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I'm a student of mathematics in Germany.


1d
comment Every sequence in $\mathbb{R}$ has a monotonic subsequence
Let us continue this discussion in chat.
1d
comment Every sequence in $\mathbb{R}$ has a monotonic subsequence
@TonyK Okay, thanks. As I said, I’m experimenting with writing styles. I’m glad for the feedback: Thank you. I hope it’s okay now. If you still think the answer needs improvement we should probably move on to chat.
1d
comment Every sequence in $\mathbb{R}$ has a monotonic subsequence
@TonyK No, just any one of them, of course. But for that, you can use choice or pick the first one – why does it matter?
1d
comment Every sequence in $\mathbb{R}$ has a monotonic subsequence
@TonyK Why should it matter whether I take the first index $k$ for a given $n$ or not?
1d
comment Every sequence in $\mathbb{R}$ has a monotonic subsequence
@TonyK I get you criticism. I’m experimenting with different writing styles. I hope it’s clearer now. The argument should work. (Although I see a little gap. Let me fix this as well.)
Dec
19
comment Simple question on diagrams in a category
No, it doesn’t.
Dec
18
comment Products of homology groups
Isn’t the embedding rather induced by the projection $\star → S^n$?
Dec
18
comment What's the definition of a “local property”?
I think they are not equivalent in general. But I agree with you in that I prefer definition (3) over the others.
Dec
18
comment What's the definition of a “local property”?
Good answer! Sadly, the second description of how “locally” is used and understood isn’t always applicable: Many topologists think of a locally compact space as one with each point having a compact neighbourhood, not an entire local base of compact neighbourhoods.
Dec
16
comment So bad at maths sometimes I feel depressed about it
“You will always find someone who knows more math than you, and this is ok. This goes for all of us.” – I think this is false by Zorn’s lemma or something.
Dec
16
comment So bad at maths sometimes I feel depressed about it
For the emotional part of your situation: Nothing’s lost, I think. I wouldn’t worry too much. Don’t be embarassed, don’t feel ashamed, don’t feel bad: Everything’s fine. For the actual advice on studying calculus: I don’t know.
Dec
16
comment Prove that this element is nonzero in a tensor product
Just mentioning it because I needed a moment to see exactly how your answer applies to the question.
Dec
16
comment iPod Shuffle question
Please replace "iPod" by "MP3 player" or "tracklist".
Dec
16
comment Prove that this element is nonzero in a tensor product
If you use flatness of $ℚ$, then it already suffices to say $ℤ → \prod_{n ∈ ℕ} ℤ/nℤ$ has trivial kernel for $\bigcap_{n ∈ ℕ} nℤ = 0$, so tensoring by $ℚ$ we get an injective map $ℚ \otimes_ℤ ℤ → ℚ \otimes_ℤ \prod_{n ∈ ℕ} ℤ/nℤ$.
Dec
15
comment Extending a uniformly continous function to the closure of its domain
By “closer” you mean “closure”, right?
Dec
14
comment Are the polynomial functions on $S^1$ dense in $C(S^1,ℂ)$?
You don’t need Morera’s theorem, but only Cauchy’s integral theorem, do you? Or am I missing something? Anyway, really nice proof, I appreciate it a lot! Edit: Oh, from your last paragraph, it is actually clear that you understand Morera’s theorem to be the equivalence.
Dec
14
comment Can $ℂ$ be viewed as a (nontrivial) field of fractions?
@MartinBrandenburg I should have written “nontrivial”. I wanted to avoid “Yeah, well $S = ℂ$.” but I feared there might be some other nontrivial subrings I overlooked which are equally pointless, so I wrote “interesting”. The answer given by Hurkyl is absolutely satisfactory to me.
Dec
14
comment Can $ℂ$ be viewed as a (nontrivial) field of fractions?
@Sal: Why not? Since the continuation of the $p$-adic absolute value has to be unique for algebraic extensions of $ℚ_p$, it makes sense to define a continued $p$-adic absolute value on the algebraic closure $\overline{ℚ_p}$. Because this value defines the metric on $\overline{ℚ_p}$, it has to be uniformly continuous and can therefore be extended to its own metric completion. The metric on the completion is still an ultrametric and so the ring given by Hurkyl is still a valid ring whose field of fractions must be all of $ℂ_p$.
Dec
14
comment Can $ℂ$ be viewed as a (nontrivial) field of fractions?
Can we extend this to the reals as well? Is $S' = O_ℂ ∩ ℝ$ a subring with $ℝ = Q(S')$ (where $O_ℂ = \{z ∈ ℂ;~|z|_p ≤ 1\}$)?
Dec
13
comment Finding polynomial generators in a subspace
I edited your tex code. Let me know/unedit if it doesn’t fit with your intentions. What is $x$? Which base field are you considering, $ℝ$, $ℂ$? Does $\mathrm{gen} A$ denote the span of $A$ for that base field? Is $P_3$ the space of polynomials of the degree (exactly) $3$ over the base field? You should clarify.