If $X$ is an infinite set, then $X$ with the discrete topology is not compact
$begingroup$
I am new to compact sets, and I had some hard time trying to solve this:
Let $X$ be an infinite set; $T$ the discrete topology on $X$ .Prove that $(X,T)$ is not compact.
I know that I may consider the open cover ${{x},x in X}$ and prove that it doesn't have a finite subcover,but I don't know how to write it exactly.
Moreover, I seriously can't understant why $X$ itself cannot be considered as a finite subcover?
Any help is appreciated.
general-topology
asked Dec 7 '18 at 3:12
FifiFifi
256
$endgroup$
add a comment |
$begingroup$
I am new to compact sets, and I had some hard time trying to solve this:
Let $X$ be an infinite set; $T$ the discrete topology on $X$ .Prove that $(X,T)$ is not compact.
I know that I may consider the open cover ${{x},x in X}$ and prove that it doesn't have a finite subcover,but I don't know how to write it exactly.
Moreover, I seriously can't understant why $X$ itself cannot be considered as a finite subcover?
Any help is appreciated.
general-topology
asked Dec 7 '18 at 3:12
FifiFifi
256
$endgroup$
1
$begingroup$
${ { x } : x in X}$ can't be a finite subcover because $X$ is infinite.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 3:14
1
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 Also, $Xnotin{{x}:xin X}$.
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 3:14
2
$begingroup$
Each singleton set is open. You would have to find a finite number of them whose union is $X$ -- clearly impossible. ${X}$ is not a subcover because $X$ is not one of the sets in the original cover.
$endgroup$
– saulspatz
Dec 7 '18 at 3:15
$begingroup$
Thank you very much.I find it more clear now.
$endgroup$
– Fifi
Dec 7 '18 at 3:18
add a comment |
$begingroup$
I am new to compact sets, and I had some hard time trying to solve this:
Let $X$ be an infinite set; $T$ the discrete topology on $X$ .Prove that $(X,T)$ is not compact.
I know that I may consider the open cover ${{x},x in X}$ and prove that it doesn't have a finite subcover,but I don't know how to write it exactly.
Moreover, I seriously can't understant why $X$ itself cannot be considered as a finite subcover?
Any help is appreciated.
general-topology
asked Dec 7 '18 at 3:12
FifiFifi
256
$endgroup$
I am new to compact sets, and I had some hard time trying to solve this:
Let $X$ be an infinite set; $T$ the discrete topology on $X$ .Prove that $(X,T)$ is not compact.
I know that I may consider the open cover ${{x},x in X}$ and prove that it doesn't have a finite subcover,but I don't know how to write it exactly.
Moreover, I seriously can't understant why $X$ itself cannot be considered as a finite subcover?
Any help is appreciated.
general-topology
general-topology
asked Dec 7 '18 at 3:12
FifiFifi
256
asked Dec 7 '18 at 3:12
FifiFifi
256
edited Dec 7 '18 at 19:19
Fifi
asked Dec 7 '18 at 3:12
FifiFifi
256
asked Dec 7 '18 at 3:12
FifiFifi
256
asked Dec 7 '18 at 3:12
FifiFifi
256
256
1
$begingroup$
${ { x } : x in X}$ can't be a finite subcover because $X$ is infinite.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 3:14
1
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 Also, $Xnotin{{x}:xin X}$.
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 3:14
2
$begingroup$
Each singleton set is open. You would have to find a finite number of them whose union is $X$ -- clearly impossible. ${X}$ is not a subcover because $X$ is not one of the sets in the original cover.
$endgroup$
– saulspatz
Dec 7 '18 at 3:15
$begingroup$
Thank you very much.I find it more clear now.
$endgroup$
– Fifi
Dec 7 '18 at 3:18
add a comment |
1
$begingroup$
${ { x } : x in X}$ can't be a finite subcover because $X$ is infinite.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 3:14
1
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 Also, $Xnotin{{x}:xin X}$.
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 3:14
2
$begingroup$
Each singleton set is open. You would have to find a finite number of them whose union is $X$ -- clearly impossible. ${X}$ is not a subcover because $X$ is not one of the sets in the original cover.
$endgroup$
– saulspatz
Dec 7 '18 at 3:15
$begingroup$
Thank you very much.I find it more clear now.
$endgroup$
– Fifi
Dec 7 '18 at 3:18
1
1
$begingroup$
${ { x } : x in X}$ can't be a finite subcover because $X$ is infinite.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 3:14
$begingroup$
${ { x } : x in X}$ can't be a finite subcover because $X$ is infinite.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 3:14
1
1
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 Also, $Xnotin{{x}:xin X}$.
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 3:14
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 Also, $Xnotin{{x}:xin X}$.
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 3:14
2
2
$begingroup$
Each singleton set is open. You would have to find a finite number of them whose union is $X$ -- clearly impossible. ${X}$ is not a subcover because $X$ is not one of the sets in the original cover.
$endgroup$
– saulspatz
Dec 7 '18 at 3:15
$begingroup$
Each singleton set is open. You would have to find a finite number of them whose union is $X$ -- clearly impossible. ${X}$ is not a subcover because $X$ is not one of the sets in the original cover.
$endgroup$
– saulspatz
Dec 7 '18 at 3:15
$begingroup$
Thank you very much.I find it more clear now.
$endgroup$
– Fifi
Dec 7 '18 at 3:18
$begingroup$
Thank you very much.I find it more clear now.
$endgroup$
– Fifi
Dec 7 '18 at 3:18
add a comment |
1 Answer
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Indeed take the cover $mathcal{U}={{x}: x in X}$. This is clearly a cover of $X$ as every $p in X$ is in ${p} in mathcal{U}$ and in fact only in that set.
It's an open cover because the topology is discrete and so all subsets of $X$ are open.
A finite subcover is a finite subset of $mathcal{U}$ (so these are all singleton sets too) that together cover $X$ too. But this cannot exist: if $mathcal{F}$ is a finite subset of $mathcal{U}$ then it consists of finitely many singletons ${x_1},{x_2},ldots, {x_N}$ for some finite $N$. But $X$ has infinitely many points so there are infinitely many $x notin {x_1,ldots,x_N}$, say $p$ is one of those. Then $p$ is in none of the sets of $mathcal{F}$, and so this finite subset of $mathcal{U}$ is not a cover of $X$.
Hence $mathcal{U}$ has no finite subcover. In fact, if you think about it, we cannot remove a single element ${p}$ from $mathcal{U}$ or $p$ would no longer be covered at all; every point is only covered once... The set $X$ is not even an element of $mathcal{U}$ so does not qualify as a possible subcover. A subcover is a selection of open sets from the cover, and $X$ is not one of them, there are just singletons.
To think about: the open cover of all doubletons ${x,y}, x neq y$ would also have worked: a finite subset (say of size $N$) of that cover can only cover $2N$ points at most and so must also miss points of $X$, etc. In this cover every point is covered infinitely many times, and still we cannot reduce it to a finite subcover.
To show non-compactness we only have to find one cover without a finite subcover, because compactness is the strong statement that every open cover of $X$ must have a finite subcover. So one counterexample suffices to kill it.
answered Dec 7 '18 at 7:15
Henno BrandsmaHenno Brandsma
107k347114
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$begingroup$
Indeed take the cover $mathcal{U}={{x}: x in X}$. This is clearly a cover of $X$ as every $p in X$ is in ${p} in mathcal{U}$ and in fact only in that set.
It's an open cover because the topology is discrete and so all subsets of $X$ are open.
A finite subcover is a finite subset of $mathcal{U}$ (so these are all singleton sets too) that together cover $X$ too. But this cannot exist: if $mathcal{F}$ is a finite subset of $mathcal{U}$ then it consists of finitely many singletons ${x_1},{x_2},ldots, {x_N}$ for some finite $N$. But $X$ has infinitely many points so there are infinitely many $x notin {x_1,ldots,x_N}$, say $p$ is one of those. Then $p$ is in none of the sets of $mathcal{F}$, and so this finite subset of $mathcal{U}$ is not a cover of $X$.
Hence $mathcal{U}$ has no finite subcover. In fact, if you think about it, we cannot remove a single element ${p}$ from $mathcal{U}$ or $p$ would no longer be covered at all; every point is only covered once... The set $X$ is not even an element of $mathcal{U}$ so does not qualify as a possible subcover. A subcover is a selection of open sets from the cover, and $X$ is not one of them, there are just singletons.
To think about: the open cover of all doubletons ${x,y}, x neq y$ would also have worked: a finite subset (say of size $N$) of that cover can only cover $2N$ points at most and so must also miss points of $X$, etc. In this cover every point is covered infinitely many times, and still we cannot reduce it to a finite subcover.
To show non-compactness we only have to find one cover without a finite subcover, because compactness is the strong statement that every open cover of $X$ must have a finite subcover. So one counterexample suffices to kill it.
answered Dec 7 '18 at 7:15
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
add a comment |
$begingroup$
Indeed take the cover $mathcal{U}={{x}: x in X}$. This is clearly a cover of $X$ as every $p in X$ is in ${p} in mathcal{U}$ and in fact only in that set.
It's an open cover because the topology is discrete and so all subsets of $X$ are open.
A finite subcover is a finite subset of $mathcal{U}$ (so these are all singleton sets too) that together cover $X$ too. But this cannot exist: if $mathcal{F}$ is a finite subset of $mathcal{U}$ then it consists of finitely many singletons ${x_1},{x_2},ldots, {x_N}$ for some finite $N$. But $X$ has infinitely many points so there are infinitely many $x notin {x_1,ldots,x_N}$, say $p$ is one of those. Then $p$ is in none of the sets of $mathcal{F}$, and so this finite subset of $mathcal{U}$ is not a cover of $X$.
Hence $mathcal{U}$ has no finite subcover. In fact, if you think about it, we cannot remove a single element ${p}$ from $mathcal{U}$ or $p$ would no longer be covered at all; every point is only covered once... The set $X$ is not even an element of $mathcal{U}$ so does not qualify as a possible subcover. A subcover is a selection of open sets from the cover, and $X$ is not one of them, there are just singletons.
To think about: the open cover of all doubletons ${x,y}, x neq y$ would also have worked: a finite subset (say of size $N$) of that cover can only cover $2N$ points at most and so must also miss points of $X$, etc. In this cover every point is covered infinitely many times, and still we cannot reduce it to a finite subcover.
To show non-compactness we only have to find one cover without a finite subcover, because compactness is the strong statement that every open cover of $X$ must have a finite subcover. So one counterexample suffices to kill it.
answered Dec 7 '18 at 7:15
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
add a comment |
$begingroup$
Indeed take the cover $mathcal{U}={{x}: x in X}$. This is clearly a cover of $X$ as every $p in X$ is in ${p} in mathcal{U}$ and in fact only in that set.
It's an open cover because the topology is discrete and so all subsets of $X$ are open.
A finite subcover is a finite subset of $mathcal{U}$ (so these are all singleton sets too) that together cover $X$ too. But this cannot exist: if $mathcal{F}$ is a finite subset of $mathcal{U}$ then it consists of finitely many singletons ${x_1},{x_2},ldots, {x_N}$ for some finite $N$. But $X$ has infinitely many points so there are infinitely many $x notin {x_1,ldots,x_N}$, say $p$ is one of those. Then $p$ is in none of the sets of $mathcal{F}$, and so this finite subset of $mathcal{U}$ is not a cover of $X$.
Hence $mathcal{U}$ has no finite subcover. In fact, if you think about it, we cannot remove a single element ${p}$ from $mathcal{U}$ or $p$ would no longer be covered at all; every point is only covered once... The set $X$ is not even an element of $mathcal{U}$ so does not qualify as a possible subcover. A subcover is a selection of open sets from the cover, and $X$ is not one of them, there are just singletons.
To think about: the open cover of all doubletons ${x,y}, x neq y$ would also have worked: a finite subset (say of size $N$) of that cover can only cover $2N$ points at most and so must also miss points of $X$, etc. In this cover every point is covered infinitely many times, and still we cannot reduce it to a finite subcover.
To show non-compactness we only have to find one cover without a finite subcover, because compactness is the strong statement that every open cover of $X$ must have a finite subcover. So one counterexample suffices to kill it.
answered Dec 7 '18 at 7:15
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
Indeed take the cover $mathcal{U}={{x}: x in X}$. This is clearly a cover of $X$ as every $p in X$ is in ${p} in mathcal{U}$ and in fact only in that set.
It's an open cover because the topology is discrete and so all subsets of $X$ are open.
A finite subcover is a finite subset of $mathcal{U}$ (so these are all singleton sets too) that together cover $X$ too. But this cannot exist: if $mathcal{F}$ is a finite subset of $mathcal{U}$ then it consists of finitely many singletons ${x_1},{x_2},ldots, {x_N}$ for some finite $N$. But $X$ has infinitely many points so there are infinitely many $x notin {x_1,ldots,x_N}$, say $p$ is one of those. Then $p$ is in none of the sets of $mathcal{F}$, and so this finite subset of $mathcal{U}$ is not a cover of $X$.
Hence $mathcal{U}$ has no finite subcover. In fact, if you think about it, we cannot remove a single element ${p}$ from $mathcal{U}$ or $p$ would no longer be covered at all; every point is only covered once... The set $X$ is not even an element of $mathcal{U}$ so does not qualify as a possible subcover. A subcover is a selection of open sets from the cover, and $X$ is not one of them, there are just singletons.
To think about: the open cover of all doubletons ${x,y}, x neq y$ would also have worked: a finite subset (say of size $N$) of that cover can only cover $2N$ points at most and so must also miss points of $X$, etc. In this cover every point is covered infinitely many times, and still we cannot reduce it to a finite subcover.
To show non-compactness we only have to find one cover without a finite subcover, because compactness is the strong statement that every open cover of $X$ must have a finite subcover. So one counterexample suffices to kill it.
answered Dec 7 '18 at 7:15
Henno BrandsmaHenno Brandsma
107k347114
answered Dec 7 '18 at 7:15
Henno BrandsmaHenno Brandsma
107k347114
answered Dec 7 '18 at 7:15
Henno BrandsmaHenno Brandsma
107k347114
answered Dec 7 '18 at 7:15
Henno BrandsmaHenno Brandsma
107k347114
107k347114
add a comment |
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$begingroup$
${ { x } : x in X}$ can't be a finite subcover because $X$ is infinite.
$endgroup$
– RandomMathGuy
Dec 7 '18 at 3:14
1
$begingroup$
Please see math.meta.stackexchange.com/questions/5020 Also, $Xnotin{{x}:xin X}$.
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 3:14
2
$begingroup$
Each singleton set is open. You would have to find a finite number of them whose union is $X$ -- clearly impossible. ${X}$ is not a subcover because $X$ is not one of the sets in the original cover.
$endgroup$
– saulspatz
Dec 7 '18 at 3:15
$begingroup$
Thank you very much.I find it more clear now.
$endgroup$
– Fifi
Dec 7 '18 at 3:18