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8h
comment Why is the cartesian product so categorically robust?
I don’t know all of these categories, but – at least as far as I know – for most of them the corresponding forgetful functor has a left adjoint, thus is continuous. Morally speaking, I’d guess that so many of these forgetful functors have left adjoints because mathematicians (for historic reasons alone) tend to build stuff from sets and that already suggests a forgetful functor as well as an left adjoint for it.
1d
comment How to not feel bad for doing math?
@AsafKaragila Why not? It’s a real concern to some and I guess it can be a real mood killer if you’re considering spending 3–5 years working hard on one particular problem/question. Even though I don’t pursue a career as a mathematician, I can relate to this concern. Why not take it seriously?
1d
comment How to not feel bad for doing math?
@Did I added a comma and read the sentence as “It’s the only job I can think of, which doesn’t contribute to bettering the world somehow.”.
May
25
comment What mathematician you would have liked to know?
I’m no moderator, so I have not been adressed by your appeal. But let me say – why should the community make an exception and show flexibility for a question which (as far as I’m concerned) could serve as the example par excellence for provoking primarily-opinion based answers? (And by the way: Hilbert, I guess.)
May
23
comment How to insert Gothic letters in Word?
That’s not a math question, though.
May
23
accepted Are binomial series multiplicative in their bases?
May
23
asked Are binomial series multiplicative in their bases?
May
21
answered Proving that the ball is converx
May
21
comment Proving that the ball is converx
Invoke the definition of convexity and the triangle inequality.
May
15
comment If a field extension contains a cyclotomic extension is it solvable?
Recall that a solvable group requires a subnormal series in which all factors are abelian. What can you really say about $\mathrm{Gal}(L/M)$? As an extreme case, take any non-solvable extension $L/K$ and choose $M = K$, which trivially is a cyclotomic extension.
May
9
comment How to teach Mathematical Induction mathematically?
Unfortunately, in order to make it possible for others to help, you would have to list everything that you have tried so far, forcing you to relive your painful experience.
May
9
comment Some very short clarification on quotient groups
@elDin0 No. Your problem is on set-theoretic level: It seems like you are not differentiating between an element of a set and an element of an element of a set. Take your favorite mathematical starting object $x$ (if you’re a minimalist, choose $x = ∅$, the empty set) and then consider $y = \{x\}$ and $z = \{y\}$. Then $z ≠ \{x\}$ because the only element of $z$ is $y = \{x\}$ and $x ≠ \{x\}$. Likewise, $\{A_3, (1~2)A_3\} ≠ \{ \text{elements of $A_3$}, \text{elements of $(1~2)A_3$}\}$.
May
9
comment Some very short clarification on quotient groups
Do you know the isomorphism theorem for groups? How is $A_3$ defined for you? Do you know how many elements $A_3$ and$S_3$ have?
May
8
comment Looking for Open Source Math Software with Poor Documentation
I upvoted for your charity intentions solely. Felt right.
May
7
comment Existence of a holomorphic function with the desired property
@learnmore Linear maps are holomorphic, what are the linear maps $ℂ → ℂ$? Which one of these restrict to $D → D$?
May
7
comment Existence of a holomorphic function with the desired property
@Blake $f(D) ⊂ (\overline D)° = D$.
May
7
answered Existence of a holomorphic function with the desired property
May
7
comment Existence of a holomorphic function with the desired property
$\frac{3}{4}·\frac{1}{3} = \frac{1}{4}$ and $\lvert\frac{3}{4}\rvert < 1$. Otherwise, use the open mapping theorem and the Schwarz lemma.
May
7
revised Connected spaces where all subsets are either open or closed
added 44 characters in body
May
7
comment Connected spaces where all subsets are either open or closed
@bof Ah, neat. Thanks. One-point spaces being the least interesting examples imaginable, though. But the link actually proves there are no other.