A subset of a metric space is open iff its complement is closed
$begingroup$
I have been reading baby Rudin and I am currently in the topology chapter. I came across the proof that a metric space is open iff its complement is closed and I liked his slick and direct style but I also wanted to prove it on my own. So this is my attempt
The definitions that I use here are:
$p$ is a limit point of a subset $E$ of a metric space iff every open neighborhood of $p$ contains a point in $E$ which is not equal to $p$
$E$ is closed iff it contains all its limit points.
A point $p$ is an interior point of $E$ if there exists an open neighborhood $N(p)$ of $p$ such that $N(p)$ is contained within $E$
$E$ is open iff all its points are interior points.
Proof: Suppose $E$ open and $E^c$ not closed. Then there is a limit point $x$ of $E^c$ which is not in $E^c$. Then $x in E$. But $E$ is open so there is some open neighborhood of $x$ which is contained within $E$ and hence has no element in $E^c$. But this is impossible since $x$ is a limit point of $E^c$.
On the converse, suppose $E^c$ closed and $E$ not open. Then there is a point $p$ in $E$ which is not an interior point of $E$. Then every open neighborhood of $p$ contains a point in $E^c$. Then $p$ is clearly a limit point of $E^c$. But $E^c$ is closed so $p in E^c$ which is impossible.
$square$
I'm making this post also for other people who are studying real analysis and to see different possibilities for the solutions as well as getting some feedback on my proof writing. Any comment is appreciated.
real-analysis general-topology metric-spaces
asked Dec 23 '18 at 23:17
Sei SakataSei Sakata
10810
$endgroup$
add a comment |
$begingroup$
I have been reading baby Rudin and I am currently in the topology chapter. I came across the proof that a metric space is open iff its complement is closed and I liked his slick and direct style but I also wanted to prove it on my own. So this is my attempt
The definitions that I use here are:
$p$ is a limit point of a subset $E$ of a metric space iff every open neighborhood of $p$ contains a point in $E$ which is not equal to $p$
$E$ is closed iff it contains all its limit points.
A point $p$ is an interior point of $E$ if there exists an open neighborhood $N(p)$ of $p$ such that $N(p)$ is contained within $E$
$E$ is open iff all its points are interior points.
Proof: Suppose $E$ open and $E^c$ not closed. Then there is a limit point $x$ of $E^c$ which is not in $E^c$. Then $x in E$. But $E$ is open so there is some open neighborhood of $x$ which is contained within $E$ and hence has no element in $E^c$. But this is impossible since $x$ is a limit point of $E^c$.
On the converse, suppose $E^c$ closed and $E$ not open. Then there is a point $p$ in $E$ which is not an interior point of $E$. Then every open neighborhood of $p$ contains a point in $E^c$. Then $p$ is clearly a limit point of $E^c$. But $E^c$ is closed so $p in E^c$ which is impossible.
$square$
I'm making this post also for other people who are studying real analysis and to see different possibilities for the solutions as well as getting some feedback on my proof writing. Any comment is appreciated.
real-analysis general-topology metric-spaces
asked Dec 23 '18 at 23:17
Sei SakataSei Sakata
10810
$endgroup$
2
$begingroup$
The only problem here is you are talking about a metric space being open/closed. You should be talking about a subset of a metric space being open or closed, not the entire metric space. The entire metric space is always both open and closed. Otherwise, your proof is fine.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:24
$begingroup$
Thanks for the suggestion. edit made for the future viewers
$endgroup$
– Sei Sakata
Dec 23 '18 at 23:30
$begingroup$
You have not fixed the problem identified by Kavi in the question but only i in the title. Your list of definitions does not include a definition for $E^c$.
$endgroup$
– Rob Arthan
Dec 24 '18 at 0:18
$begingroup$
By convention, $E^c$ usually denotes complement of a set and I thought that it was quite obvious from the context. And for $E$ not being the entire metric space, I first define $E$ to be a subset of a metric space for the definition of a limit point. And proceed to use that notation but is that not okay?
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:16
add a comment |
$begingroup$
I have been reading baby Rudin and I am currently in the topology chapter. I came across the proof that a metric space is open iff its complement is closed and I liked his slick and direct style but I also wanted to prove it on my own. So this is my attempt
The definitions that I use here are:
$p$ is a limit point of a subset $E$ of a metric space iff every open neighborhood of $p$ contains a point in $E$ which is not equal to $p$
$E$ is closed iff it contains all its limit points.
A point $p$ is an interior point of $E$ if there exists an open neighborhood $N(p)$ of $p$ such that $N(p)$ is contained within $E$
$E$ is open iff all its points are interior points.
Proof: Suppose $E$ open and $E^c$ not closed. Then there is a limit point $x$ of $E^c$ which is not in $E^c$. Then $x in E$. But $E$ is open so there is some open neighborhood of $x$ which is contained within $E$ and hence has no element in $E^c$. But this is impossible since $x$ is a limit point of $E^c$.
On the converse, suppose $E^c$ closed and $E$ not open. Then there is a point $p$ in $E$ which is not an interior point of $E$. Then every open neighborhood of $p$ contains a point in $E^c$. Then $p$ is clearly a limit point of $E^c$. But $E^c$ is closed so $p in E^c$ which is impossible.
$square$
I'm making this post also for other people who are studying real analysis and to see different possibilities for the solutions as well as getting some feedback on my proof writing. Any comment is appreciated.
real-analysis general-topology metric-spaces
asked Dec 23 '18 at 23:17
Sei SakataSei Sakata
10810
$endgroup$
I have been reading baby Rudin and I am currently in the topology chapter. I came across the proof that a metric space is open iff its complement is closed and I liked his slick and direct style but I also wanted to prove it on my own. So this is my attempt
The definitions that I use here are:
$p$ is a limit point of a subset $E$ of a metric space iff every open neighborhood of $p$ contains a point in $E$ which is not equal to $p$
$E$ is closed iff it contains all its limit points.
A point $p$ is an interior point of $E$ if there exists an open neighborhood $N(p)$ of $p$ such that $N(p)$ is contained within $E$
$E$ is open iff all its points are interior points.
Proof: Suppose $E$ open and $E^c$ not closed. Then there is a limit point $x$ of $E^c$ which is not in $E^c$. Then $x in E$. But $E$ is open so there is some open neighborhood of $x$ which is contained within $E$ and hence has no element in $E^c$. But this is impossible since $x$ is a limit point of $E^c$.
On the converse, suppose $E^c$ closed and $E$ not open. Then there is a point $p$ in $E$ which is not an interior point of $E$. Then every open neighborhood of $p$ contains a point in $E^c$. Then $p$ is clearly a limit point of $E^c$. But $E^c$ is closed so $p in E^c$ which is impossible.
$square$
I'm making this post also for other people who are studying real analysis and to see different possibilities for the solutions as well as getting some feedback on my proof writing. Any comment is appreciated.
real-analysis general-topology metric-spaces
real-analysis general-topology metric-spaces
asked Dec 23 '18 at 23:17
Sei SakataSei Sakata
10810
asked Dec 23 '18 at 23:17
Sei SakataSei Sakata
10810
edited Dec 23 '18 at 23:29
Sei Sakata
asked Dec 23 '18 at 23:17
Sei SakataSei Sakata
10810
asked Dec 23 '18 at 23:17
Sei SakataSei Sakata
10810
asked Dec 23 '18 at 23:17
Sei SakataSei Sakata
10810
10810
2
$begingroup$
The only problem here is you are talking about a metric space being open/closed. You should be talking about a subset of a metric space being open or closed, not the entire metric space. The entire metric space is always both open and closed. Otherwise, your proof is fine.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:24
$begingroup$
Thanks for the suggestion. edit made for the future viewers
$endgroup$
– Sei Sakata
Dec 23 '18 at 23:30
$begingroup$
You have not fixed the problem identified by Kavi in the question but only i in the title. Your list of definitions does not include a definition for $E^c$.
$endgroup$
– Rob Arthan
Dec 24 '18 at 0:18
$begingroup$
By convention, $E^c$ usually denotes complement of a set and I thought that it was quite obvious from the context. And for $E$ not being the entire metric space, I first define $E$ to be a subset of a metric space for the definition of a limit point. And proceed to use that notation but is that not okay?
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:16
add a comment |
2
$begingroup$
The only problem here is you are talking about a metric space being open/closed. You should be talking about a subset of a metric space being open or closed, not the entire metric space. The entire metric space is always both open and closed. Otherwise, your proof is fine.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:24
$begingroup$
Thanks for the suggestion. edit made for the future viewers
$endgroup$
– Sei Sakata
Dec 23 '18 at 23:30
$begingroup$
You have not fixed the problem identified by Kavi in the question but only i in the title. Your list of definitions does not include a definition for $E^c$.
$endgroup$
– Rob Arthan
Dec 24 '18 at 0:18
$begingroup$
By convention, $E^c$ usually denotes complement of a set and I thought that it was quite obvious from the context. And for $E$ not being the entire metric space, I first define $E$ to be a subset of a metric space for the definition of a limit point. And proceed to use that notation but is that not okay?
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:16
2
2
$begingroup$
The only problem here is you are talking about a metric space being open/closed. You should be talking about a subset of a metric space being open or closed, not the entire metric space. The entire metric space is always both open and closed. Otherwise, your proof is fine.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:24
$begingroup$
The only problem here is you are talking about a metric space being open/closed. You should be talking about a subset of a metric space being open or closed, not the entire metric space. The entire metric space is always both open and closed. Otherwise, your proof is fine.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:24
$begingroup$
Thanks for the suggestion. edit made for the future viewers
$endgroup$
– Sei Sakata
Dec 23 '18 at 23:30
$begingroup$
Thanks for the suggestion. edit made for the future viewers
$endgroup$
– Sei Sakata
Dec 23 '18 at 23:30
$begingroup$
You have not fixed the problem identified by Kavi in the question but only i in the title. Your list of definitions does not include a definition for $E^c$.
$endgroup$
– Rob Arthan
Dec 24 '18 at 0:18
$begingroup$
You have not fixed the problem identified by Kavi in the question but only i in the title. Your list of definitions does not include a definition for $E^c$.
$endgroup$
– Rob Arthan
Dec 24 '18 at 0:18
$begingroup$
By convention, $E^c$ usually denotes complement of a set and I thought that it was quite obvious from the context. And for $E$ not being the entire metric space, I first define $E$ to be a subset of a metric space for the definition of a limit point. And proceed to use that notation but is that not okay?
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:16
$begingroup$
By convention, $E^c$ usually denotes complement of a set and I thought that it was quite obvious from the context. And for $E$ not being the entire metric space, I first define $E$ to be a subset of a metric space for the definition of a limit point. And proceed to use that notation but is that not okay?
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:16
add a comment |
1 Answer
1
active
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$begingroup$
You're a little unclear with your use of open sets and neighbourhoods. As such, your definition of an open set reads as circular. You've defined an open set as the collection of all interior points, which you've defined as a point with an open neighbourhood within the set. What do you mean by open neighborhood? Again an open set?
answered Dec 24 '18 at 1:29
BMcNallyBMcNally
112
$endgroup$
$begingroup$
I am not sure how these definitions yield any circularity. A neighborhood in a metric space is just a set containing an open ball. I probably should have just said neighborhood instead of open neighborhood(but every neighborhood defined this way is open). But there is no circularity here.
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:45
$begingroup$
I think my problem is that a neighbourhood of point in point set topology is defined as sometimes differently, usually as the open set that contains a point or a just as a set containing an open set that contains a point, even for some treatments of a purely metric space context. If this was meant as a self contained treatment of this idea for people studying real analysis then perhaps a definition of open neighbourhood and open ball might make things clearer.
$endgroup$
– BMcNally
Dec 24 '18 at 19:17
add a comment |
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1 Answer
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$begingroup$
You're a little unclear with your use of open sets and neighbourhoods. As such, your definition of an open set reads as circular. You've defined an open set as the collection of all interior points, which you've defined as a point with an open neighbourhood within the set. What do you mean by open neighborhood? Again an open set?
answered Dec 24 '18 at 1:29
BMcNallyBMcNally
112
$endgroup$
$begingroup$
I am not sure how these definitions yield any circularity. A neighborhood in a metric space is just a set containing an open ball. I probably should have just said neighborhood instead of open neighborhood(but every neighborhood defined this way is open). But there is no circularity here.
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:45
$begingroup$
I think my problem is that a neighbourhood of point in point set topology is defined as sometimes differently, usually as the open set that contains a point or a just as a set containing an open set that contains a point, even for some treatments of a purely metric space context. If this was meant as a self contained treatment of this idea for people studying real analysis then perhaps a definition of open neighbourhood and open ball might make things clearer.
$endgroup$
– BMcNally
Dec 24 '18 at 19:17
add a comment |
$begingroup$
You're a little unclear with your use of open sets and neighbourhoods. As such, your definition of an open set reads as circular. You've defined an open set as the collection of all interior points, which you've defined as a point with an open neighbourhood within the set. What do you mean by open neighborhood? Again an open set?
answered Dec 24 '18 at 1:29
BMcNallyBMcNally
112
$endgroup$
$begingroup$
I am not sure how these definitions yield any circularity. A neighborhood in a metric space is just a set containing an open ball. I probably should have just said neighborhood instead of open neighborhood(but every neighborhood defined this way is open). But there is no circularity here.
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:45
$begingroup$
I think my problem is that a neighbourhood of point in point set topology is defined as sometimes differently, usually as the open set that contains a point or a just as a set containing an open set that contains a point, even for some treatments of a purely metric space context. If this was meant as a self contained treatment of this idea for people studying real analysis then perhaps a definition of open neighbourhood and open ball might make things clearer.
$endgroup$
– BMcNally
Dec 24 '18 at 19:17
add a comment |
$begingroup$
You're a little unclear with your use of open sets and neighbourhoods. As such, your definition of an open set reads as circular. You've defined an open set as the collection of all interior points, which you've defined as a point with an open neighbourhood within the set. What do you mean by open neighborhood? Again an open set?
answered Dec 24 '18 at 1:29
BMcNallyBMcNally
112
$endgroup$
You're a little unclear with your use of open sets and neighbourhoods. As such, your definition of an open set reads as circular. You've defined an open set as the collection of all interior points, which you've defined as a point with an open neighbourhood within the set. What do you mean by open neighborhood? Again an open set?
answered Dec 24 '18 at 1:29
BMcNallyBMcNally
112
answered Dec 24 '18 at 1:29
BMcNallyBMcNally
112
answered Dec 24 '18 at 1:29
BMcNallyBMcNally
112
answered Dec 24 '18 at 1:29
BMcNallyBMcNally
112
112
$begingroup$
I am not sure how these definitions yield any circularity. A neighborhood in a metric space is just a set containing an open ball. I probably should have just said neighborhood instead of open neighborhood(but every neighborhood defined this way is open). But there is no circularity here.
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:45
$begingroup$
I think my problem is that a neighbourhood of point in point set topology is defined as sometimes differently, usually as the open set that contains a point or a just as a set containing an open set that contains a point, even for some treatments of a purely metric space context. If this was meant as a self contained treatment of this idea for people studying real analysis then perhaps a definition of open neighbourhood and open ball might make things clearer.
$endgroup$
– BMcNally
Dec 24 '18 at 19:17
add a comment |
$begingroup$
I am not sure how these definitions yield any circularity. A neighborhood in a metric space is just a set containing an open ball. I probably should have just said neighborhood instead of open neighborhood(but every neighborhood defined this way is open). But there is no circularity here.
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:45
$begingroup$
I think my problem is that a neighbourhood of point in point set topology is defined as sometimes differently, usually as the open set that contains a point or a just as a set containing an open set that contains a point, even for some treatments of a purely metric space context. If this was meant as a self contained treatment of this idea for people studying real analysis then perhaps a definition of open neighbourhood and open ball might make things clearer.
$endgroup$
– BMcNally
Dec 24 '18 at 19:17
$begingroup$
I am not sure how these definitions yield any circularity. A neighborhood in a metric space is just a set containing an open ball. I probably should have just said neighborhood instead of open neighborhood(but every neighborhood defined this way is open). But there is no circularity here.
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:45
$begingroup$
I am not sure how these definitions yield any circularity. A neighborhood in a metric space is just a set containing an open ball. I probably should have just said neighborhood instead of open neighborhood(but every neighborhood defined this way is open). But there is no circularity here.
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:45
$begingroup$
I think my problem is that a neighbourhood of point in point set topology is defined as sometimes differently, usually as the open set that contains a point or a just as a set containing an open set that contains a point, even for some treatments of a purely metric space context. If this was meant as a self contained treatment of this idea for people studying real analysis then perhaps a definition of open neighbourhood and open ball might make things clearer.
$endgroup$
– BMcNally
Dec 24 '18 at 19:17
$begingroup$
I think my problem is that a neighbourhood of point in point set topology is defined as sometimes differently, usually as the open set that contains a point or a just as a set containing an open set that contains a point, even for some treatments of a purely metric space context. If this was meant as a self contained treatment of this idea for people studying real analysis then perhaps a definition of open neighbourhood and open ball might make things clearer.
$endgroup$
– BMcNally
Dec 24 '18 at 19:17
add a comment |
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$begingroup$
The only problem here is you are talking about a metric space being open/closed. You should be talking about a subset of a metric space being open or closed, not the entire metric space. The entire metric space is always both open and closed. Otherwise, your proof is fine.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:24
$begingroup$
Thanks for the suggestion. edit made for the future viewers
$endgroup$
– Sei Sakata
Dec 23 '18 at 23:30
$begingroup$
You have not fixed the problem identified by Kavi in the question but only i in the title. Your list of definitions does not include a definition for $E^c$.
$endgroup$
– Rob Arthan
Dec 24 '18 at 0:18
$begingroup$
By convention, $E^c$ usually denotes complement of a set and I thought that it was quite obvious from the context. And for $E$ not being the entire metric space, I first define $E$ to be a subset of a metric space for the definition of a limit point. And proceed to use that notation but is that not okay?
$endgroup$
– Sei Sakata
Dec 24 '18 at 1:16