Thomas E.
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 Jun5 comment About a continuous function Yeah, that does the trick. +1 for nice answer. Jun5 comment About a continuous function Why is $f(t,x)=\max_{y\in X}f(t,y)$? Jun5 comment About a continuous function Are you sure that $m$ is well defined? Should there be $\sup_{x}$ instad of $\max_{x}$? Or are you assuming that $X$ is compact? Jun5 comment (Un-)Countable union of open sets @DaveL.Renfro: can you give some hints how to prove this? Jun5 comment Limit of $\frac{x^{x^x}}{x}$ as $x\to 0^+$ Maybe the last equality should be $...=-2\cdot 0\cdot 1=0$ instead? Jun4 comment Continuous function on metric space @JasonDeVito: You probably mean that the sum equals $0$? Jun4 comment Continuous function on metric space @Hiperion: Disjointess is not enough (but it is required), you have to use the property that $A$ and $B$ are closed. Hint: $\bar{A}=\{x\in X:d(x,A)=0\}$, where $\bar{A}$ is the closure of $A$ in $X$. Jun4 revised Continuous function on metric space edited tags Jun4 comment Checking for completeness of $\mathbb{R}$ with metric defined by $d_1(x,y) =\mid e^x - e^y \mid$ @srijan: yes, that's correct. Jun4 comment Point set topology question: compact Hausdorff topologies You probably mean "tau", which is \tau and looks the following: $\tau$. Jun4 comment Checking for completeness of $\mathbb{R}$ with metric defined by $d_1(x,y) =\mid e^x - e^y \mid$ For $d_{1}$, note that you must start with an arbitrary Cauchy sequence in $(\mathbb{R},d_{1})$ and not in the standard metric. The current argumentation does not show that $d_{1}$ is complete. Jun4 comment $\varphi\colon M\to N$ continuous and open. Then $f$ continuous iff $f\circ\varphi$ continuous. How does this proof show that $f^{-1}A$ is open in $N$? Jun4 revised Sequential characterization of closedness of the set edited tags Jun4 comment Almost sure convergence for sequence of function If the sequence $(f_{t})$ is indexed by $t\in (0,1)$, how can you observe limiting when $t\to\infty$? Jun3 comment Sequential characterization of closedness of the set @xan: Sure; you're welcome. I'm glad I was able to help you figure this out. Jun3 answered Sequential characterization of closedness of the set Jun3 comment Showing a subset of $C([0,1])$ is compact. What topology does $C([0,1])$ have? The uniform one? If yes, then work with sequences instead of covers. Try showing that any sequence has a uniformly convergent subsequence. Jun1 comment Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral Sure; I corrected it and re-opened the answer. Jun1 revised Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral added 161 characters in body Jun1 comment Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral ... by taking sup over all such $s$ we obtain $\int_{\mathbb{R}} f\leq \int_{A}f+\int_{A^{c}}f$. Can you show the other direction similarly? By choosing arbitrary simple functions $0\leq s_{1}\leq f|_{A}$ and $0\leq s_{2}\leq f|_{A^{c}}$, and showing $\int_{A}s_{1}+\int_{A}s_{2}\leq \int_{\mathbb{R}}f$, and taking supremum over all such $s_{1}$ and $s_{2}$ obtaining the other inequality.