Limits in a dcpo with the Scott topology
This is exercise 4.7.7 in Non-Hausdorff Topology and Domain Theory by Jean Goubault-Larrecq.
Consider a dcpo $(X,leq)$ and equip it with the Scott topology where the opens are the upward-closed sets $U$ such that for any directed family $(x_i)_{i in I}$ with $sup_{i in I}x_i in U$ we can find some $i in I$ such that $x_i in U$.
Now let $(x_i)_{i in I}$ be a directed family, and let $N = (x_i)_{i in I, sqsubseteq}$ be the net given by $i sqsubseteq j$ if and only if $x_i leq x_j$.
I want to show that
a point $y$ is a limit of the net $N$ $iff$ $y leq sup_{i in I} x_i$
(this supremum exists because we assumed $X$ was a dcpo). I have been able to show that if $y leq sup_{i in I} x_i$, then $y$ is a limit of the net $N$.
However, the other direction is causing me trouble.
If we assume that $y$ is a limit point of $N$, then by definition this means that for every neighborhood $U$ of $y$, we can find some $x_i in U$ and hence it follows that $x_j in U$ for all $x_j geq x_i$. Because $U$ is Scott-open, it follows that $sup_{i in I} x_i$, so we know that $sup_{i in I} x_i$ is in any neighborhood of $y$. However, I don't see how that allows me to conclude anything about the ordering $leq$ between $y$ and $sup_{i in I} x_i$.
Could somebody help me prove this? Do I need a different approach to the problem?
general-topology order-theory domain-theory
asked Mar 24 '17 at 13:08
mrp
3,77251537
add a comment |
This is exercise 4.7.7 in Non-Hausdorff Topology and Domain Theory by Jean Goubault-Larrecq.
Consider a dcpo $(X,leq)$ and equip it with the Scott topology where the opens are the upward-closed sets $U$ such that for any directed family $(x_i)_{i in I}$ with $sup_{i in I}x_i in U$ we can find some $i in I$ such that $x_i in U$.
Now let $(x_i)_{i in I}$ be a directed family, and let $N = (x_i)_{i in I, sqsubseteq}$ be the net given by $i sqsubseteq j$ if and only if $x_i leq x_j$.
I want to show that
a point $y$ is a limit of the net $N$ $iff$ $y leq sup_{i in I} x_i$
(this supremum exists because we assumed $X$ was a dcpo). I have been able to show that if $y leq sup_{i in I} x_i$, then $y$ is a limit of the net $N$.
However, the other direction is causing me trouble.
If we assume that $y$ is a limit point of $N$, then by definition this means that for every neighborhood $U$ of $y$, we can find some $x_i in U$ and hence it follows that $x_j in U$ for all $x_j geq x_i$. Because $U$ is Scott-open, it follows that $sup_{i in I} x_i$, so we know that $sup_{i in I} x_i$ is in any neighborhood of $y$. However, I don't see how that allows me to conclude anything about the ordering $leq$ between $y$ and $sup_{i in I} x_i$.
Could somebody help me prove this? Do I need a different approach to the problem?
general-topology order-theory domain-theory
asked Mar 24 '17 at 13:08
mrp
3,77251537
add a comment |
This is exercise 4.7.7 in Non-Hausdorff Topology and Domain Theory by Jean Goubault-Larrecq.
Consider a dcpo $(X,leq)$ and equip it with the Scott topology where the opens are the upward-closed sets $U$ such that for any directed family $(x_i)_{i in I}$ with $sup_{i in I}x_i in U$ we can find some $i in I$ such that $x_i in U$.
Now let $(x_i)_{i in I}$ be a directed family, and let $N = (x_i)_{i in I, sqsubseteq}$ be the net given by $i sqsubseteq j$ if and only if $x_i leq x_j$.
I want to show that
a point $y$ is a limit of the net $N$ $iff$ $y leq sup_{i in I} x_i$
(this supremum exists because we assumed $X$ was a dcpo). I have been able to show that if $y leq sup_{i in I} x_i$, then $y$ is a limit of the net $N$.
However, the other direction is causing me trouble.
If we assume that $y$ is a limit point of $N$, then by definition this means that for every neighborhood $U$ of $y$, we can find some $x_i in U$ and hence it follows that $x_j in U$ for all $x_j geq x_i$. Because $U$ is Scott-open, it follows that $sup_{i in I} x_i$, so we know that $sup_{i in I} x_i$ is in any neighborhood of $y$. However, I don't see how that allows me to conclude anything about the ordering $leq$ between $y$ and $sup_{i in I} x_i$.
Could somebody help me prove this? Do I need a different approach to the problem?
general-topology order-theory domain-theory
asked Mar 24 '17 at 13:08
mrp
3,77251537
This is exercise 4.7.7 in Non-Hausdorff Topology and Domain Theory by Jean Goubault-Larrecq.
Consider a dcpo $(X,leq)$ and equip it with the Scott topology where the opens are the upward-closed sets $U$ such that for any directed family $(x_i)_{i in I}$ with $sup_{i in I}x_i in U$ we can find some $i in I$ such that $x_i in U$.
Now let $(x_i)_{i in I}$ be a directed family, and let $N = (x_i)_{i in I, sqsubseteq}$ be the net given by $i sqsubseteq j$ if and only if $x_i leq x_j$.
I want to show that
a point $y$ is a limit of the net $N$ $iff$ $y leq sup_{i in I} x_i$
(this supremum exists because we assumed $X$ was a dcpo). I have been able to show that if $y leq sup_{i in I} x_i$, then $y$ is a limit of the net $N$.
However, the other direction is causing me trouble.
If we assume that $y$ is a limit point of $N$, then by definition this means that for every neighborhood $U$ of $y$, we can find some $x_i in U$ and hence it follows that $x_j in U$ for all $x_j geq x_i$. Because $U$ is Scott-open, it follows that $sup_{i in I} x_i$, so we know that $sup_{i in I} x_i$ is in any neighborhood of $y$. However, I don't see how that allows me to conclude anything about the ordering $leq$ between $y$ and $sup_{i in I} x_i$.
Could somebody help me prove this? Do I need a different approach to the problem?
general-topology order-theory domain-theory
general-topology order-theory domain-theory
asked Mar 24 '17 at 13:08
mrp
3,77251537
asked Mar 24 '17 at 13:08
mrp
3,77251537
asked Mar 24 '17 at 13:08
mrp
3,77251537
asked Mar 24 '17 at 13:08
mrp
3,77251537
asked Mar 24 '17 at 13:08
mrp
3,77251537
3,77251537
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
That $x$ is in any neighborhood of $y$ means that $y$ is in the closure of $x$, i. e. $yleqslant x$ in the specialization order of the topology. But it is well known that the specialization order of the Scott topology recovers the original dcpo order.
For completeness, here is a proof. Note that closed subsets of the Scott topology are precisely those downsets which are sub-dcpos. It follows that the closure of a point $x$ is its principal downset ${ymid yle x}$. (Indeed this is the smallest downset containing $x$, and it happens to be a sub-dcpo.)
Thus by the above, $yleqslant x$ in the specialization order iff $y$ is in the closure of $x$ iff $yle x$ in the original order.
answered Nov 20 at 5:40
მამუკა ჯიბლაძე
737817
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2201257%2flimits-in-a-dcpo-with-the-scott-topology%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
That $x$ is in any neighborhood of $y$ means that $y$ is in the closure of $x$, i. e. $yleqslant x$ in the specialization order of the topology. But it is well known that the specialization order of the Scott topology recovers the original dcpo order.
For completeness, here is a proof. Note that closed subsets of the Scott topology are precisely those downsets which are sub-dcpos. It follows that the closure of a point $x$ is its principal downset ${ymid yle x}$. (Indeed this is the smallest downset containing $x$, and it happens to be a sub-dcpo.)
Thus by the above, $yleqslant x$ in the specialization order iff $y$ is in the closure of $x$ iff $yle x$ in the original order.
answered Nov 20 at 5:40
მამუკა ჯიბლაძე
737817
add a comment |
That $x$ is in any neighborhood of $y$ means that $y$ is in the closure of $x$, i. e. $yleqslant x$ in the specialization order of the topology. But it is well known that the specialization order of the Scott topology recovers the original dcpo order.
For completeness, here is a proof. Note that closed subsets of the Scott topology are precisely those downsets which are sub-dcpos. It follows that the closure of a point $x$ is its principal downset ${ymid yle x}$. (Indeed this is the smallest downset containing $x$, and it happens to be a sub-dcpo.)
Thus by the above, $yleqslant x$ in the specialization order iff $y$ is in the closure of $x$ iff $yle x$ in the original order.
answered Nov 20 at 5:40
მამუკა ჯიბლაძე
737817
add a comment |
That $x$ is in any neighborhood of $y$ means that $y$ is in the closure of $x$, i. e. $yleqslant x$ in the specialization order of the topology. But it is well known that the specialization order of the Scott topology recovers the original dcpo order.
For completeness, here is a proof. Note that closed subsets of the Scott topology are precisely those downsets which are sub-dcpos. It follows that the closure of a point $x$ is its principal downset ${ymid yle x}$. (Indeed this is the smallest downset containing $x$, and it happens to be a sub-dcpo.)
Thus by the above, $yleqslant x$ in the specialization order iff $y$ is in the closure of $x$ iff $yle x$ in the original order.
answered Nov 20 at 5:40
მამუკა ჯიბლაძე
737817
That $x$ is in any neighborhood of $y$ means that $y$ is in the closure of $x$, i. e. $yleqslant x$ in the specialization order of the topology. But it is well known that the specialization order of the Scott topology recovers the original dcpo order.
For completeness, here is a proof. Note that closed subsets of the Scott topology are precisely those downsets which are sub-dcpos. It follows that the closure of a point $x$ is its principal downset ${ymid yle x}$. (Indeed this is the smallest downset containing $x$, and it happens to be a sub-dcpo.)
Thus by the above, $yleqslant x$ in the specialization order iff $y$ is in the closure of $x$ iff $yle x$ in the original order.
answered Nov 20 at 5:40
მამუკა ჯიბლაძე
737817
answered Nov 20 at 5:40
მამუკა ჯიბლაძე
737817
answered Nov 20 at 5:40
მამუკა ჯიბლაძე
737817
answered Nov 20 at 5:40
მამუკა ჯიბლაძე
737817
737817
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2201257%2flimits-in-a-dcpo-with-the-scott-topology%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
