Thoth
Reputation
2,061
Next privilege 2,500 Rep.
Create tag synonyms
 Mar 26 comment Let $\mathfrak{m}$ be maximal in $R$. Show that $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ is an $R/\mathfrak{m}$-vector space for all $n\geq 0$ This is a bit beyond my knowledge level, so I can revisit it once I have the requisite algebra, I'd still like to know if there were in any flaws in my reasoning above. Mar 26 asked Let $\mathfrak{m}$ be maximal in $R$. Show that $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ is an $R/\mathfrak{m}$-vector space for all $n\geq 0$ Mar 24 comment For a family of functions $F\subset C(X)$ in the metric space $(C(X),d)$, if $F$ is compact on compact subsets of $X$, then $F$ is compact on $X$ @Jose27, but I don't think the family you gave is equicontinuous Mar 24 comment For a family of functions $F\subset C(X)$ in the metric space $(C(X),d)$, if $F$ is compact on compact subsets of $X$, then $F$ is compact on $X$ @Jose27, oh you're completely right, so $(f_n)$ doesn't converge at all with the $d$ metric. Mar 24 revised For a family of functions $F\subset C(X)$ in the metric space $(C(X),d)$, if $F$ is compact on compact subsets of $X$, then $F$ is compact on $X$ added 1 characters in body Mar 24 comment For a family of functions $F\subset C(X)$ in the metric space $(C(X),d)$, if $F$ is compact on compact subsets of $X$, then $F$ is compact on $X$ @Jose27, there seems to be two things going on here. First, I don't think a lack of uniform convergence poses a problem, as the $\frac{1}{2^n}$ in my metric allows one to find a fixed $N$ independent of the point chosen from $X$. However your counterexample does work to disprove the problem as stated in the title for a different reason: $f=0$ is not in your family, and thus your family isn't closed (read: not compact). However in the problem as stated in the body of the post, $F$ is required to be closed, so in that case your counterexample doesn't satisfy the assumptions. Mar 24 revised For a family of functions $F\subset C(X)$ in the metric space $(C(X),d)$, if $F$ is compact on compact subsets of $X$, then $F$ is compact on $X$ added 47 characters in body Mar 24 revised For a family of functions $F\subset C(X)$ in the metric space $(C(X),d)$, if $F$ is compact on compact subsets of $X$, then $F$ is compact on $X$ edited body Mar 24 asked For a family of functions $F\subset C(X)$ in the metric space $(C(X),d)$, if $F$ is compact on compact subsets of $X$, then $F$ is compact on $X$ Mar 23 accepted Prove that the closed unit ball of $L^2[a,b]$ is closed in $L^1[a,b]$ Mar 23 comment Prove that the closed unit ball of $L^2[a,b]$ is closed in $L^1[a,b]$ The subsequence can just be $f_n$ itself right? Mar 23 asked Prove that the closed unit ball of $L^2[a,b]$ is closed in $L^1[a,b]$ Mar 20 comment Why is the Euclidean metric called the prime at infinity? @ZhenLin: The answer in that question is sufficiently far above my head that I can't quite tell whether it would answer my question, but it seems like it might, thanks. Mar 20 asked Why is the Euclidean metric called the prime at infinity? Mar 14 awarded Yearling Mar 5 comment Prove that the sequence satisfying $a^{p^n}\equiv a^{p^{n-1}}$mod $p^n$ for $n\geq 1$, converges to the Teichmuller Representative I guess continuity was the issue that was stumping me. Since the only way to represent this Teichmuller Representative is with an infinite sequence (or series), the best I could hope to do was to show better and better approximations of the number were evaluated closer and closer to zero by $x^p-x$, but that would rest on the continuity of polynomials in $\mathbb{Q}_p$, which I'm not sure I was suppose to know. Mar 5 revised Prove that the sequence satisfying $a^{p^n}\equiv a^{p^{n-1}}$mod $p^n$ for $n\geq 1$, converges to the Teichmuller Representative added 209 characters in body Mar 5 asked Prove that the sequence satisfying $a^{p^n}\equiv a^{p^{n-1}}$mod $p^n$ for $n\geq 1$, converges to the Teichmuller Representative Mar 3 comment Prove that $\displaystyle\mathbb{Q}_p$ always contains $p$ solutions to the equation $x^p-x=0$ This is a really enlightening answer, thanks for this. Mar 3 accepted Prove that $\displaystyle\mathbb{Q}_p$ always contains $p$ solutions to the equation $x^p-x=0$