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Jun
1
comment Please verify this definition of locally convex vector space
@luka5z If you found this answer helpful then you might consider upvoting it (in addition to accepting).
Jun
1
comment Please verify this definition of locally convex vector space
I think so. ${}$
Jun
1
comment Please verify this definition of locally convex vector space
No, I think it's because a norm is a seminorm. Then the family consists of just one thing: the norm. As for your definition: No, I think the definition states that the topology on the space is induced by the separating family of seminorms. But maybe that's what you mean?
Jun
1
comment Please verify this definition of locally convex vector space
Sorry, I misread your question. Note that every normed space is locally convex.
Jun
1
comment Please verify this definition of locally convex vector space
Can you include the definition of locally convex vector space in your question please?
Jun
1
answered Separating family of seminorms
Jun
1
comment $V$ is open , then $V=\{x\in \mathbb R:f(x)>0\}$ for some continuous function $f$
@Etienne Too bad : ) Thank you for your comment!
May
31
comment $V$ is open , then $V=\{x\in \mathbb R:f(x)>0\}$ for some continuous function $f$
@Etienne Would it also work to take the convolution of the characteristic function of $V$ with a bump function?
May
31
comment Compact Operator <=> Separable Range
@Freeze_S Compact operators have much nicer properties than arbitrary operators. You can think of them as behaving a bit like operators on finite dimensional vector spaces.
May
31
answered Compact Operator <=> Separable Range
May
31
comment Compact Operator <=> Separable Range
It's true that compact implies separable range. The other direction I suspect does not hold. Why not try to find a counterexample?
May
28
revised Uniform convergence on compact sets
edited body; edited tags; edited title
May
28
comment How to write this as one matrix?
What is the notation $(\cdot, \cdot)$?
May
28
comment Show that $Z_i\sim N(\mu_i,V_{ii})$
Why the downvote?
May
28
comment Show that $E(Z)=\mu, Cov(Z)=V$
Why the downvote?
May
28
answered Ordinal $10^\omega$
May
27
comment Prove or disprove: If L : V → U and M : U → V are linear mappings such that (M ◦ L)(x) = x for all x ∈ V, then M is onto.
I think you're absolutely right, it's obvious and there is nothing to prove. Given any $x$ in $V$, $Lx$ maps to it.
May
27
comment Math Major: How to read textbooks in better style or method ? And how to select best books?
This is the second time that I felt frustrated that my reading is far too slow and I read your answer and then didn't feel so bad anymore. Wonderful.
May
27
comment Diagonal convergence
Oh I see, it's in the title: $x_n \subseteq x_r$.
May
27
comment Diagonal convergence
Is $x_r$ a subsequence of $x_n$? I assume $f \in C(X)$?