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7h
comment Does the open mapping theorem imply the Baire category theorem?
I've taken a quick look, and I think it looks correct! A couple comments: (a) I guess you want to take $y_0 = 0$ or something like that, to get the induction started. (2) The notation $y_{n+1} \in 3^{-(n+1)} S_{n+1}$ is a little hard to sort out. I guess a simpler way of stating the desired statement is simply that $\|T_n y_n\| \ge 2 \cdot 3^{-(n+1)} \|T_n\|$ for all $n$.
8h
comment Proof that $C^\infty(0,1)$ is a subring
It's a subring of lots of different rings - you'll have to decide which one you want to use. As for showing it's a subring - write down the definition of "subring" and start checking the required conditions. Where do you get stuck?
8h
comment Every $f\in\omega^\omega$ is bounded by the “increasing enumeration” of the intersection of a countable dense set and a dense open set in $\mathbb{R}$
meta.math.stackexchange.com/a/9960/822
23h
comment A Vitali set is non-measurable, direct proof, without using countable additivity
Is this a back-door proof that $m^*$ is finitely additive on measurable sets? Anyway, that's a significant improvement, because it's certainly a lot less work than countable additivity.
1d
comment A Vitali set is non-measurable, direct proof, without using countable additivity
@AsafKaragila: I might get to teach the class again in 2017... :-)
1d
revised A Vitali set is non-measurable, direct proof, without using countable additivity
added 357 characters in body
1d
asked A Vitali set is non-measurable, direct proof, without using countable additivity
1d
revised Looking for example of topological spaces where sequential continuity does not imply continuity
comment at end
1d
comment Does every compact simply-connected subset of $\mathbb{R}^n$ have an efficient $r$-covering path for all $r>0$?
"Differentiable" or any stronger assumption is going to complicate things, for sure. Maybe you would rather specify "Lipschitz" or "rectifiable"? Anyway, my instinct would be to take a sequence of $r$-covering paths whose lengths approach the infimum, and try to use Arzela-Ascoli to produce a subsequential limit that should be a minimizer.
1d
answered If $\lambda=$ measure of a set and all $G_k$'s are open sets, then : $\lambda ( \cup_{k=1}^{\infty} G_k ) \le \sum _{k=1}^{\infty}\lambda ( G_k)$
1d
comment If $\lambda=$ measure of a set and all $G_k$'s are open sets, then : $\lambda ( \cup_{k=1}^{\infty} G_k ) \le \sum _{k=1}^{\infty}\lambda ( G_k)$
I assume you meant to write $I_1, I_2 \subseteq G_3$, etc, not $\in$.
1d
comment A conjecture about generating algebras on a probability space
Oh, I see now. I was confused because $n$ was used in two different ways. I would denote your "one single set" as $F_N$ or something like that.
1d
revised Looking for example of topological spaces where sequential continuity does not imply continuity
added 137 characters in body
1d
comment If a functional is bounded below must it have a minimizing sequence?
Then you have to amend your problem statement, because you said $\mathcal{I} : X \to \mathbb{R}$, and $\infty$ is not a real number. But the idea will still work, since then $\{\mathcal{I}(u) \mid u \in X\} = \{\infty\}$, whose infimum is also $\infty$. So $\alpha = \infty$ and every sequence is minimizing.
1d
comment A compact $n$-manifold is orientable iff there is an everywhere nonzero $n$-form
As a suggestion: Your original title was "Let M be a compact differentiable manifold of dimension n without boundary." It's not very helpful to just use the first sentence of your problem statement as a title; readers can't tell from that what your question is actually about. (If everyone followed that, just think of the number of questions whose title would be "Let $p$ be a prime".) It's also not good style to have the title contain information which is not also in the body. I changed it to something that seems better to me, but you are welcome to make further edits.
1d
revised A compact $n$-manifold is orientable iff there is an everywhere nonzero $n$-form
better title
1d
answered Looking for example of topological spaces where sequential continuity does not imply continuity
1d
comment Is a maximal open simply connected subset $U$ of a manifold $M$, necessarily dense?
At a minimum, you want to assume $M$ is connected, right?
1d
comment If a functional is bounded below must it have a minimizing sequence?
This is an immediate consequence of the definition of infimum.
1d
comment Folland, Real Analysis Theorem 1.19
The proof seems to be backward. You should explain how the sets $V_j$, $H_j$ are chosen before trying to take unions and intersections of them.