angry

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"I'm thinking about starting a math blog"

cue the shaking of heads and the omfgs

compactness lost

	1,376 reputation	bio	website		visits	member for	1 year, 3 months
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Apr 14	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!
Apr 14	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?
Apr 14	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 $\large\tau$
Apr 7	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?
Apr 7	comment	Given that $(X_n)_{n\geq 0}$ is a Markov Chain, prove that $(X_{kn})_{n\geq 0}$ is a Markov Chain Well 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?
Apr 7	asked	Given that $(X_n)_{n\geq 0}$ is a Markov Chain, prove that $(X_{kn})_{n\geq 0}$ is a Markov Chain
Mar 27	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?
Mar 27	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.
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 $L^1[a,b]$