user63123
Reputation
Top tag
Next privilege 75 Rep.
Set bounties
 Nov19 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 $d(n)$? Nov19 accepted Compact metric spaces is second countable and axiom of countable choice Nov19 asked Compact metric spaces is second countable and axiom of countable choice Oct24 awarded Tumbleweed Sep9 awarded Editor Sep9 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". Sep9 revised Van der Waerden type numbers (for geometric progressions) remove false statement Sep6 revised Van der Waerden type numbers (for geometric progressions) edited tags Sep6 asked Van der Waerden type numbers (for geometric progressions) Feb20 awarded Supporter Feb20 comment $\omega(x)=\mathbb{R}^2$ is false. Yes_______________ Feb20 comment $\omega(x)=\mathbb{R}^2$ is false. Great! Thank you very much. Feb20 awarded Scholar Feb20 accepted $\omega(x)=\mathbb{R}^2$ is false. Feb20 awarded Student Feb20 asked $\omega(x)=\mathbb{R}^2$ is false.