A compact open set
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Is there an open set which is compact ? I would say that $emptyset$ is an open set compact because it's bounded and closed too. Is it correct ?
general-topology
asked Aug 20 '15 at 12:12
idmidm
8,61521446
$endgroup$
add a comment |
$begingroup$
Is there an open set which is compact ? I would say that $emptyset$ is an open set compact because it's bounded and closed too. Is it correct ?
general-topology
asked Aug 20 '15 at 12:12
idmidm
8,61521446
$endgroup$
1
$begingroup$
It is
$endgroup$
– A.Γ.
Aug 20 '15 at 12:20
1
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Yes. Are you after a nontrivial open compact set?
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– Gary Moore
Aug 20 '15 at 12:20
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math.stackexchange.com/questions/304567/… Here you will find an answer
$endgroup$
– avati91
Aug 20 '15 at 12:26
add a comment |
$begingroup$
Is there an open set which is compact ? I would say that $emptyset$ is an open set compact because it's bounded and closed too. Is it correct ?
general-topology
asked Aug 20 '15 at 12:12
idmidm
8,61521446
$endgroup$
Is there an open set which is compact ? I would say that $emptyset$ is an open set compact because it's bounded and closed too. Is it correct ?
general-topology
general-topology
asked Aug 20 '15 at 12:12
idmidm
8,61521446
asked Aug 20 '15 at 12:12
idmidm
8,61521446
asked Aug 20 '15 at 12:12
idmidm
8,61521446
asked Aug 20 '15 at 12:12
idmidm
8,61521446
asked Aug 20 '15 at 12:12
idmidm
8,61521446
8,61521446
1
$begingroup$
It is
$endgroup$
– A.Γ.
Aug 20 '15 at 12:20
1
$begingroup$
Yes. Are you after a nontrivial open compact set?
$endgroup$
– Gary Moore
Aug 20 '15 at 12:20
$begingroup$
math.stackexchange.com/questions/304567/… Here you will find an answer
$endgroup$
– avati91
Aug 20 '15 at 12:26
add a comment |
1
$begingroup$
It is
$endgroup$
– A.Γ.
Aug 20 '15 at 12:20
1
$begingroup$
Yes. Are you after a nontrivial open compact set?
$endgroup$
– Gary Moore
Aug 20 '15 at 12:20
$begingroup$
math.stackexchange.com/questions/304567/… Here you will find an answer
$endgroup$
– avati91
Aug 20 '15 at 12:26
1
1
$begingroup$
It is
$endgroup$
– A.Γ.
Aug 20 '15 at 12:20
$begingroup$
It is
$endgroup$
– A.Γ.
Aug 20 '15 at 12:20
1
1
$begingroup$
Yes. Are you after a nontrivial open compact set?
$endgroup$
– Gary Moore
Aug 20 '15 at 12:20
$begingroup$
Yes. Are you after a nontrivial open compact set?
$endgroup$
– Gary Moore
Aug 20 '15 at 12:20
$begingroup$
math.stackexchange.com/questions/304567/… Here you will find an answer
$endgroup$
– avati91
Aug 20 '15 at 12:26
$begingroup$
math.stackexchange.com/questions/304567/… Here you will find an answer
$endgroup$
– avati91
Aug 20 '15 at 12:26
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The empty set is certainly compact, all finite spaces are.
A compact subset of a Hausdorff space (for example a metric space) is always closed, so your set would be closed and open. Now that can happen, but in a connected space the only open and closed subsets are the empty set and the space itself.
answered Aug 20 '15 at 12:33
Carsten SCarsten S
7,31711436
$endgroup$
add a comment |
$begingroup$
Consider a set $X$ with more than two elements, equipped with discrete metric $d(x,y)=1$ if $x not = y$ and $d(x,y)=0$ if $x=y$. I will leave it to you as an exercises to prove that a set is compact if and only if it is finite. Pick an element $a in X$. Consider the ball $D(a,frac{1}{2})$. Clearly,$D(a,frac{1}{2})=lbrace a rbrace $ and hence $lbrace a rbrace $ is open and compact at the same time.
answered Aug 20 '15 at 12:24
mich95mich95
7,02211226
$endgroup$
add a comment |
$begingroup$
In a proper space a closed bounded set is compact but in general topological space it is not true that a closed bounded set is compact. There are plenty of example like spaces of infinite sequence. Take two disjoint open bounded close disc give it a subspace topology of $R^2$ you will have open set which is compact.
answered Aug 20 '15 at 12:25
GGTGGT
485311
$endgroup$
add a comment |
$begingroup$
If you look at $X=[0,1]cup(4,5)$ as a topological subspace of $Bbb R$, then $[0,1]$ is still compact, and it is open since it is equal to $(-1,2)cap X$.
Going to a completely different direction, if you look at a set with the co-finite topology, then every set is compact. In particular every co-finite set, which is open. If the set is finite, this is just the discrete topology, but if the set is infinite, then this ends up being a stranger topology than you might be used to.
answered Aug 20 '15 at 12:38
Asaf Karagila♦Asaf Karagila
308k33441775
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add a comment |
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4 Answers
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active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The empty set is certainly compact, all finite spaces are.
A compact subset of a Hausdorff space (for example a metric space) is always closed, so your set would be closed and open. Now that can happen, but in a connected space the only open and closed subsets are the empty set and the space itself.
answered Aug 20 '15 at 12:33
Carsten SCarsten S
7,31711436
$endgroup$
add a comment |
$begingroup$
The empty set is certainly compact, all finite spaces are.
A compact subset of a Hausdorff space (for example a metric space) is always closed, so your set would be closed and open. Now that can happen, but in a connected space the only open and closed subsets are the empty set and the space itself.
answered Aug 20 '15 at 12:33
Carsten SCarsten S
7,31711436
$endgroup$
add a comment |
$begingroup$
The empty set is certainly compact, all finite spaces are.
A compact subset of a Hausdorff space (for example a metric space) is always closed, so your set would be closed and open. Now that can happen, but in a connected space the only open and closed subsets are the empty set and the space itself.
answered Aug 20 '15 at 12:33
Carsten SCarsten S
7,31711436
$endgroup$
The empty set is certainly compact, all finite spaces are.
A compact subset of a Hausdorff space (for example a metric space) is always closed, so your set would be closed and open. Now that can happen, but in a connected space the only open and closed subsets are the empty set and the space itself.
answered Aug 20 '15 at 12:33
Carsten SCarsten S
7,31711436
edited Aug 20 '15 at 13:25
answered Aug 20 '15 at 12:33
Carsten SCarsten S
7,31711436
answered Aug 20 '15 at 12:33
Carsten SCarsten S
7,31711436
answered Aug 20 '15 at 12:33
Carsten SCarsten S
7,31711436
7,31711436
add a comment |
add a comment |
$begingroup$
Consider a set $X$ with more than two elements, equipped with discrete metric $d(x,y)=1$ if $x not = y$ and $d(x,y)=0$ if $x=y$. I will leave it to you as an exercises to prove that a set is compact if and only if it is finite. Pick an element $a in X$. Consider the ball $D(a,frac{1}{2})$. Clearly,$D(a,frac{1}{2})=lbrace a rbrace $ and hence $lbrace a rbrace $ is open and compact at the same time.
answered Aug 20 '15 at 12:24
mich95mich95
7,02211226
$endgroup$
add a comment |
$begingroup$
Consider a set $X$ with more than two elements, equipped with discrete metric $d(x,y)=1$ if $x not = y$ and $d(x,y)=0$ if $x=y$. I will leave it to you as an exercises to prove that a set is compact if and only if it is finite. Pick an element $a in X$. Consider the ball $D(a,frac{1}{2})$. Clearly,$D(a,frac{1}{2})=lbrace a rbrace $ and hence $lbrace a rbrace $ is open and compact at the same time.
answered Aug 20 '15 at 12:24
mich95mich95
7,02211226
$endgroup$
add a comment |
$begingroup$
Consider a set $X$ with more than two elements, equipped with discrete metric $d(x,y)=1$ if $x not = y$ and $d(x,y)=0$ if $x=y$. I will leave it to you as an exercises to prove that a set is compact if and only if it is finite. Pick an element $a in X$. Consider the ball $D(a,frac{1}{2})$. Clearly,$D(a,frac{1}{2})=lbrace a rbrace $ and hence $lbrace a rbrace $ is open and compact at the same time.
answered Aug 20 '15 at 12:24
mich95mich95
7,02211226
$endgroup$
Consider a set $X$ with more than two elements, equipped with discrete metric $d(x,y)=1$ if $x not = y$ and $d(x,y)=0$ if $x=y$. I will leave it to you as an exercises to prove that a set is compact if and only if it is finite. Pick an element $a in X$. Consider the ball $D(a,frac{1}{2})$. Clearly,$D(a,frac{1}{2})=lbrace a rbrace $ and hence $lbrace a rbrace $ is open and compact at the same time.
answered Aug 20 '15 at 12:24
mich95mich95
7,02211226
answered Aug 20 '15 at 12:24
mich95mich95
7,02211226
answered Aug 20 '15 at 12:24
mich95mich95
7,02211226
answered Aug 20 '15 at 12:24
mich95mich95
7,02211226
7,02211226
add a comment |
add a comment |
$begingroup$
In a proper space a closed bounded set is compact but in general topological space it is not true that a closed bounded set is compact. There are plenty of example like spaces of infinite sequence. Take two disjoint open bounded close disc give it a subspace topology of $R^2$ you will have open set which is compact.
answered Aug 20 '15 at 12:25
GGTGGT
485311
$endgroup$
add a comment |
$begingroup$
In a proper space a closed bounded set is compact but in general topological space it is not true that a closed bounded set is compact. There are plenty of example like spaces of infinite sequence. Take two disjoint open bounded close disc give it a subspace topology of $R^2$ you will have open set which is compact.
answered Aug 20 '15 at 12:25
GGTGGT
485311
$endgroup$
add a comment |
$begingroup$
In a proper space a closed bounded set is compact but in general topological space it is not true that a closed bounded set is compact. There are plenty of example like spaces of infinite sequence. Take two disjoint open bounded close disc give it a subspace topology of $R^2$ you will have open set which is compact.
answered Aug 20 '15 at 12:25
GGTGGT
485311
$endgroup$
In a proper space a closed bounded set is compact but in general topological space it is not true that a closed bounded set is compact. There are plenty of example like spaces of infinite sequence. Take two disjoint open bounded close disc give it a subspace topology of $R^2$ you will have open set which is compact.
answered Aug 20 '15 at 12:25
GGTGGT
485311
answered Aug 20 '15 at 12:25
GGTGGT
485311
answered Aug 20 '15 at 12:25
GGTGGT
485311
answered Aug 20 '15 at 12:25
GGTGGT
485311
485311
add a comment |
add a comment |
$begingroup$
If you look at $X=[0,1]cup(4,5)$ as a topological subspace of $Bbb R$, then $[0,1]$ is still compact, and it is open since it is equal to $(-1,2)cap X$.
Going to a completely different direction, if you look at a set with the co-finite topology, then every set is compact. In particular every co-finite set, which is open. If the set is finite, this is just the discrete topology, but if the set is infinite, then this ends up being a stranger topology than you might be used to.
answered Aug 20 '15 at 12:38
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
add a comment |
$begingroup$
If you look at $X=[0,1]cup(4,5)$ as a topological subspace of $Bbb R$, then $[0,1]$ is still compact, and it is open since it is equal to $(-1,2)cap X$.
Going to a completely different direction, if you look at a set with the co-finite topology, then every set is compact. In particular every co-finite set, which is open. If the set is finite, this is just the discrete topology, but if the set is infinite, then this ends up being a stranger topology than you might be used to.
answered Aug 20 '15 at 12:38
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
add a comment |
$begingroup$
If you look at $X=[0,1]cup(4,5)$ as a topological subspace of $Bbb R$, then $[0,1]$ is still compact, and it is open since it is equal to $(-1,2)cap X$.
Going to a completely different direction, if you look at a set with the co-finite topology, then every set is compact. In particular every co-finite set, which is open. If the set is finite, this is just the discrete topology, but if the set is infinite, then this ends up being a stranger topology than you might be used to.
answered Aug 20 '15 at 12:38
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
If you look at $X=[0,1]cup(4,5)$ as a topological subspace of $Bbb R$, then $[0,1]$ is still compact, and it is open since it is equal to $(-1,2)cap X$.
Going to a completely different direction, if you look at a set with the co-finite topology, then every set is compact. In particular every co-finite set, which is open. If the set is finite, this is just the discrete topology, but if the set is infinite, then this ends up being a stranger topology than you might be used to.
answered Aug 20 '15 at 12:38
Asaf Karagila♦Asaf Karagila
308k33441775
answered Aug 20 '15 at 12:38
Asaf Karagila♦Asaf Karagila
308k33441775
answered Aug 20 '15 at 12:38
Asaf Karagila♦Asaf Karagila
308k33441775
answered Aug 20 '15 at 12:38
Asaf Karagila♦Asaf Karagila
308k33441775
308k33441775
add a comment |
add a comment |
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1
$begingroup$
It is
$endgroup$
– A.Γ.
Aug 20 '15 at 12:20
1
$begingroup$
Yes. Are you after a nontrivial open compact set?
$endgroup$
– Gary Moore
Aug 20 '15 at 12:20
$begingroup$
math.stackexchange.com/questions/304567/… Here you will find an answer
$endgroup$
– avati91
Aug 20 '15 at 12:26