't intersect and contain and respectively. Then is an open set containing and not meeting the diagonal.
 
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