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seen Aug 19 at 16:00

Aug
5
comment The configuration space of a compact space is not compact
So to show that $\{(x,y,z)|x= y\}$ is closed we can show that $\{(x,y,z)|x\not = y\}$ is open and then we use neighborhoods to separate $x$ and $y$
Aug
5
revised The configuration space of a compact space is not compact
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Aug
5
comment The configuration space of a compact space is not compact
And by the way I'm talking about only one approach :) I edited the post to be clearer.. and thanks alot for your help..
Aug
5
comment The configuration space of a compact space is not compact
It is excluded of course: all entries $x,y$ and $z$ are all distinct in $C_3(X)$. That is why its complement $\Delta$ is what we call the fat diagonal consisting of entries where at least two of them are equal.
Aug
5
asked The configuration space of a compact space is not compact
Aug
2
comment Can the boundary of a subset be open?
you mean the descrete topology?
Aug
2
comment Can the boundary of a subset be open?
Dear @Darth Geek I did not understand your notation.
Aug
2
comment Can the boundary of a subset be open?
The only example given other than the trivial discrete case is your above condition, so apart from this situation we can say that the boundary is never open ?
Aug
2
asked Can the boundary of a subset be open?
Aug
2
comment Open set whose boundary is not a null set
Isn't $(0,1)$ a simpler example than those given so far !! indeed, $(0,1)$ is an open subset of $\mathbb R$ such that $\partial (0,1)=\{0,1\}\not = \emptyset$.
Aug
1
comment Spaces where all singletons are closed
By stronger I mean a statement that insures closedness of the most general subset of a $T_1$ space, for example as I said we know that any finite subset of $T_1$ space is closed. but what about infinite subsets do we have a statement on them under the $T_1$ condition? otherwise what do we gain from defining a condition on singletons to be all closed ? Thank you for your help!
Aug
1
comment Spaces where all singletons are closed
Given that any union of finitely many closed sets is closed, what does this imply on subsets of $T_1$ spaces ? It means that any finite subset of a $T_1$ space is a closed subset ? do we have a stronger statement?
Aug
1
accepted Spaces where all singletons are closed
Aug
1
comment Spaces where all singletons are closed
I see in the wikipedia page that the examples of non $T_1$ spaces are not "familiar" spaces, does this mean that being $T_1$ is a rather weak condition that all familiar spaces have ?
Aug
1
asked Spaces where all singletons are closed
Jul
21
comment The cone minus its apex deformation retracts onto its basis
What prevents the argument in my post to be valid for all $C(X)$ insted of $C(X)-P$ is that the retraction $r$ I gave above is not well defined at $[x,0]$ because there are more than one $x$ and there is no continuous way to choose one of them to send it to $(x,1)$ is that correct ?
Jul
21
accepted The cone minus its apex deformation retracts onto its basis
Jul
21
comment The cone minus its apex deformation retracts onto its basis
Thank you Stefan. I edited my post following the comments and your answer. If what I've wrote is correct then i don't see why using the homeomorphism $q'$. Isn't simpler to do it directly as in my edited post ?
Jul
21
revised The cone minus its apex deformation retracts onto its basis
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Jul
21
revised The cone minus its apex deformation retracts onto its basis
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