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visits member for 3 years, 10 months
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Oct
16
comment Proof of that there is no metric on $\mathbb{R}$ which is equivalent to the natural metric and which induces a metric on $(0,1)$
Please can you provide your definition of when two metrics are equivalent?
Oct
15
comment Meaning of a discrete topological sub-space?
@Eric_ Yes, that's right. And sorry, I couldn't make sense of it before.
Oct
15
comment Meaning of a discrete topological sub-space?
Duh, beat me by 52 seconds and I didn't even get a notification that someone else posted an answer.
Oct
14
comment Density of sets
@learningmaths I see you got an answer along the same vein from Yiorgos so I'll assume your question has been answered to your satisfaction. There's nothing I can add : )
Oct
14
comment Density of sets
I have to leave my keyboard right now and I might add an answer to the second question when I return.
Oct
14
comment Density of sets
+1 for providing a question I like thinking about.
Oct
11
comment Metric topology induced by the sum of two metrics
Yay! You're back. Cool.
Oct
9
comment Follow up on a previous question of mine (characters in star algebra)
@MartinArgerami True but I still don't understand what's going on here. Concretely, what is OP doing wrong in the proof in the question? In the question the C star algebra is not zero so you can pick any non-zero element $a$ and use that $r(a^\ast a) = \|a^\ast a\|>0$, no?
Oct
9
comment Follow up on a previous question of mine (characters in star algebra)
Also every proper modular ideal is contained in a maximal ideal so even if there is no $1$ it seems plausible that maximal ideals exist. (?)
Oct
9
comment Follow up on a previous question of mine (characters in star algebra)
Please could you enlighten me as to what's going on here? It's a theorem in Murphy's book (see thm. 1.3.4. on page 14) that in a non-unital abelian Banach algebra $A$ we have $\sigma (a) = \{\tau (a) \mid \tau \in \Omega (A)\} \cup \{0\}$, so surely there cannot be a counterexample. Thank you in advance.
Oct
5
comment Every homomorphism on a C*-algebra is a *-homomorphism
Okay, but I didn't understand why it was not correct...(Sorry, I only saw your reply now)
Sep
26
comment Every homomorphism on a C*-algebra is a *-homomorphism
@niki What was the mistake?
Sep
1
comment Homology of disjoint union is direct sum of homologies
@Vrouvrou Sorry this is way too long ago and I think I never really understood Mayer-Vietoris, sadly.
Aug
18
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.