Thomas E.
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 Mar24 comment $E$ compact, real-valued $f : E \to \mathbb{R}$ continuous iff graph is compact - is real valued necessary? @SantiagoCanez. You're right. $g^{-1}$ is clearly continuous as it's the restriction of a projection map on the first coordinate, but this detail is essential in order to conclude the continuity of $g$ given the other assumptions. Mar24 answered Are $l^p$ spaces compact? Mar24 comment infinite Union of compact sets @pratik: Each $I_{n}$ is compact. But is $\mathbb{R}$ compact? Mar24 answered infinite Union of compact sets Mar24 comment $E$ compact, real-valued $f : E \to \mathbb{R}$ continuous iff graph is compact - is real valued necessary? Can you elaborate why in the second direction of your proof: $g$ is continuous knowing that $G$ is compact and $g$ is bijective? This doesn't seem right. Mar19 revised Compact Projections to $S^1$ added 190 characters in body Mar19 comment Compact Projections to $S^1$ $\mathbb{R}^{2}$ is certainly not compact. Mar19 answered Compact Projections to $S^1$ Mar19 awarded Popular Question Mar12 answered Countability of certain subset of $\mathbb{R}$ Mar12 comment what can you conclude about the number of solutions of the linear system Ax = b? Why would $A$ have a right inverse? It could be that $a_{1},...,a_{4}$ span something lower dimensional that $\mathbb{R}^{3}$. In fact $A$ could be anything from the null matrix to full rank matrix here. Mar12 comment If $\| f \|_p \leq 1$, then $|f(x)| \leq 1$ for some $x$ Yeah, that would do the job. Mar12 comment what can you conclude about the number of solutions of the linear system Ax = b? @sovon: Which part did you not understand? Mar12 comment If $\| f \|_p \leq 1$, then $|f(x)| \leq 1$ for some $x$ @kobe: If $f>g$ then in general $\int f\geq \int g$. The pointwise consideration is not taking in to account any properties of the measure and the integration depends on the measure. I guess his question is how do you conclude here that the inequality is strict with the integrals? Might be trivial in the case of Lebesgue measure, but what is the argument? Mar12 revised what can you conclude about the number of solutions of the linear system Ax = b? added 98 characters in body Mar12 comment what can you conclude about the number of solutions of the linear system Ax = b? @YourAdHere: If you assume that $a_{1},...,a_{4}$ are the columns of $A$, then yes. But in that case there are infinitely many solutions as I pointed out. Mar12 answered what can you conclude about the number of solutions of the linear system Ax = b? Mar12 comment what can you conclude about the number of solutions of the linear system Ax = b? What do we know about $a_{1},...,a_{4}$? Are those the columns of $A$? Mar10 comment A doubt in Probability essentials by Protter @user3503589: You're welcome. Mar10 comment A doubt in Probability essentials by Protter @jdods: $X$ is an indicator function in part $(v)$ and this seems to be a discontinuation of it, and rather a general statement. In either case it makes no difference, if one leaves the absolute values the argument goes through with same steps.