General topology proving something for all of its points Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraIf a set contains all its limit points must it be closed?Topology and limit pointsTwo trivial questions in general topologypoint set topology: closed points denseLimit points TopologyFinite point set has limit points for general topological spaces?General topologyGeneral topology: looking for a brief and clear proof that the two main definitions of “closure” are the same.A subset of a topological space is closed iff it contains all its limit points - true or false?Questions on general topology
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General topology proving something for all of its points
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraIf a set contains all its limit points must it be closed?Topology and limit pointsTwo trivial questions in general topologypoint set topology: closed points denseLimit points TopologyFinite point set has limit points for general topological spaces?General topologyGeneral topology: looking for a brief and clear proof that the two main definitions of “closure” are the same.A subset of a topological space is closed iff it contains all its limit points - true or false?Questions on general topology
$begingroup$
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
456
$endgroup$
|
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
456
$endgroup$
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
456
$endgroup$
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
456
asked Mar 25 at 14:14
JensensJensens
456
edited Mar 25 at 14:16
Jensens
asked Mar 25 at 14:14
JensensJensens
456
asked Mar 25 at 14:14
JensensJensens
456
asked Mar 25 at 14:14
JensensJensens
456
456
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
177k24138250
$endgroup$
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
123k7122211
$endgroup$
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
24.1k23482
$endgroup$
$begingroup$
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
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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
177k24138250
$endgroup$
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
177k24138250
$endgroup$
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
177k24138250
$endgroup$
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
177k24138250
answered Mar 25 at 14:21
José Carlos SantosJosé Carlos Santos
177k24138250
answered Mar 25 at 14:21
José Carlos SantosJosé Carlos Santos
177k24138250
answered Mar 25 at 14:21
José Carlos SantosJosé Carlos Santos
177k24138250
177k24138250
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
123k7122211
$endgroup$
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
123k7122211
$endgroup$
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
123k7122211
$endgroup$
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
123k7122211
answered Mar 25 at 14:20
ArthurArthur
123k7122211
answered Mar 25 at 14:20
ArthurArthur
123k7122211
answered Mar 25 at 14:20
ArthurArthur
123k7122211
123k7122211
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
24.1k23482
$endgroup$
$begingroup$
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
24.1k23482
$endgroup$
$begingroup$
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
24.1k23482
$endgroup$
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
24.1k23482
answered Mar 25 at 17:11
tomasztomasz
24.1k23482
answered Mar 25 at 17:11
tomasztomasz
24.1k23482
answered Mar 25 at 17:11
tomasztomasz
24.1k23482
24.1k23482
$begingroup$
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 |
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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
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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.
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– Arthur
Mar 25 at 14:18
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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.
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– Robert Thingum
Mar 25 at 14:19
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It's because finite union of closed sets is closed
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– YuiTo Cheng
Mar 25 at 14:22