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seen Jan 7 at 12:25

Nov
19
comment Compact metric spaces is second countable and axiom of countable choice
Thank you! But in reason nr 2. if $ \ \mathcal V_n = \{B(x_{n,i},\frac{1}{n}), i=1,...,d(n)\} $ for some function $ d $ and if we take $ f_{n}:\mathcal V_n \rightarrow \mathbb{N} $ such that $ f_{n}(B(x_{n,i}))=i$ we cannot use this functions (injections) to prove that $ \bigcup\mathcal V_n $ is countable? Is that because we have to know more about choosing $ x_{n,i} $ and $ d(n) $?
Nov
19
accepted Compact metric spaces is second countable and axiom of countable choice
Nov
19
asked Compact metric spaces is second countable and axiom of countable choice
Oct
24
awarded  Tumbleweed
Sep
9
awarded  Editor
Sep
9
comment Van der Waerden type numbers (for geometric progressions)
Thank you. I've seen this, there are some facts about van der Waerden theorem for geometric progression, but there's nothing about "van der Waerden geometric numbers".
Sep
9
revised Van der Waerden type numbers (for geometric progressions)
remove false statement
Sep
6
revised Van der Waerden type numbers (for geometric progressions)
edited tags
Sep
6
asked Van der Waerden type numbers (for geometric progressions)
Feb
20
awarded  Supporter
Feb
20
comment $\omega(x)=\mathbb{R}^2$ is false.
Yes_______________
Feb
20
comment $\omega(x)=\mathbb{R}^2$ is false.
Great! Thank you very much.
Feb
20
awarded  Scholar
Feb
20
accepted $\omega(x)=\mathbb{R}^2$ is false.
Feb
20
awarded  Student
Feb
20
asked $\omega(x)=\mathbb{R}^2$ is false.