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Aug
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awarded  Yearling
Aug
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Aug
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Aug
3
comment Why not define 'limits' to include isolated points?
@FreshAir: Yes if the domain is first countable, e.g., a metric space, e.g., a subspace of $\mathbb R^n$, and if $a$ is not an isolated point.
Aug
3
comment Why not define 'limits' to include isolated points?
@FreshAir, I think it means $\lim_nf(x_n)=f(\lim_nx_n)$.
Jul
26
reviewed Looks OK Dominating a Four Dimensional Chessboard with Rooks
Jul
26
reviewed Leave Closed Integral of reciprocal absolute value function
Jul
26
reviewed Reopen Vector inequation problem
Jul
26
revised Alternative ways to prove $\{f:f(0)=\sum_k f(\frac{k}{\sqrt{n}})g_n (k)\}$ is dense in $\{f\in C^2 (\mathbb{R}) : f(0)=\int_{\mathbb{R}} f(u)g(u)du\}$
edited title
Jul
26
comment The sequence $f_n=x^n$ is not weakly convergent in $C[0,1]$
@reuns: The fact you quoted doesn't help here. Knowing that the sequence does not converge strongly isn't enough to conclude that it doesn't converge weakly.
Jul
26
comment Prob. 3, Sec. 3.2 in Kreyszig's Functional Analysis Book: Is the space of all polynomials of a fixed degree complete?
math.stackexchange.com/questions/168275/…, math.stackexchange.com/questions/42663/…
Jul
26
comment Prove that the linear transformations are the same.
See math.stackexchange.com/questions/57350/…
Jul
25
reviewed Close Cauchy Schwartz Inequalities
Jul
25
reviewed Reviewed Confused about a well-ordering lemma
Jul
25
reviewed Leave Open I want to find a topologicaly embedding $f : X \rightarrow Y$ and $g: Y \rightarrow X$, yet $X$ is not homeomorphic to $Y$.
Jul
25
reviewed Leave Open Prove that there exist infinite many integers $m$ such that $\gcd\left(\binom{m}{k},l\right)=1$
Jul
24
comment $A v = \lambda v \implies A^* v = \bar{\lambda} v$ if $A$ is normal
math.stackexchange.com/q/879787, math.stackexchange.com/q/436318
Jul
24
comment $A v = \lambda v \implies A^* v = \bar{\lambda} v$ if $A$ is normal
If we are using your first line, then we can note that $A-\lambda I$ is also normal, hence if $Av=\lambda v$ then $\|(A-\lambda I)^*v\|=\|(A-\lambda I) v\|=0$.
Jul
24
comment An example of Lebesgue measurable set but not Borel measurable besides the “subset of Cantor set” example.
I understand what you were asking, but you were stating stuff about $\aleph$s that depend on the (generalized) CH without relevance to the argument. Sorry I was unclear.
Jul
24
comment An example of Lebesgue measurable set but not Borel measurable besides the “subset of Cantor set” example.
It is $\aleph_2$ if you assume the Continuum Hypothesis, but you do not need CH for Cantor's theorem that $2^n>n$ even when $n$ is an infinite cardinal number. So $2^{2^{\aleph_0}}>2^{\aleph_0}$