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2d
comment Every subsequence of $x_n$ has a further subsequence which converges to $x$.Then the sequence $x_n$ converges to $x$.
What's the setting here? Metric spaces?
Jul
22
comment Showing that an inclusion is null homotopic
@t.b. It's funny. Reading this I barely recognise my old self from 4 years ago.
Jul
22
comment Group of Unitaries: Strong Continuity
@Freeze_S Yes. The only exception that comes to mind now is when you post a comment on someone else's post like e.g. this answer. Then the author will get notified of the comments whether they include pings or not. No problem!
Jul
21
comment Group of Unitaries: Strong Continuity
@Freeze_S Wonderful. Also note that unless you ping users with "@username" they won't get pinged and won't notice it if you replied. I only saw it because I came here to check.
Jul
21
comment Group of Unitaries: Strong Continuity
@Freeze_S So why is the downvote still there? I'm going to make a trivial edit in case you don't have enough privileges.
Jul
15
comment Solve Burgers' Equation with side condition.
en.wikipedia.org/wiki/Burgers'_equation
Jul
14
comment Topology of uniform convergence?
I get it. Whoever wrote the Wikipedia article misuses "topology of uniform convergence" to mean "topology induced by $\|\cdot\|_\infty$". How terrible. This individual should be banned from writing about mathematics.
Jul
14
comment Topology of uniform convergence?
But the "topology of uniform convergence" seems to mean something different. Does it mean that the same word is used to mean two different things or does it mean that the "topology of uniform convergence" coincides with the topology induced by the sup-norm in certain cases?
Jul
3
comment Show that a space is separable.
@DanielRust But the product space does not only consist of continuous maps. What am I missing?
Jul
2
comment Topology of Normed Space
What Lost1 said but with $x \in X \setminus \{0\}$.
Jul
1
comment Center of $GL_n(\mathbb R)$ is the set of matrices $\lambda I$
@gniourf_gniourf Yeah. Except the question is not asking for a proof. It's asking for a proof-verification.
Jun
28
comment The special orthogonal group is a manifold
palio, shouldn't you correct the $n^2$ as pointed out to you in the comments?
Jun
28
comment Hahn Banach and separation of points
Fast : ) And I kept re-reading your answer, wondering how this works when suddenly a wild edit appeared : )
Jun
26
comment Best Less-Famous Texts for Forcing
Haim, note that Halbeisen does not contain any problems, neither solved nor unsolved.
Jun
24
comment Density of linear span of idempotents in $L^{\infty}$
@DanielFischer Could you tell me if I got it? I posted a tentative proof here.
Jun
24
comment Density of linear span of idempotents in $L^{\infty}$
@DanielFischer Oh, "essential range", I see (new word added to my vocabulary!). And now I also understand the second half of your comment. I thought we use compactness to cover the range using finitely many $\varepsilon$-balls. Then the inverse image of each ball yields a measurable set $S_k$. Pick $c_k$ to be any value in $f(S_k)$. If we choose $\varepsilon$ small enough it will make the error small enough. But boundedness is of course enough to do that.
Jun
24
comment Density of linear span of idempotents in $L^{\infty}$
@DanielFischer But why is the image of an (essentially) bounded function compact? I can only see that it's bounded (obviously).
Jun
24
comment Density of linear span of idempotents in $L^{\infty}$
@DanielFischer Oh, I see that you already gave this hint in your very first comment to this question! Thank you : )
Jun
24
comment Density of linear span of idempotents in $L^{\infty}$
@DanielFischer Could you give me a hint on how to show that the simple functions are dense in $L^\infty$? What I have is that if $\varepsilon > 0$ then the goal is to find measurable sets $S_1,\dots, S_n$ and coefficients $c_1,\dots, c_n$ such that $$ \|f- \sum_{k=1}^n c_k \chi_{S_k}\|_\infty < \varepsilon$$ Now I'm not sure how to actually construct the sets and determine the coefficients.
Jun
24
comment Density of linear span of idempotents in $L^{\infty}$
@DanielFischer Thank you! And: Yes, you're right (we are pedantic : ))