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Doing math.


Jun
5
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?
Jun
4
comment Continuous function on metric space
@JasonDeVito: You probably mean that the sum equals $0$?
Jun
4
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 $X$.
Jun
4
revised Continuous function on metric space
edited tags
Jun
4
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.
Jun
4
comment Point set topology question: compact Hausdorff topologies
You probably mean "tau", which is \tau and looks the following: $\tau$.
Jun
4
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.
Jun
4
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 $N$?
Jun
4
revised Sequential characterization of closedness of the set
edited tags
Jun
4
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 $t\to\infty$?
Jun
3
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.
Jun
3
answered Sequential characterization of closedness of the set
Jun
3
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.
Jun
1
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.
Jun
1
revised Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral
added 161 characters in body
Jun
1
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.
Jun
1
comment Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral
With this lemma it works. If you want to avoid working suprumems over sets in equalities, you may do something like the following too (you already got the idea). $"\Rightarrow"$: Let $0\leq s\leq f$ be arbitrary simple function, and define $s_{1}=s 1_{A}$ and $s_{2}=s 1_{A^{c}}$, whence $s=s_{1}+s_{2}$ and $s_{1}(x)\leq f(x)$ for $x\in A$ and $s_{2}(x)\leq f(x)$ for $x\in A^{c}$. Now using what you had already proven $\int_{R} s=\int_{R} s 1_{A}+s 1_{A^{c}}=\int_{R}s 1_{A}+\int_{R}s 1_{A^{c}}=\int_{A} s_{1}+\int_{A^{c}} s_{2}\leq \int_{A}f+\int_{A^{c}}f$. (Continues below)
Jun
1
comment Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral
There was a tiny flaw, which I should've fixed, and I saw you posted your own:) I will read it soon.
Jun
1
answered Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral
Jun
1
comment Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral
Have you proved that for any non-negative measurable function $f$ there exists a nondecreasing sequence of simple functions $(\psi_{i})$ so that $\psi_{i}\to f$ point-wise? With this and monotone convergence theorem you may conclude the result. Or you may want to notice that $\int_{A} f=\int_{A}f+\int_{A^{c}}0=\int_{A}f1_{A}+\int_{A^{c}}f1_{A}=\int_{\mathbb{R}}f 1_{A}$.