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bio website tancast.com/wp-content/…
location Where it's always Christmas
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visits member for 4 years
seen 10 hours ago

Oct
21
comment Using Regular Value Theorem to Show that $S^2$ is a $2$-dimensional Manifold.
There is not much to say but would one of you post an answer? I would do it but I don't think I should get upvotes for other people's work.
Oct
20
comment What is $\bigcup_{n=1}^{\infty}[0,1-\frac{1}{n}]$?
@AsafKaragila One can never be suspicious enough though : ) I apologize, my bad!
Oct
20
comment What is $\bigcup_{n=1}^{\infty}[0,1-\frac{1}{n}]$?
@AsafKaragila Ok. (Edit: True, the two downvoters seem to have changed their minds...)
Oct
20
comment What is $\bigcup_{n=1}^{\infty}[0,1-\frac{1}{n}]$?
@AsafKaragila Hence the downvote or what?
Oct
20
comment What is $\bigcup_{n=1}^{\infty}[0,1-\frac{1}{n}]$?
Can you make up your mind whether you want to down vote or up vote?
Oct
20
comment Show that the directional derivative is linear by definition
Could you help me understand what you're doing? In the title you ask about showing that the directional derivative is linear but then in the question you don't actually use the expression for the directional derivative (which would include the direction vector).
Oct
20
comment Show that the directional derivative is linear by definition
Thank you but I still don't understand. I think I'm confused by the question. I'll post a comment there...
Oct
20
comment If every subspace of V is T-invariant, prove that T is a multiple of the identity map.
Shouldn't it be $Tw = a_v w$? And given that you use $c_v$ further down, maybe you want to change $a_v$ into $c_v$ at the beginning? Or am misunderstanding your answer?
Oct
20
comment Sequence of bounded linear functionals on $C^1[0,1]$ that shows Principle of Uniform Boundedness fails without completion.
Why "hint"? I consider this a full answer.
Oct
20
comment $X,Y$ are Banach spaces, $T$ is linear, $x_n\to 0$ and $Tx_n\to y$, then $y=0$ and $T$ is continuous.
I don't understand the stuff after "Then, ...": you seem to be showing that $y=0$ but that is one of the assumptions.
Oct
20
comment Show that the directional derivative is linear by definition
I'm sorry but could you help me understand this? What I understand so far: In higher dimension the definition of the derivative involves a matrix so it is linear. On the other hand, the directional derivative does not seem to include such a matrix. Why does it nonetheless follow from the definition of directional derivative that it is linear? Thanks a lot in advance.
Oct
19
comment Partitions without 2
Not as active as before: Even with 50% activity you'd still be one of the more active users. Is less activity due to less interesting questions? Or is it my imagination that there are less interesting questions compared to 3 years ago?
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?