# angry

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"I'm thinking about starting a math blog"

cue the shaking of heads and the omfgs

compactness lost

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 Apr14 comment Let $(X,\large\tau)$ be a normal topology, then show that the weak topology induced by the cont. real-valued functions on $X$ is $\large\tau$That did it, thank you! Apr14 comment Let $(X,\large\tau)$ be a normal topology, then show that the weak topology induced by the cont. real-valued functions on $X$ is $\large\tau$do I need to use one of the theorems/lemmas I mentioned? Apr14 asked Let $(X,\large\tau)$ be a normal topology, then show that the weak topology induced by the cont. real-valued functions on $X$ is $\large\tau$ Apr7 comment Given that $(X_n)_{n\geq 0}$ is a Markov Chain, prove that $(X_{kn})_{n\geq 0}$ is a Markov Chain@Donkey_2009: Could you describe the general procedure? Apr7 comment Given that $(X_n)_{n\geq 0}$ is a Markov Chain, prove that $(X_{kn})_{n\geq 0}$ is a Markov ChainWell because any power of the matrix still has rows (or columns) which add up to 1. But the Markov chain condition is more restrictive than that isn't it? Apr7 asked Given that $(X_n)_{n\geq 0}$ is a Markov Chain, prove that $(X_{kn})_{n\geq 0}$ is a Markov Chain Mar27 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 $n\geq 0$how do I check mark my question as the answer? Mar27 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 $n\geq 0$@MartinBrandenburg, I swore on the pajama name never again to use a homomorphism. Mar26 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. Mar26 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 $n\geq 0$ Mar24 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 Mar24 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. Mar24 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 Mar24 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. Mar24 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 Mar24 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 Mar24 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 $X$ Mar23 accepted Prove that the closed unit ball of $L^2[a,b]$ is closed in $L^1[a,b]$ Mar23 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? Mar23 asked Prove that the closed unit ball of $L^2[a,b]$ is closed in $L^1[a,b]$