Is this a basis for the Scott topology on the collection of open sets of a topology?
$begingroup$
If $mathcal{O}(X)$ is the collection of open sets in the topological space $X$ ordered by inclusion and $Ksubseteq X$ is a compact subset of $X$ then it is easy to see that:
$$mathscr{U}_K = {Oinmathcal{O}(X) : Ksubseteq O}$$
is open in the Scott topology on $mathcal{O}(X)$. If $X$ is finite then the collection ${mathscr{U}_K:K mbox{compact in} X}$, is clearly a basis for this topology.
My question is: Is this collection always a basis for the Scott topology on $mathcal{O}(X)$?
general-topology order-theory
asked Jan 23 at 16:28
Bernard HurleyBernard Hurley
1787
$endgroup$
add a comment |
$begingroup$
If $mathcal{O}(X)$ is the collection of open sets in the topological space $X$ ordered by inclusion and $Ksubseteq X$ is a compact subset of $X$ then it is easy to see that:
$$mathscr{U}_K = {Oinmathcal{O}(X) : Ksubseteq O}$$
is open in the Scott topology on $mathcal{O}(X)$. If $X$ is finite then the collection ${mathscr{U}_K:K mbox{compact in} X}$, is clearly a basis for this topology.
My question is: Is this collection always a basis for the Scott topology on $mathcal{O}(X)$?
general-topology order-theory
asked Jan 23 at 16:28
Bernard HurleyBernard Hurley
1787
$endgroup$
add a comment |
$begingroup$
If $mathcal{O}(X)$ is the collection of open sets in the topological space $X$ ordered by inclusion and $Ksubseteq X$ is a compact subset of $X$ then it is easy to see that:
$$mathscr{U}_K = {Oinmathcal{O}(X) : Ksubseteq O}$$
is open in the Scott topology on $mathcal{O}(X)$. If $X$ is finite then the collection ${mathscr{U}_K:K mbox{compact in} X}$, is clearly a basis for this topology.
My question is: Is this collection always a basis for the Scott topology on $mathcal{O}(X)$?
general-topology order-theory
asked Jan 23 at 16:28
Bernard HurleyBernard Hurley
1787
$endgroup$
If $mathcal{O}(X)$ is the collection of open sets in the topological space $X$ ordered by inclusion and $Ksubseteq X$ is a compact subset of $X$ then it is easy to see that:
$$mathscr{U}_K = {Oinmathcal{O}(X) : Ksubseteq O}$$
is open in the Scott topology on $mathcal{O}(X)$. If $X$ is finite then the collection ${mathscr{U}_K:K mbox{compact in} X}$, is clearly a basis for this topology.
My question is: Is this collection always a basis for the Scott topology on $mathcal{O}(X)$?
general-topology order-theory
general-topology order-theory
asked Jan 23 at 16:28
Bernard HurleyBernard Hurley
1787
asked Jan 23 at 16:28
Bernard HurleyBernard Hurley
1787
asked Jan 23 at 16:28
Bernard HurleyBernard Hurley
1787
asked Jan 23 at 16:28
Bernard HurleyBernard Hurley
1787
asked Jan 23 at 16:28
Bernard HurleyBernard Hurley
1787
1787
add a comment |
add a comment |
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