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May
26
comment Dimension of vector space (count 0 or not)
@1234 No, the $0$ vector is strictly required by the definition of a vector space. If a space does not contain $0$ then it's not a vector space. But any set of vectors that form a basis for the space cannot contain $0$ otherwise they are not a basis.
May
26
comment Dimension of vector space (count 0 or not)
@1234 Yes, $0$ is always an element of a vector space. But it's not part of any basis and the basis is what is used to defined the dimension of the space.
May
26
comment How to prove uniform continuity problem!
@yyzzer1234 Here is how I did it: I realised that the function is continuous and that continuous functions map compact sets to compact sets and are therefore bounded on compact sets. Then $[-K,K]$ is compact for any $K$ and all I needed to do to finish was to choose $K$ conveniently. In this case, I chose $1$, because it does the trick (you see it by trying different $K$ in the fraction you're trying to bound). $K=10$ or $K=13$ will work, too.
May
26
comment Show $M_1\cap M_2$ submanifold iff $N_x(M_1)\cap N_x(M_2) = \{0\}$ and dimensions
If the manifolds are two dimension as in your question, are the normal spaces of dimension one? (the surface normal at that point) Or am I missing something?
May
26
comment Show $M_1\cap M_2$ submanifold iff $N_x(M_1)\cap N_x(M_2) = \{0\}$ and dimensions
Great thanks for the clarification! Just for the record: I did not downvote your question.
May
26
comment How to prove uniform continuity problem!
@yyzzer1234 I added more detail, I hope it's clearer now.
May
26
comment Show $M_1\cap M_2$ submanifold iff $N_x(M_1)\cap N_x(M_2) = \{0\}$ and dimensions
What?s $N_x (M_i)$?
May
26
comment How to figure whether it is a compact operator
I intended to post this when I voted to reopen this question but somehow missed the chance.
May
25
comment How to prove uniform continuity problem!
Okay, I'll try to add some stuff later (when I have time).
May
24
comment Rings having the same characters but not isomorphic.
I understand that $k[[t]]$ is probably the ring of formal power series but what is $k[[t]]t^{8 \nu}$?
May
24
comment Rings having the same characters but not isomorphic.
I'm sorry but I don't even understand the notation (I'm sure it's standard). Is $k$ a field and does $+$ denote direct sum?
May
22
comment If $g:V \rightarrow V$ is an injective linear transformation. Prove if $V$ is finite dimensional then $g$ is surjective.
Yes. That's the part I understand. Everything after "and so as..." is what I cannot follow.
May
22
comment If $g:V \rightarrow V$ is an injective linear transformation. Prove if $V$ is finite dimensional then $g$ is surjective.
I'm sorry but I don't understand your argument in $\implies$. It appears to me that you're using the rank nullity theorem in $\Longleftarrow$. And from your title I think it sounds that you're only aske to prove $\implies$.
May
15
comment Extending a functional with same norm
@MartinArgerami Thank you very much for your comment! For some reason I didn't get pinged and I only discovered it now by coincidence.
May
13
comment Reflexive normed spaces are Banach
@DanielFischer That's what I thought: nothing to prove at all.
May
13
comment Reflexive normed spaces are Banach
@DanielFischer But on Wikipedia condition (iii) would make $J$ an isomorphism of normed spaces...?
May
13
comment Reflexive normed spaces are Banach
@DanielFischer I don't think it's needed. It's enough that it's a normed vector space isomorphism, it doesn't need to be an isometry. Or am I missing something?
May
13
comment Extending a functional with same norm
@user23791 But you can. I'll add a second way of solving the question.
May
13
comment Extending a functional with same norm
@user23791 You don't need to construct it explicitly.
May
13
comment Extending a functional with same norm
@user23791 No, that's your sublinear map bounding $\phi_0$. Hahn-Banach doesn't construct an extension, it gives you its existence.