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Jun
16
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.
Jun
14
comment “Every linear mapping on a finite dimensional space is continuous”
Isn't $T(E)$ of dimension $\le n$? (assuming $E$ is of dimension $n$)
Jun
13
comment In topology class, continuous and surjective problem
@topy It was a pleasure!
Jun
13
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.
Jun
13
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).
Jun
13
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 $X$?
Jun
7
comment Find the minimum distance that equal maximum inner product
Yes, I'm still trying to understand.
Jun
7
comment Find the minimum distance that equal maximum inner product
But left side of what?
Jun
7
comment Find the minimum distance that equal maximum inner product
Is there a condition missing or am I misunderstanding something?
Jun
7
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 $0$.
Jun
7
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?
Jun
7
comment How to prove uniform continuity problem!
@SwapnilTri Thank you for your kind words : )
Jun
6
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?
Jun
4
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.
Jun
4
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.
Jun
4
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.
Jun
4
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
@tomasz What's the difference between direct sum and "simple internal sum"? Doesn't $X=A \oplus B$ imply $X = A + B$ if $A,B$ are subrings of some $X$?
Jun
4
comment A question on countability of isolated points of a subset of R
@Seirios Oops, you are right. Thank you for your reply.
Jun
3
comment A question on countability of isolated points of a subset of R
I was wondering why you chose to use two rational points. Couldn't one argue that since $x$ is isolated there is an $\varepsilon$-ball such that $B(x,\varepsilon) \cap A = \{x\}$. Then there exists a rational $q$ in this ball. Define $\phi(x) = q$.
Jun
2
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
Thank you for your comments. I will get back to you, right now (for the next few hours) something else is keeping me from thinking about this.