just typed \sqcup_{\sqcup_{\sqcup_{...

cursed

I'm reading the above theorem in Rudin's PMA and this answer explaining the last sentence. The answer claims $f$ does not necessarily attain its supremum nor infimum on $[x_1,x_2]$. Why not? It's continuous and hence should attain those values on that compact interval. Grateful if someone could check this.

$I(x)$ is the unit step function.

Oh wait.

$f$ is only continuous at $s$.

Ok, then the answer makes sense.

@Thorgott If I understand correctly, an $n$-manifold is orientable iff every loop has a neighborhood homeomorphic to $S^1 \times \mathbb{R}^{n-1}$. A non-orientable $n$ manifold must contain a subset homeomorphic to the product of a Möbius band and $\mathbb{R}^{n-2}$. Is it true?

every embedded loop*, and the homeomorphism needs to respect the embeddings of course

I'm not sure about the second claim

sounds wrong

man, why is higher algebra so hard to parse

what would be the geometric meaning of the third fundamental form

@Thorgott :)))

I'm confused about a basic thing. If there's an inequality of the form $|L-M|\leq\epsilon$, where $\epsilon>0$ and $L,M$ are unknown constants, and I want to show by contradiction this inequality is true, how do I go about this? Is the correct negation that $L-M>\epsilon$ or $M-L>\epsilon$ and then I have to proceed by showing both these cases lead to a contradiction?

I.e. it doesn't suffice to show that only one of the cases lead to a contradiction?

The reason I'm asking in the first place is because of a statement in Rudin's...maybe I should just give you the full context.

The negation of the statement is $|L-M| > \varepsilon$, which means that either $L-M > \varepsilon$, or $-(L-M) < \varepsilon$. But this doesn't seem like the complete statement you want to show.

Presumably, there is going to be some "there exists some $\varepsilon > 0$ such that for any $\delta > 0$ [something something something]...".

Yes, kind of. Let me give you the full context.

I'm reading Rudin's Theorem 6.17. $\mathscr{R}$ means the set of Riemann integrable functions and $\mathscr{R}(\alpha)$ the set of Riemann-Stieltjes integrable functions. Rudin starts off the proof with fixing an $\epsilon>0$, and by Riemann's integrability criterion, there is a partition $P$ such that $$U(P,\alpha')-L(P,\alpha')<\epsilon.$$Then (I'm skipping a couple of steps), Rudin derives the inequality $$|U(P,f,\alpha)-U(P,\alpha')|\leq M\epsilon,$$where $M$ is constant.

He then claims this inequality remains true for every refinement of $P$, so that $$\left|\overline {\int_a^b}f \,d\alpha - \overline {\int_a^b}f\alpha' \,dx\right|\leq M\epsilon.\tag1$$ I don't get his logic, so I'm trying to deduce $(1)$ by contradiction, as done here. You see, they start off by saying, suppose $\overline {\int_a^b}f \,d\alpha - \overline {\int_a^b}f\alpha' \,dx>M\epsilon$.

@AlessandroCodenotti yeah that's understandable. Sorry for my behaviour earlier, I was mentally exhausted. I took so much time trying to find an example of an $F$-space whose subspace is not an $F$-space, just to be met with no recognition

From the perspective of nicely behaved spaces (metrizable for example), this theory is totally irrelevant, but it does become somewhat relevant if you care about Boolean algebras

The concept of an $F$-space/$F'$-space along with many other concepts were created by Gillman and Henrisken from what I know, in the context of the ring of continuous functions $C(X)$

they're important because of spaces like $\beta\mathbb{N}\setminus\mathbb{N}$, which is not basically disconnected, but is an $F$-space

similarly something like $\beta\mathbb{R}\setminus\mathbb{R}$ is a connected $F$-space

the concept of basically disconnected, $F$-space, $F'$-space, those can all be thought as variations of the concept of extremally disconnected space, where we weaken the definition by replacing open sets by cozero sets

so while not "nice" spaces, they are still important you see, because Stone-Cech compactification of nice spaces is important

what's not in the literature from what I've seen, is how those spaces behave with respect to subspaces, that is, examples of subspaces which would be badly behaved

the most popular one seems to be that in article by Dow, of an open dense subspace of a compact $F$-space which is not an $F$-space, but I find that example to not be satisfying because it's still an $F'$-space

that's why I went to search for an example, and I finally could see that the article by Gillman I found could be used to give an example of a basically disconnected space whose dense subspace is not an $F'$-space

moreover that example was relatively easy in comparison to the example given by Dow

then I observed that this example I found can be compared to the theorem that dense subspace of extremally disconnected space is extremally disconnected, that is, its total failure for basically disconnected spaces

so while not being fully satisfying, because its not a subspace of extremally disconnected space how I'd like it to be, it's still satisfying from the perspective of failure of that theorem

I said all that, but still, I don't think anyone cares about what I have to say

I think I'll stop sharing

@Jakobian Have you considered making a blog about the things that you are interested in?