General topology proving something for all of its points
$begingroup$
My question is: if you prove that something is true for all points in a topological space or a subset of some topological space, does that imply that this property holds for the whole topological space or the subset of the topological space?
EDIT: more concrete if you have a topological space where all of its points are closed then is this space also closed? If that even makes sense.
If this is true am I then allowed to pick an arbitrary point of the space and then show that since it holds for this one point then the topological space has this property?
general-topology
asked Mar 25 at 14:14
JensensJensens
366
$endgroup$
|
show 6 more comments
$begingroup$
My question is: if you prove that something is true for all points in a topological space or a subset of some topological space, does that imply that this property holds for the whole topological space or the subset of the topological space?
EDIT: more concrete if you have a topological space where all of its points are closed then is this space also closed? If that even makes sense.
If this is true am I then allowed to pick an arbitrary point of the space and then show that since it holds for this one point then the topological space has this property?
general-topology
asked Mar 25 at 14:14
JensensJensens
366
$endgroup$
1
$begingroup$
What kind of property are we talking about?
$endgroup$
– st.math
Mar 25 at 14:15
1
$begingroup$
$X$ is closed in $X$ by definition...
$endgroup$
– YuiTo Cheng
Mar 25 at 14:17
1
$begingroup$
Well, any space is closed in itself. That's part of the definition of a topology, and doesn't really have anything to do with whether single points are closed.
$endgroup$
– Arthur
Mar 25 at 14:18
1
$begingroup$
Your question is too broad to make much sense. Every point $x$ in a topological space $X$ has the property that $xneq X$ as sets can't contain themselves, but $X$ does not have this property. I can't actually think of a property that applies to points that can also apply to spaces in any meaningful way.
$endgroup$
– Robert Thingum
Mar 25 at 14:19
2
$begingroup$
It's because finite union of closed sets is closed
$endgroup$
– YuiTo Cheng
Mar 25 at 14:22
|
show 6 more comments
$begingroup$
My question is: if you prove that something is true for all points in a topological space or a subset of some topological space, does that imply that this property holds for the whole topological space or the subset of the topological space?
EDIT: more concrete if you have a topological space where all of its points are closed then is this space also closed? If that even makes sense.
If this is true am I then allowed to pick an arbitrary point of the space and then show that since it holds for this one point then the topological space has this property?
general-topology
asked Mar 25 at 14:14
JensensJensens
366
$endgroup$
My question is: if you prove that something is true for all points in a topological space or a subset of some topological space, does that imply that this property holds for the whole topological space or the subset of the topological space?
EDIT: more concrete if you have a topological space where all of its points are closed then is this space also closed? If that even makes sense.
If this is true am I then allowed to pick an arbitrary point of the space and then show that since it holds for this one point then the topological space has this property?
general-topology
general-topology
asked Mar 25 at 14:14
JensensJensens
366
asked Mar 25 at 14:14
JensensJensens
366
edited Mar 25 at 14:16
Jensens
asked Mar 25 at 14:14
JensensJensens
366
asked Mar 25 at 14:14
JensensJensens
366
asked Mar 25 at 14:14
JensensJensens
366
366
1
$begingroup$
What kind of property are we talking about?
$endgroup$
– st.math
Mar 25 at 14:15
1
$begingroup$
$X$ is closed in $X$ by definition...
$endgroup$
– YuiTo Cheng
Mar 25 at 14:17
1
$begingroup$
Well, any space is closed in itself. That's part of the definition of a topology, and doesn't really have anything to do with whether single points are closed.
$endgroup$
– Arthur
Mar 25 at 14:18
1
$begingroup$
Your question is too broad to make much sense. Every point $x$ in a topological space $X$ has the property that $xneq X$ as sets can't contain themselves, but $X$ does not have this property. I can't actually think of a property that applies to points that can also apply to spaces in any meaningful way.
$endgroup$
– Robert Thingum
Mar 25 at 14:19
2
$begingroup$
It's because finite union of closed sets is closed
$endgroup$
– YuiTo Cheng
Mar 25 at 14:22
|
show 6 more comments
1
$begingroup$
What kind of property are we talking about?
$endgroup$
– st.math
Mar 25 at 14:15
1
$begingroup$
$X$ is closed in $X$ by definition...
$endgroup$
– YuiTo Cheng
Mar 25 at 14:17
1
$begingroup$
Well, any space is closed in itself. That's part of the definition of a topology, and doesn't really have anything to do with whether single points are closed.
$endgroup$
– Arthur
Mar 25 at 14:18
1
$begingroup$
Your question is too broad to make much sense. Every point $x$ in a topological space $X$ has the property that $xneq X$ as sets can't contain themselves, but $X$ does not have this property. I can't actually think of a property that applies to points that can also apply to spaces in any meaningful way.
$endgroup$
– Robert Thingum
Mar 25 at 14:19
2
$begingroup$
It's because finite union of closed sets is closed
$endgroup$
– YuiTo Cheng
Mar 25 at 14:22
1
1
$begingroup$
What kind of property are we talking about?
$endgroup$
– st.math
Mar 25 at 14:15
$begingroup$
What kind of property are we talking about?
$endgroup$
– st.math
Mar 25 at 14:15
1
1
$begingroup$
$X$ is closed in $X$ by definition...
$endgroup$
– YuiTo Cheng
Mar 25 at 14:17
$begingroup$
$X$ is closed in $X$ by definition...
$endgroup$
– YuiTo Cheng
Mar 25 at 14:17
1
1
$begingroup$
Well, any space is closed in itself. That's part of the definition of a topology, and doesn't really have anything to do with whether single points are closed.
$endgroup$
– Arthur
Mar 25 at 14:18
$begingroup$
Well, any space is closed in itself. That's part of the definition of a topology, and doesn't really have anything to do with whether single points are closed.
$endgroup$
– Arthur
Mar 25 at 14:18
1
1
$begingroup$
Your question is too broad to make much sense. Every point $x$ in a topological space $X$ has the property that $xneq X$ as sets can't contain themselves, but $X$ does not have this property. I can't actually think of a property that applies to points that can also apply to spaces in any meaningful way.
$endgroup$
– Robert Thingum
Mar 25 at 14:19
$begingroup$
Your question is too broad to make much sense. Every point $x$ in a topological space $X$ has the property that $xneq X$ as sets can't contain themselves, but $X$ does not have this property. I can't actually think of a property that applies to points that can also apply to spaces in any meaningful way.
$endgroup$
– Robert Thingum
Mar 25 at 14:19
2
2
$begingroup$
It's because finite union of closed sets is closed
$endgroup$
– YuiTo Cheng
Mar 25 at 14:22
$begingroup$
It's because finite union of closed sets is closed
$endgroup$
– YuiTo Cheng
Mar 25 at 14:22
|
show 6 more comments
3 Answers
3
active
oldest
votes
$begingroup$
The answer, as far as your specific example is concerned, is negative. Every topological space $X$ is a closed subset of itself. However, there are topological spaces in which not all points are closed.
A better example would be: a set which consists of a single point is always compact and connected, but lots of topological spaces are neither compact nor connected.
answered Mar 25 at 14:21
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
add a comment |
$begingroup$
As your example property "is closed" illustrates, the properties of single points in a space and the space as a whole are not entirely linked together. At the very least, it's not something you can count on in general. I would personally suggest you instead as a general rule assume they are not connected, and make note of the times it does happen.
answered Mar 25 at 14:20
ArthurArthur
122k7122210
$endgroup$
add a comment |
$begingroup$
Taken literally, your question is ill-posed.
This is because a topological space and a point in the topological space are different kinds of things. When we say that a point is closed in a topological space, what we really mean is that its singleton is closed. This is literally very different, but because "a point is closed" taken literally is, in general, nonsensical, this short of shorthand is acceptable.
Having this in mind, you could rephrase your question to a more meaningful (not nonsensical) one:
Let $X$ be a topological space, and let $Asubseteq X$. If $P$ is a topological property of a subset of $X$ and for every $ain A$, the singleton ${a}$ has the property $P$, does $A$ also have the property $P$?
The answer is trivially no. If you consider "not being a singleton" a topological property, then it fails spectacularly. Otherwise, the property you consider, "being closed" (definitely a topological property) also fails: for example, if you consider $(0,1)subseteq {mathbf R}$, then (the singleton of) every point in $(0,1)$ is closed in the reals, but $(0,1)$ is not.
You might ask for what properties $P$ this is true. One such property is being open: if for every $ain A$, the singleton ${a}$ is open in $X$, then $A$ itself is open (as a union of open sets). I strongly suspect that this is just about the only interesting and nontrivial property for which this is true (for suitable notions of "interesting" and "nontrivial").
A related, far more interesting question is about what topological properties are local, or in other words, what properties of a topological space are true for a space if and only if every point has a neighbourhood with the same property.
answered Mar 25 at 17:11
tomasztomasz
24k23482
$endgroup$
$begingroup$
Very good answer appreciate it, yes it was difficult for me to ask the question properly I think know it makes a lot more sense
$endgroup$
– Jensens
Mar 25 at 18:14
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%2f3161827%2fgeneral-topology-proving-something-for-all-of-its-points%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer, as far as your specific example is concerned, is negative. Every topological space $X$ is a closed subset of itself. However, there are topological spaces in which not all points are closed.
A better example would be: a set which consists of a single point is always compact and connected, but lots of topological spaces are neither compact nor connected.
answered Mar 25 at 14:21
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
add a comment |
$begingroup$
The answer, as far as your specific example is concerned, is negative. Every topological space $X$ is a closed subset of itself. However, there are topological spaces in which not all points are closed.
A better example would be: a set which consists of a single point is always compact and connected, but lots of topological spaces are neither compact nor connected.
answered Mar 25 at 14:21
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
add a comment |
$begingroup$
The answer, as far as your specific example is concerned, is negative. Every topological space $X$ is a closed subset of itself. However, there are topological spaces in which not all points are closed.
A better example would be: a set which consists of a single point is always compact and connected, but lots of topological spaces are neither compact nor connected.
answered Mar 25 at 14:21
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
The answer, as far as your specific example is concerned, is negative. Every topological space $X$ is a closed subset of itself. However, there are topological spaces in which not all points are closed.
A better example would be: a set which consists of a single point is always compact and connected, but lots of topological spaces are neither compact nor connected.
answered Mar 25 at 14:21
José Carlos SantosJosé Carlos Santos
172k22132239
answered Mar 25 at 14:21
José Carlos SantosJosé Carlos Santos
172k22132239
answered Mar 25 at 14:21
José Carlos SantosJosé Carlos Santos
172k22132239
answered Mar 25 at 14:21
José Carlos SantosJosé Carlos Santos
172k22132239
172k22132239
add a comment |
add a comment |
$begingroup$
As your example property "is closed" illustrates, the properties of single points in a space and the space as a whole are not entirely linked together. At the very least, it's not something you can count on in general. I would personally suggest you instead as a general rule assume they are not connected, and make note of the times it does happen.
answered Mar 25 at 14:20
ArthurArthur
122k7122210
$endgroup$
add a comment |
$begingroup$
As your example property "is closed" illustrates, the properties of single points in a space and the space as a whole are not entirely linked together. At the very least, it's not something you can count on in general. I would personally suggest you instead as a general rule assume they are not connected, and make note of the times it does happen.
answered Mar 25 at 14:20
ArthurArthur
122k7122210
$endgroup$
add a comment |
$begingroup$
As your example property "is closed" illustrates, the properties of single points in a space and the space as a whole are not entirely linked together. At the very least, it's not something you can count on in general. I would personally suggest you instead as a general rule assume they are not connected, and make note of the times it does happen.
answered Mar 25 at 14:20
ArthurArthur
122k7122210
$endgroup$
As your example property "is closed" illustrates, the properties of single points in a space and the space as a whole are not entirely linked together. At the very least, it's not something you can count on in general. I would personally suggest you instead as a general rule assume they are not connected, and make note of the times it does happen.
answered Mar 25 at 14:20
ArthurArthur
122k7122210
answered Mar 25 at 14:20
ArthurArthur
122k7122210
answered Mar 25 at 14:20
ArthurArthur
122k7122210
answered Mar 25 at 14:20
ArthurArthur
122k7122210
122k7122210
add a comment |
add a comment |
$begingroup$
Taken literally, your question is ill-posed.
This is because a topological space and a point in the topological space are different kinds of things. When we say that a point is closed in a topological space, what we really mean is that its singleton is closed. This is literally very different, but because "a point is closed" taken literally is, in general, nonsensical, this short of shorthand is acceptable.
Having this in mind, you could rephrase your question to a more meaningful (not nonsensical) one:
Let $X$ be a topological space, and let $Asubseteq X$. If $P$ is a topological property of a subset of $X$ and for every $ain A$, the singleton ${a}$ has the property $P$, does $A$ also have the property $P$?
The answer is trivially no. If you consider "not being a singleton" a topological property, then it fails spectacularly. Otherwise, the property you consider, "being closed" (definitely a topological property) also fails: for example, if you consider $(0,1)subseteq {mathbf R}$, then (the singleton of) every point in $(0,1)$ is closed in the reals, but $(0,1)$ is not.
You might ask for what properties $P$ this is true. One such property is being open: if for every $ain A$, the singleton ${a}$ is open in $X$, then $A$ itself is open (as a union of open sets). I strongly suspect that this is just about the only interesting and nontrivial property for which this is true (for suitable notions of "interesting" and "nontrivial").
A related, far more interesting question is about what topological properties are local, or in other words, what properties of a topological space are true for a space if and only if every point has a neighbourhood with the same property.
answered Mar 25 at 17:11
tomasztomasz
24k23482
$endgroup$
$begingroup$
Very good answer appreciate it, yes it was difficult for me to ask the question properly I think know it makes a lot more sense
$endgroup$
– Jensens
Mar 25 at 18:14
add a comment |
$begingroup$
Taken literally, your question is ill-posed.
This is because a topological space and a point in the topological space are different kinds of things. When we say that a point is closed in a topological space, what we really mean is that its singleton is closed. This is literally very different, but because "a point is closed" taken literally is, in general, nonsensical, this short of shorthand is acceptable.
Having this in mind, you could rephrase your question to a more meaningful (not nonsensical) one:
Let $X$ be a topological space, and let $Asubseteq X$. If $P$ is a topological property of a subset of $X$ and for every $ain A$, the singleton ${a}$ has the property $P$, does $A$ also have the property $P$?
The answer is trivially no. If you consider "not being a singleton" a topological property, then it fails spectacularly. Otherwise, the property you consider, "being closed" (definitely a topological property) also fails: for example, if you consider $(0,1)subseteq {mathbf R}$, then (the singleton of) every point in $(0,1)$ is closed in the reals, but $(0,1)$ is not.
You might ask for what properties $P$ this is true. One such property is being open: if for every $ain A$, the singleton ${a}$ is open in $X$, then $A$ itself is open (as a union of open sets). I strongly suspect that this is just about the only interesting and nontrivial property for which this is true (for suitable notions of "interesting" and "nontrivial").
A related, far more interesting question is about what topological properties are local, or in other words, what properties of a topological space are true for a space if and only if every point has a neighbourhood with the same property.
answered Mar 25 at 17:11
tomasztomasz
24k23482
$endgroup$
$begingroup$
Very good answer appreciate it, yes it was difficult for me to ask the question properly I think know it makes a lot more sense
$endgroup$
– Jensens
Mar 25 at 18:14
add a comment |
$begingroup$
Taken literally, your question is ill-posed.
This is because a topological space and a point in the topological space are different kinds of things. When we say that a point is closed in a topological space, what we really mean is that its singleton is closed. This is literally very different, but because "a point is closed" taken literally is, in general, nonsensical, this short of shorthand is acceptable.
Having this in mind, you could rephrase your question to a more meaningful (not nonsensical) one:
Let $X$ be a topological space, and let $Asubseteq X$. If $P$ is a topological property of a subset of $X$ and for every $ain A$, the singleton ${a}$ has the property $P$, does $A$ also have the property $P$?
The answer is trivially no. If you consider "not being a singleton" a topological property, then it fails spectacularly. Otherwise, the property you consider, "being closed" (definitely a topological property) also fails: for example, if you consider $(0,1)subseteq {mathbf R}$, then (the singleton of) every point in $(0,1)$ is closed in the reals, but $(0,1)$ is not.
You might ask for what properties $P$ this is true. One such property is being open: if for every $ain A$, the singleton ${a}$ is open in $X$, then $A$ itself is open (as a union of open sets). I strongly suspect that this is just about the only interesting and nontrivial property for which this is true (for suitable notions of "interesting" and "nontrivial").
A related, far more interesting question is about what topological properties are local, or in other words, what properties of a topological space are true for a space if and only if every point has a neighbourhood with the same property.
answered Mar 25 at 17:11
tomasztomasz
24k23482
$endgroup$
Taken literally, your question is ill-posed.
This is because a topological space and a point in the topological space are different kinds of things. When we say that a point is closed in a topological space, what we really mean is that its singleton is closed. This is literally very different, but because "a point is closed" taken literally is, in general, nonsensical, this short of shorthand is acceptable.
Having this in mind, you could rephrase your question to a more meaningful (not nonsensical) one:
Let $X$ be a topological space, and let $Asubseteq X$. If $P$ is a topological property of a subset of $X$ and for every $ain A$, the singleton ${a}$ has the property $P$, does $A$ also have the property $P$?
The answer is trivially no. If you consider "not being a singleton" a topological property, then it fails spectacularly. Otherwise, the property you consider, "being closed" (definitely a topological property) also fails: for example, if you consider $(0,1)subseteq {mathbf R}$, then (the singleton of) every point in $(0,1)$ is closed in the reals, but $(0,1)$ is not.
You might ask for what properties $P$ this is true. One such property is being open: if for every $ain A$, the singleton ${a}$ is open in $X$, then $A$ itself is open (as a union of open sets). I strongly suspect that this is just about the only interesting and nontrivial property for which this is true (for suitable notions of "interesting" and "nontrivial").
A related, far more interesting question is about what topological properties are local, or in other words, what properties of a topological space are true for a space if and only if every point has a neighbourhood with the same property.
answered Mar 25 at 17:11
tomasztomasz
24k23482
answered Mar 25 at 17:11
tomasztomasz
24k23482
answered Mar 25 at 17:11
tomasztomasz
24k23482
answered Mar 25 at 17:11
tomasztomasz
24k23482
24k23482
$begingroup$
Very good answer appreciate it, yes it was difficult for me to ask the question properly I think know it makes a lot more sense
$endgroup$
– Jensens
Mar 25 at 18:14
add a comment |
$begingroup$
Very good answer appreciate it, yes it was difficult for me to ask the question properly I think know it makes a lot more sense
$endgroup$
– Jensens
Mar 25 at 18:14
$begingroup$
Very good answer appreciate it, yes it was difficult for me to ask the question properly I think know it makes a lot more sense
$endgroup$
– Jensens
Mar 25 at 18:14
$begingroup$
Very good answer appreciate it, yes it was difficult for me to ask the question properly I think know it makes a lot more sense
$endgroup$
– Jensens
Mar 25 at 18:14
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.
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%2f3161827%2fgeneral-topology-proving-something-for-all-of-its-points%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
1
$begingroup$
What kind of property are we talking about?
$endgroup$
– st.math
Mar 25 at 14:15
1
$begingroup$
$X$ is closed in $X$ by definition...
$endgroup$
– YuiTo Cheng
Mar 25 at 14:17
1
$begingroup$
Well, any space is closed in itself. That's part of the definition of a topology, and doesn't really have anything to do with whether single points are closed.
$endgroup$
– Arthur
Mar 25 at 14:18
1
$begingroup$
Your question is too broad to make much sense. Every point $x$ in a topological space $X$ has the property that $xneq X$ as sets can't contain themselves, but $X$ does not have this property. I can't actually think of a property that applies to points that can also apply to spaces in any meaningful way.
$endgroup$
– Robert Thingum
Mar 25 at 14:19
2
$begingroup$
It's because finite union of closed sets is closed
$endgroup$
– YuiTo Cheng
Mar 25 at 14:22