Comments on: 245B, Notes 9: The Baire category theorem and its Banach space consequences

By: The closed graph theorem in various categories « What’s new

The closed graph theorem in various categories « What’s new — Wed, 21 Nov 2012 02:30:53 +0000

[...] qualitative and quantitative notions of regularity preservation properties of an operator ; see this blog post for further [...]

By: Rex

Rex — Sat, 25 Feb 2012 23:22:43 +0000

As a pedagogical question, how many in-class lectures does it usually require for you to cover the content of one of your blog posts such as this one? Surely it is not possible to pack such an immense amount of material into a single 90-minute lecture.

By: An elementry proof of Schur’s Theorem « regularize

An elementry proof of Schur’s Theorem « regularize — Fri, 07 Oct 2011 08:27:26 +0000

[...] this alternative proof with Exercise 11 (right after Theorem 5 [Nikodym convergence theorem]) in this blog post by Terry where one shall take a similar path to proof Schur’s [...]

By: Dirk

Dirk — Fri, 07 Oct 2011 07:41:19 +0000

In the proof of the Nikodym convergence theorem the notation changes from to . Unfortunately this typo is also present in the book where I noticed it…

[Corrected, thanks - T.]

By: 254A, Notes 2: Building Lie structure from representations and metrics « What’s new

Thu, 08 Sep 2011 22:10:54 +0000

[...] (Hint: mimic the proof of the open mapping theorem for Banach spaces, as discussed for instance in these notes. In particular, take advantage of the Baire category [...]

By: Jack

Jack — Sun, 21 Aug 2011 04:27:37 +0000

Hmm, the complement of a nowhere dense set is an everywhere dense set. But the complement of everywhere dense set is not necessarily nowhere dense set, right? So the relationship is not the exact the same as that of closed and open sets.

By: Terence Tao

Terence Tao — Sat, 20 Aug 2011 17:00:05 +0000

A closed ball in a complete metric space is still complete.

By: Jack

Jack — Sat, 20 Aug 2011 16:30:49 +0000

Hmm. In your notes, you use that proposition to prove that a countable number of nowhere dense sets cannot cover a complete metric space. But in Exercise 1, one needs to show that “a countable number of nowhere dense sets cannot cover a ball”, which seems a stronger conclusion. I am wondering if one needs some special technique here.

By: Terence Tao

Terence Tao — Fri, 19 Aug 2011 16:07:05 +0000

Yes, one can use that proposition to establish Exercise 1. I think you may be conflating two different notions of "opposite", namely "logical negation" and "set complement". The negation of "E is nowhere dense" is "E is somewhere dense". But the complement of a nowhere dense set is an everywhere dense set. This is similar to how the complement of a closed set is an open set, but how the negation of "E is closed" is _not_ "E is open" (since there exist sets that are neither open nor closed, and there also exist sets that are both open and closed).

By: Jack

Jack — Fri, 19 Aug 2011 15:11:40 +0000

Prof. Tao,

For exercise 1, do we have to use the proposition that a set is nowhere dense if and only if is open and dense? This is what I learned from your topology 121 notes(http://www.math.ucla.edu/~tao/resource/general/121.1.00s/compact.pdf).

It seems that De Morgan will not help anything here.

Besides, it’s a little surprising that the opposite of “nowhere dense” is “dense” but not “somewhere dense” according to your lecture notes.