# Matt N.

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 Jun1 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). Jun1 comment Please verify this definition of locally convex vector space I think so. ${}$ Jun1 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? Jun1 comment Please verify this definition of locally convex vector space Sorry, I misread your question. Note that every normed space is locally convex. Jun1 comment Please verify this definition of locally convex vector space Can you include the definition of locally convex vector space in your question please? Jun1 answered Separating family of seminorms Jun1 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! May31 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? May31 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. May31 answered Compact Operator <=> Separable Range May31 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? May28 revised Uniform convergence on compact sets edited body; edited tags; edited title May28 comment How to write this as one matrix? What is the notation $(\cdot, \cdot)$? May28 comment Show that $Z_i\sim N(\mu_i,V_{ii})$ Why the downvote? May28 comment Show that $E(Z)=\mu, Cov(Z)=V$ Why the downvote? May28 answered Ordinal $10^\omega$ May27 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. May27 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. May27 comment Diagonal convergence Oh I see, it's in the title: $x_n \subseteq x_r$. May27 comment Diagonal convergence Is $x_r$ a subsequence of $x_n$? I assume $f \in C(X)$?