Intermediate Math Challenge - September 2018

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  • #26
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Pardon the offtopic.
I find the introduction in the first post a bit strange. Summer is long gone now :DD Or you could be talking about summer 2019 or the folks that live their lives upside down.
Indian Summer :-p
 
  • #27
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Let [itex](X,d)[/itex] be a metric space. Denote [itex]*U = \mbox{Int}(U^c)[/itex], where [itex]U\subseteq X[/itex] is open. Then [itex]***U = *U[/itex].
Proof. We have
[tex]
***U = \mbox{Int}\left ( \mbox{Int} \left ( \mbox{Int}(U^c)\right )^c\right )^c
[/tex]
It holds that [itex](\mbox{Int}V)^c = \mbox{Cl}(V^c)[/itex]. Thus we have
[tex]
\mbox{Int}\left ( \mbox{Int} \left ( \mbox{Int}(U^c)\right )^c\right )^c = \mbox{Int} \left (\mbox{ClCl}(U^c)\right ) = \mbox{Int}(U^c),
[/tex]
since the closure map is idempotent and [itex]U^c[/itex] is closed.
A work in progress. I've worked out that
[tex]
U^c \subseteq \mbox{Int}(U^c) \Rightarrow (\mbox{Int}(U^c))^c \subseteq U\Rightarrow \mbox{Int}\left (\mbox{Int}(U^c)\right )^c\subseteq U
[/tex]
(inclusion in the other direction always holds) So I think it suffices to find a metrisable topology (not normable?) with an open set whose complement is not open. I tried messing around in [itex]\mathbb R[/itex] with the usual topology, but it's normable, I think that's too strong. Probably some embarrasingly trivial example. I am too tired now :(
 
Last edited:
  • #28
Infrared
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Let [itex](X,d)[/itex] be a metric space. Denote [itex]*U = \mbox{Int}(U^c)[/itex], where [itex]U\subseteq X[/itex] is open. Then [itex]***U = *U[/itex].
Proof. We have
[tex]
***U = \mbox{Int}\left ( \mbox{Int} \left ( \mbox{Int}(U^c)\right )^c\right )^c
[/tex]
It holds that [itex](\mbox{Int}V)^c = \mbox{Cl}(V^c)[/itex]. Thus we have
[tex]
\mbox{Int}\left ( \mbox{Int} \left ( \mbox{Int}(U^c)\right )^c\right )^c = \mbox{Int} \left (\mbox{ClCl}(U^c)\right ) = \mbox{Int}(U^c),
[/tex]
since the closure map is idempotent and [itex]U^c[/itex] is closed.
Good, this is correct.
A work in progress. I've worked out that
[tex]
U^c \subseteq \mbox{Int}(U^c) \Rightarrow (\mbox{Int}(U^c))^c \subseteq U\Rightarrow \mbox{Int}\left (\mbox{Int}(U^c)\right )^c\subseteq U
[/tex]
So I think it suffices to find a metrisable topology (not normable?) with an open set whose complement is not open. I tried messing around in [itex]\mathbb R[/itex] with the usual topology, but it's normable, I think that's too strong. Probably some embarrasingly trivial example. I am too tired now :(
Keep looking. Simple counterexamples exist.
 
Last edited:
  • #29
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Graah, I said it is probably something embarrasingly trivial
Take [itex]\mathbb R[/itex] with the standard topology. Then [itex]U := \mathbb R\setminus\{0\}[/itex] is open and [itex]**U = \mathbb R[/itex].
The previous proof contains calculation errors, fortunately the result doesn't change. For the sake of clarity I will (hopefully) give a better proof.
We use the identity [itex](\mbox{Int}V)^c = \mbox{Cl}(V^c)[/itex]
Firstly, take [itex]V = (\mbox{Int}(U^c))^c[/itex], then
[tex]
***U = \mbox{Int} (\mbox{Int}V)^c = \mbox{Int}\mbox{Cl}(V^c) = \mbox{Int}\mbox{Cl}(\mbox{Int}(U^c)) = \mbox{Int}\mbox{Cl}\mbox{Int}(U^c).
[/tex]
Due to closedness [itex]U^c = \mbox{Cl}(U^c)[/itex]. The composition map [itex]\mbox{Int}\mbox{Cl}[/itex] is idempotent thus we have
[tex]
***U = \mbox{Int}\mbox{Cl}\mbox{Int}(U^c) = \mbox{Int}\mbox{Cl}\mbox{Int}\mbox{Cl}(U^c) = \mbox{Int}\mbox{Cl}(U^c) = \mbox{Int}(U^c) = *U.
[/tex]

The [itex]***U = *U[/itex] identity for open sets is more general. In fact, I didn't use any kind of specific topological property (such as the space being Hausdorff or even metrisable) to prove it. Hmm.

I'm interested in the motivation of the map ##*## (other than it being a more compact way of writing interior of complement). Is it used in development of theory somewhere?
 
Last edited:

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