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Sep
4
comment How to show that the $k$th return to $y$ is a stopping time
Sorry, I'm new to Markov chains, can you clarify what you mean by this?
Sep
4
revised How to show that the $k$th return to $y$ is a stopping time
edited title
Sep
4
revised Are all compact sets in $ \Bbb R^n$, $G_\delta$ sets?
formatting
Sep
4
suggested approved edit on Are all compact sets in $ \Bbb R^n$, $G_\delta$ sets?
Sep
4
asked How to show that the $k$th return to $y$ is a stopping time
Sep
1
revised Proving that $a,n$ and $b, n$ relatively prime implies $ab,n$ relatively prime
corrected question
Sep
1
accepted Proving that $a,n$ and $b, n$ relatively prime implies $ab,n$ relatively prime
Sep
1
suggested rejected edit on Which of the following subsets of $\mathbb{R}^2$ are compact
Aug
31
comment Proving that $a,n$ and $b, n$ relatively prime implies $ab,n$ relatively prime
Sorry everyone! The body is a typo. I mean the $\gcd(ab,n) = 1$
Aug
31
comment Proving that $a,n$ and $b, n$ relatively prime implies $ab,n$ relatively prime
Well...this is embarrassing. Thanks!
Aug
31
asked Proving that $a,n$ and $b, n$ relatively prime implies $ab,n$ relatively prime
Aug
30
accepted Proof With and Without Truth Tables
Aug
30
asked Proof With and Without Truth Tables
Aug
27
accepted Proving lack of convergence in $\sup$-norm
Aug
27
comment Proving lack of convergence in $\sup$-norm
Ah, thank you Ilya
Aug
27
accepted Checking Uniform Convergence
Aug
27
asked Proving lack of convergence in $\sup$-norm
Aug
26
comment Convergence in $\sup$ norm $\Rightarrow$ Cauchy in $\sup$
I see. So since $(f_n)$ is convergent, I have that for $\epsilon >0$ there is an $N$, respectively an $M$, so that for $n \ge N$, $m \ge M$, we have $\|f_n -f \|_{\infty} < {\epsilon \over 2}$ and $\|f_m -f \|_{\infty} < {\epsilon \over 2}$. Therefore we have $\|f_n - f_m\|_{\infty} < \epsilon$. Thank you for this succinct explanation
Aug
26
accepted Convergence in $\sup$ norm $\Rightarrow$ Cauchy in $\sup$
Aug
26
asked Convergence in $\sup$ norm $\Rightarrow$ Cauchy in $\sup$