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Dec
25
reviewed Reviewed integral of $\ln x$ from 0 to 1
Dec
25
revised Difference between $\lim_{n\to\infty}P(|X_n-X|<\epsilon)=1$ and $\lim_{n\to\infty}P(|X_n-X|=0)=1$
added 29 characters in body
Dec
25
revised Difference between $\lim_{n\to\infty}P(|X_n-X|<\epsilon)=1$ and $\lim_{n\to\infty}P(|X_n-X|=0)=1$
added 29 characters in body
Dec
25
answered Difference between $\lim_{n\to\infty}P(|X_n-X|<\epsilon)=1$ and $\lim_{n\to\infty}P(|X_n-X|=0)=1$
Dec
25
comment Proving $\lim_{n\to\infty}P\,(X_n=X) = 1$ when the sequence of random variables $X_n$ converges to $X$ a.s.
This is not true. Take $X_{n}=\frac{1}{n}$ for all $n\in\mathbb{N}$. Then $X_{n}\to 0=:X$‚ but $P(X_{n}=X)=P(\frac{1}{n}=0)=0$ for all $n\in\mathbb{N}$. Hence $\lim_{n\to\infty}P(X_{n}=X)=0$.
Dec
25
answered A question about the contractibility of the Sierpinski space
Dec
24
answered Norm of vector in $\mathbb{R}^3$ with multiple
Dec
24
answered Compact and open subspaces using products
Dec
24
comment Compact and open subspaces using products
It's true that $C\times K$ is a compact subspace of $X\times Y$, but taking $\{U\}$ as its open cover, the only finite subcover would be $\{U\}$ again.
Dec
24
comment Sets of second category-topology
Do you mean a countable union?
Dec
22
comment Urysohn's metrization theorem and Borel image
@Martin By $N_{2}$ I meant second countability and by $T_{3}$ regularity. And thanks a lot, your comments have been a valuable input to this question.
Dec
22
revised Urysohn's metrization theorem and Borel image
edited body
Dec
22
revised Urysohn's metrization theorem and Borel image
Rephrased the question a little.
Dec
22
comment Urysohn's metrization theorem and Borel image
@Martin. Thanks. I knew the connection of $G_{\delta}$ sets and completely metrizable sets, but it didn't occur to me to use it in this context. This is in fact a very nice a way of proving this. Do you think that $N_{2}$ and $T_{3}$ are alone enough for $f(X)$ to be Borel? I think I will slightly rephrase my question in this thread.
Dec
22
revised Set of all finite subsets of $\mathbb{N}$ is a countable set
Edited the word {N} in title and in the question.
Dec
22
comment Urysohn's metrization theorem and Borel image
@SamL. Very nice point that you're saying. It also answers one part of my question. Namely, that it does not follow from the topological properties as they stand, but explicitly from the choice of $f$. And the proof that I am looking at uses Urysohn's lemma countably many times appropriately with second countability of $X$, and does not give the explicit function. However, I am looking at some other results in measure theory that use this property quite freely. E.g. when embedding a Polish space to the Hilbert cube and assuming that $f(X)$ is a measurable set respect to a strict Borel measure.
Dec
22
revised Urysohn's metrization theorem and Borel image
deleted 7 characters in body
Dec
21
asked Urysohn's metrization theorem and Borel image
Dec
20
revised A continuity example
Added LaTeX
Dec
20
revised Set of measure zero and limsup of its covering
edited LaTeX.