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answered Is every $\sigma$-algebra generated by a partition?
Aug
22
comment if i know that every Cauchy sequence converge then exists an unique $x\in \mathop{\cap}_{n=1}^\infty I_n $?
No. And the intervals have to be nested: $I_1\supseteq I_2\supseteq I_3\supseteq\ldots$
Aug
22
comment Prove a theorem involving normed spaces and completeness.
@user162343 If there is a linear homeomorphism, you can just use the easily proven fact that $\langle x_n\rangle$ is Cauchy if for every neihborhood $O$ of $0$, there is $N$ such that $x_n-x_m\in O$ for all $n,m\geq N$.
Jul
10
comment Showing that $[0,1]$ is compact
@StanShunpike That is basically the definition of an open set. If $O_s$ is open, there must be some "wiggle room" around each of its elements. In particular, if $s\in O_s$ then for some $\epsilon>0$, $(s-\epsilon, s+\epsilon)\subseteq O_s$. The result follows now from $(s-\epsilon,s]\subseteq (s-\epsilon,s+\epsilon)\subseteq O_s$.
Jun
14
awarded  Favorite Question
May
31
comment Advantage of accepting non-measurable sets
@RonMaimon I'm looking forward to seeing your definition of "arbitrary shape".
May
30
comment Signaling game : response to zero-probability message
@Guilro This game has no proper subgame, if you take account of the choice of nature.
May
10
comment Is Riesz measure an extension of product measure?
I guess $f$ should be in $C(X\times Y)$.
Apr
3
answered Axioms Conditional Probability
Apr
1
answered Decomposition of measures as a product measures
Apr
1
revised Convergence of nested converging measures
added 1 character in body
Apr
1
answered Convergence of nested converging measures
Apr
1
comment Equivalent Conditions for Weak Convergence of Signed Radon Measures
What does $|\nu_{k}| = 1$ mean? Is $|\cdot|$ some norm?
Mar
31
awarded  Guru
Mar
27
comment Application of Riesz Spaces (k/a $K$-Lineals or Vector Lattices ) to Mathematical-Economics?
See here for the first paper to use Riesz spaces in economics.
Mar
24
awarded  Good Answer
Mar
24
comment normed space is complete if and only if the closed unit ball is complete
A closed subset of a complete metric space is complete. For the other direction, take a Cauchy sequence and shift the values by some vector so that they all lie eventually in the closed unit ball.
Mar
22
comment Compactness, continuity and the discrete topology
1. Just let $f$ be continuous and bounded. 2. Even with non-Hausdorff topologies on a finite set, the resulting $\sigma$-algebra has a very simple form. Let $x\equiv y$ if $x$ and $y$ are members of exactly the same open sets. If $M$ is finit or even countably infinite, the Borel $\sigma$-algebra will consists of all arbitrary unions of $\equiv$-equivalence classes. This equivalence classes are like points. If you quotient out, you can pretend to be working in a discrete setting.
Mar
22
comment Compactness, continuity and the discrete topology
@Kolmin Are you sure you want $f$ to be linear?
Mar
22
answered $\mu$ is a $f$-invariant measure