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I'm a student of mathematics in Germany.


43m
awarded  Enlightened
1h
awarded  Nice Answer
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
revised Every sequence in $\mathbb{R}$ has a monotonic subsequence
added 49 characters in body
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
revised Every sequence in $\mathbb{R}$ has a monotonic subsequence
added 21 characters in body
1d
revised Every sequence in $\mathbb{R}$ has a monotonic subsequence
added 21 characters in body
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.)
1d
revised Every sequence in $\mathbb{R}$ has a monotonic subsequence
tried to clarify the argument
1d
revised Every sequence in $\mathbb{R}$ has a monotonic subsequence
edited body
1d
revised Every sequence in $\mathbb{R}$ has a monotonic subsequence
added a possibility to fix the argument
1d
answered Every sequence in $\mathbb{R}$ has a monotonic subsequence
Dec
19
awarded  Constituent
Dec
19
comment Simple question on diagrams in a category
No, it doesn’t.
Dec
19
revised How to show $\mathbf{Q} $ is not free
added the non-finitely-generated part
Dec
19
revised How to show $\mathbf{Q} $ is not free
added 8 characters in body
Dec
19
answered How to show $\mathbf{Q} $ is not free
Dec
18
comment Products of homology groups
Isn’t the embedding rather induced by the projection $\star → S^n$?