# Matt N.

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# 2,854 Comments

 Jun24 comment Density of linear span of idempotents in $L^{\infty}$ @DanielFischer Am I right that $f\in L^\infty$ is idempotent if and only if $f=\chi_S$ (characteristic function) for $S\subseteq \Omega$ measurable? Jun19 comment Coarsest and Finest Topology Then it depends on whether the smallest generated topology is a subset of the given family or not. Maybe you could add the new assumption to the question and then someone more knowledgable than myself will write an answer. Jun19 comment Coarsest and Finest Topology No. For example if the family does not contain the empty set. Jun18 comment Distribution of $\log X$ @t.b. There is a smoke signal for you. And a cave painting. Not sure you get pinged if I send you a comment but I'm going to try. Jun16 comment Another limit to evaluate: $\lim\limits_{x \to \infty}\frac{x-\sin x }{x-\tan x}$ Copy pasting from the question into wolfram alpha gives $1$ for me. Jun14 comment “Every linear mapping on a finite dimensional space is continuous” Isn't $T(E)$ of dimension $\le n$? (assuming $E$ is of dimension $n$) Jun13 comment In topology class, continuous and surjective problem @topy It was a pleasure! Jun13 comment In topology class, continuous and surjective problem @topy Regarding your question in the comment: If you endow $[a,b]$ with the subspace topology then $[a,y) = [a,b]\cap (-\infty,y)$ hence $[a,y)$ is open in $[a,b]$. It is not closed since if it was both closed and open we could write $[a,b]$ as a disjoint union of open sets which would contradict the fact that $[a,b]$ is connected. Jun13 comment In topology class, continuous and surjective problem @topy They are open by the definition of the subspace topology: If $Y$ is any subset of $\mathbb R$ then a set $S\subseteq Y$ in the subspace topology on $Y$ is defined to be open if there exists an open set $O \subseteq \mathbb R$ such that $S= Y \cap O$. Now in your question $Y$ is $f([0,1])$. Since $(y,\infty)$ is open in $\mathbb R$ the set $f([0,1]) \cap (y,\infty)$ is open in $f([0,1])$ (in the subspace topology). Jun13 comment A problem about general topology. @DanielFischer What I don't understand is: how can one check something is a base for the topology on $X$ when one is not given a topology on $X$? Jun7 comment Find the minimum distance that equal maximum inner product Yes, I'm still trying to understand. Jun7 comment Find the minimum distance that equal maximum inner product But left side of what? Jun7 comment Find the minimum distance that equal maximum inner product Is there a condition missing or am I misunderstanding something? Jun7 comment Find the minimum distance that equal maximum inner product Also, I'm not sure this holds. If $H=\mathbb R^2$ and $x$ is $(1,0)$, $M$ is the x-axis then the minimal distance between $x$ and the x-axis is zero but the minimal distance between points on the y-axis with $\|y\|=1$ is strictly greater than $0$. Jun7 comment Find the minimum distance that equal maximum inner product What do you mean by "...LHS is $Px_0$ now how to show RHS is also $Px_0$"? I'm not being obtuse, I really don't understand. Could you elaborate a bit please? Jun7 comment How to prove uniform continuity problem! @SwapnilTri Thank you for your kind words : ) Jun6 comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$ @tomasz Yes but your first comment implies that my first comment does not pertain. Now I am thinking that it does: if you can apply the splitting lemma it is a possible answer to the question. No? Jun4 comment What does it mean to say that a forcing “collapses cardinals”? @ArthurFischer Yes, that's what I meant. Thank you for your comment. : ) But I'm still confused how this implies that some cardinals are no longer cardinals in the extension. Jun4 comment What does it mean to say that a forcing “collapses cardinals”? @AndresCaicedo I thought it meant that a cardinal will have lower cardinality in the extension. Jun4 comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$ Yes, I can prove it. That is: I don't think there is much to prove especially given your answer and your comment.