I am very concerned that future searches for "compact" will not yield properties or examples of spaces whose every open cover has a finite subcover.
October 17, 2025 at 7:47 PM
I am very concerned that future searches for "compact" will not yield properties or examples of spaces whose every open cover has a finite subcover.
every open bar with a cover has a finite subcover
November 30, 2025 at 1:26 AM
every open bar with a cover has a finite subcover
every open bar with a cover has a finite subcover
November 30, 2024 at 1:58 PM
every open bar with a cover has a finite subcover
I remember auditing a math class as a 1st yr md/phd student, and the professor looked like Walt Whitman with one arm covered in chalk dust up to the elbow. He was shouting: “Every open cover has a finite subcover! No holes!”
And that was the moment I became an experimentalist
And that was the moment I became an experimentalist
Measure Theory Thread 2:
The FIRST part of the forwards direction of the proof of the Heine-Borel Theorem (from my brain)
The first part says "If a set of reals is compact, it is closed"
(Still need to prove "if compact, it is bounded")
Using same definitions as from Thread 1, nothing new.
(1/n)
The FIRST part of the forwards direction of the proof of the Heine-Borel Theorem (from my brain)
The first part says "If a set of reals is compact, it is closed"
(Still need to prove "if compact, it is bounded")
Using same definitions as from Thread 1, nothing new.
(1/n)
Measure Theory Thread 1:
The Backwards direction of the proof of the Heine-Borel Theorem (Covered in Royden 5e chapter 1.4)
I'll do and talk about the forwards direction on Friday (which the book doesn't go over).
Here are some definitions I will be using:
(1/n)
The Backwards direction of the proof of the Heine-Borel Theorem (Covered in Royden 5e chapter 1.4)
I'll do and talk about the forwards direction on Friday (which the book doesn't go over).
Here are some definitions I will be using:
(1/n)
May 24, 2026 at 2:09 AM
I remember auditing a math class as a 1st yr md/phd student, and the professor looked like Walt Whitman with one arm covered in chalk dust up to the elbow. He was shouting: “Every open cover has a finite subcover! No holes!”
And that was the moment I became an experimentalist
And that was the moment I became an experimentalist
i'm just trying to find a finite subcover of balls
March 5, 2024 at 12:48 AM
i'm just trying to find a finite subcover of balls
That's right! And you could use the infinitely many turns in the spiral staircase to make an infinite cover with no finite subcover.
January 20, 2026 at 1:38 AM
That's right! And you could use the infinitely many turns in the spiral staircase to make an infinite cover with no finite subcover.
Chapter 14s subcover for An Adverse Adventure which is not on Namicomi as well as the other sites! links are in my linktree!
Linktree: linktr.ee/DamiDraws
Programs used: Blender and Clipstudio paint. Drakaris was created in vroid and edited in Blender and CSP. #art #b3d
Linktree: linktr.ee/DamiDraws
Programs used: Blender and Clipstudio paint. Drakaris was created in vroid and edited in Blender and CSP. #art #b3d
November 25, 2023 at 7:52 PM
Chapter 14s subcover for An Adverse Adventure which is not on Namicomi as well as the other sites! links are in my linktree!
Linktree: linktr.ee/DamiDraws
Programs used: Blender and Clipstudio paint. Drakaris was created in vroid and edited in Blender and CSP. #art #b3d
Linktree: linktr.ee/DamiDraws
Programs used: Blender and Clipstudio paint. Drakaris was created in vroid and edited in Blender and CSP. #art #b3d
Open subcover? Isn't that just free use?
December 3, 2025 at 5:43 PM
Open subcover? Isn't that just free use?
And you made crucial use of the completeness of the real numbers. Heine-Borel would fail for the rationals: there's an open cover of all rationals between 0 and 1 that has no finite subcover!
May 21, 2026 at 9:59 AM
And you made crucial use of the completeness of the real numbers. Heine-Borel would fail for the rationals: there's an open cover of all rationals between 0 and 1 that has no finite subcover!
Oh, much better! Yes, this gives an open cover of all rationals in [0,1] that has no finite subcover! 🎉
June 4, 2026 at 5:40 PM
Oh, much better! Yes, this gives an open cover of all rationals in [0,1] that has no finite subcover! 🎉
Indeed that’s what the post is about!
I was complaining here about a much more basic issue.
A topological space whose every open cover has a finite subcover is called compact, except analysts call it quasi compact.
Romulus and Remus were raised by wolves.
Grothendieck was raised by analysts.
I was complaining here about a much more basic issue.
A topological space whose every open cover has a finite subcover is called compact, except analysts call it quasi compact.
Romulus and Remus were raised by wolves.
Grothendieck was raised by analysts.
December 14, 2024 at 8:57 AM
Indeed that’s what the post is about!
I was complaining here about a much more basic issue.
A topological space whose every open cover has a finite subcover is called compact, except analysts call it quasi compact.
Romulus and Remus were raised by wolves.
Grothendieck was raised by analysts.
I was complaining here about a much more basic issue.
A topological space whose every open cover has a finite subcover is called compact, except analysts call it quasi compact.
Romulus and Remus were raised by wolves.
Grothendieck was raised by analysts.
okay so um first you gotta make an open cover of a thing, and then you gotta make it a finite subcover of the thing
September 20, 2025 at 10:09 PM
okay so um first you gotta make an open cover of a thing, and then you gotta make it a finite subcover of the thing
Okay I found my notes — compact means every open cover of the set (the set of points on the Riemann surface?) has a finite subcover, and an open cover is a collection of open sets whose union would contain the Riemann surface, while the finite subcover is a finite collection of open sets whose (1/2)
January 20, 2026 at 1:28 AM
Okay I found my notes — compact means every open cover of the set (the set of points on the Riemann surface?) has a finite subcover, and an open cover is a collection of open sets whose union would contain the Riemann surface, while the finite subcover is a finite collection of open sets whose (1/2)
Let x=(x_d)_{d in D} be a net in F. Then p(x_d)_{d in D} has an accumulation point b in B
If x has no accumulation point in F_b, each f has an open U_f s.t. x eventually stays out of U_f. The cover {U_f} has finite subcover {U_i}. It follows p(x) eventually stays out of p[\cap U_i], a nbhd of b qed
If x has no accumulation point in F_b, each f has an open U_f s.t. x eventually stays out of U_f. The cover {U_f} has finite subcover {U_i}. It follows p(x) eventually stays out of p[\cap U_i], a nbhd of b qed
July 10, 2024 at 3:40 PM
Let x=(x_d)_{d in D} be a net in F. Then p(x_d)_{d in D} has an accumulation point b in B
If x has no accumulation point in F_b, each f has an open U_f s.t. x eventually stays out of U_f. The cover {U_f} has finite subcover {U_i}. It follows p(x) eventually stays out of p[\cap U_i], a nbhd of b qed
If x has no accumulation point in F_b, each f has an open U_f s.t. x eventually stays out of U_f. The cover {U_f} has finite subcover {U_i}. It follows p(x) eventually stays out of p[\cap U_i], a nbhd of b qed
As n goes to infinity we get arbitrarily close to r so the open cover works. But if we take a finite subcover we have to stop at some n, leaving a gap around r containing infinitely many more rationals.
June 4, 2026 at 5:40 PM
As n goes to infinity we get arbitrarily close to r so the open cover works. But if we take a finite subcover we have to stop at some n, leaving a gap around r containing infinitely many more rationals.
Now that we have an open cover of the reals and consequently F1, from the Heine-Borel Theorem we know that there is a natural number N for which we can create this finite subcover of F1
(22/n)
(22/n)
June 9, 2026 at 3:21 PM
Now that we have an open cover of the reals and consequently F1, from the Heine-Borel Theorem we know that there is a natural number N for which we can create this finite subcover of F1
(22/n)
(22/n)
Closed subsets w/ qc complement are always the image of a f.p. algebra, no? Cover the complement with distinguished opens and pass to a finite subcover, now we're looking at some V(f1,...,fr), and this is the image of Spec A/(f1,...,fr)
March 11, 2025 at 8:32 PM
Closed subsets w/ qc complement are always the image of a f.p. algebra, no? Cover the complement with distinguished opens and pass to a finite subcover, now we're looking at some V(f1,...,fr), and this is the image of Spec A/(f1,...,fr)
I do think it's a nice example in the sense that it's easy to exhibit an open cover that doesn't admit a finite subcover. But first you have to get past the strangeness of the many many different zeroes
November 5, 2024 at 5:19 PM
I do think it's a nice example in the sense that it's easy to exhibit an open cover that doesn't admit a finite subcover. But first you have to get past the strangeness of the many many different zeroes
The collection of all such intervals of both types I said above will form an open cover of the interval I = [a,b] and so this collection must have a finite subcover. So there is some finite number of the kind of open intervals I defined above that covers I.
(7/n)
(7/n)
June 9, 2026 at 2:29 PM
The collection of all such intervals of both types I said above will form an open cover of the interval I = [a,b] and so this collection must have a finite subcover. So there is some finite number of the kind of open intervals I defined above that covers I.
(7/n)
(7/n)
An Analysis of the First Proofs of the Heine-Borel Theorem
(Heine-Borel: If a set S of real numbers is closed and bounded, then the set S is compact. That is, if a set S of real numbers is closed and bounded, then every open cover of the set S has a finite subcover.)
old.maa.org/press/period...
(Heine-Borel: If a set S of real numbers is closed and bounded, then the set S is compact. That is, if a set S of real numbers is closed and bounded, then every open cover of the set S has a finite subcover.)
old.maa.org/press/period...
March 8, 2025 at 9:39 PM
An Analysis of the First Proofs of the Heine-Borel Theorem
(Heine-Borel: If a set S of real numbers is closed and bounded, then the set S is compact. That is, if a set S of real numbers is closed and bounded, then every open cover of the set S has a finite subcover.)
old.maa.org/press/period...
(Heine-Borel: If a set S of real numbers is closed and bounded, then the set S is compact. That is, if a set S of real numbers is closed and bounded, then every open cover of the set S has a finite subcover.)
old.maa.org/press/period...
(Part of the joke, here, is that my father — who was a theoretical physicist — once told me the definition of paracompactness was “every open cover has a countable subcover”. Years later I pointed out to him that, no, this was “Lindelöf”, and he rolled his eyes at the distinction.)
John Andrew Madore (1938–2020)
www.madore.org
November 16, 2025 at 3:09 PM
(Part of the joke, here, is that my father — who was a theoretical physicist — once told me the definition of paracompactness was “every open cover has a countable subcover”. Years later I pointed out to him that, no, this was “Lindelöf”, and he rolled his eyes at the distinction.)
3/ I like introducing the concept of compactness as "the kinds of sets over which we know we can integrate real or complex valued functions". That makes it clear right away why we care about this concept, and also shows why the finite subcover is important.
April 10, 2024 at 12:00 PM
3/ I like introducing the concept of compactness as "the kinds of sets over which we know we can integrate real or complex valued functions". That makes it clear right away why we care about this concept, and also shows why the finite subcover is important.
1/ When first learning topology (or the topology needed for real analysis), the first exposure to the concept of a compact set is really confusing, because the definition of a compact set (every open cover of the set has a finite subcover) is pretty much entirely nonintuitive.
April 10, 2024 at 12:00 PM
1/ When first learning topology (or the topology needed for real analysis), the first exposure to the concept of a compact set is really confusing, because the definition of a compact set (every open cover of the set has a finite subcover) is pretty much entirely nonintuitive.
We can define a set E like below. If we can show that b is also in E, then it will mean the whole interval [a,b] has a finite subcover.
(6/n)
(6/n)
![Define E to be the set of numbers $x \in [a,b]$ with the property the interval $[a,x]$ can be covered by a finite number of the sets of $\mathcal{F}$.\\](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:4mx5fgnf6nx6hlfngtfivh3y/bafkreib4uikwmwtpsygmngedr7wqx44nlxnnzocwknscxtyyseu6eh7tji)
May 21, 2026 at 8:43 AM
We can define a set E like below. If we can show that b is also in E, then it will mean the whole interval [a,b] has a finite subcover.
(6/n)
(6/n)
2. Is the surreal unit interval [0,1] compact? That is, does every open cover by intervals admit a finite subcover?
July 24, 2025 at 2:21 AM
2. Is the surreal unit interval [0,1] compact? That is, does every open cover by intervals admit a finite subcover?