Intersection of two compact subsets in a Hausdorff space
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Prove that if $K$ and $L$ are compact subsets of a Hausdorff space $X$, then $K cap L$ is a compact subset of $X$.
I understand that since $K$ and $L$ are compact subsets, they each have finite coverings. Do I have to show that the intersection of $2$ finite coverings is a compact subset?
general-topology compactness
edited Dec 12 '18 at 20:33
Davide Giraudo
127k16151265
asked Dec 5 '18 at 4:09
general1597general1597
313
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add a comment |
$begingroup$
Prove that if $K$ and $L$ are compact subsets of a Hausdorff space $X$, then $K cap L$ is a compact subset of $X$.
I understand that since $K$ and $L$ are compact subsets, they each have finite coverings. Do I have to show that the intersection of $2$ finite coverings is a compact subset?
general-topology compactness
edited Dec 12 '18 at 20:33
Davide Giraudo
127k16151265
asked Dec 5 '18 at 4:09
general1597general1597
313
$endgroup$
1
$begingroup$
Hint: $K$ and $L$ must be closed sets and $K cap L subseteq K$.
$endgroup$
– Randall
Dec 5 '18 at 4:25
add a comment |
$begingroup$
Prove that if $K$ and $L$ are compact subsets of a Hausdorff space $X$, then $K cap L$ is a compact subset of $X$.
I understand that since $K$ and $L$ are compact subsets, they each have finite coverings. Do I have to show that the intersection of $2$ finite coverings is a compact subset?
general-topology compactness
edited Dec 12 '18 at 20:33
Davide Giraudo
127k16151265
asked Dec 5 '18 at 4:09
general1597general1597
313
$endgroup$
Prove that if $K$ and $L$ are compact subsets of a Hausdorff space $X$, then $K cap L$ is a compact subset of $X$.
I understand that since $K$ and $L$ are compact subsets, they each have finite coverings. Do I have to show that the intersection of $2$ finite coverings is a compact subset?
general-topology compactness
general-topology compactness
edited Dec 12 '18 at 20:33
Davide Giraudo
127k16151265
asked Dec 5 '18 at 4:09
general1597general1597
313
edited Dec 12 '18 at 20:33
Davide Giraudo
127k16151265
asked Dec 5 '18 at 4:09
general1597general1597
313
edited Dec 12 '18 at 20:33
Davide Giraudo
127k16151265
edited Dec 12 '18 at 20:33
Davide Giraudo
127k16151265
edited Dec 12 '18 at 20:33
Davide Giraudo
127k16151265
127k16151265
asked Dec 5 '18 at 4:09
general1597general1597
313
asked Dec 5 '18 at 4:09
general1597general1597
313
asked Dec 5 '18 at 4:09
general1597general1597
313
313
1
$begingroup$
Hint: $K$ and $L$ must be closed sets and $K cap L subseteq K$.
$endgroup$
– Randall
Dec 5 '18 at 4:25
add a comment |
1
$begingroup$
Hint: $K$ and $L$ must be closed sets and $K cap L subseteq K$.
$endgroup$
– Randall
Dec 5 '18 at 4:25
1
1
$begingroup$
Hint: $K$ and $L$ must be closed sets and $K cap L subseteq K$.
$endgroup$
– Randall
Dec 5 '18 at 4:25
$begingroup$
Hint: $K$ and $L$ must be closed sets and $K cap L subseteq K$.
$endgroup$
– Randall
Dec 5 '18 at 4:25
add a comment |
1 Answer
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$begingroup$
Theorem:Any compact subset of a Hausdorff space is closed.
You can find a proof here.
Let $mathcal{A}$ be an open cover of $K cap L$. Since $K$ and $L$ are closed sets ,$K cap L$ is also closed. So ${(K cap L)}^C$ is open. Then $mathcal{B}=mathcal{A}cup {(K cap L)}^C$ is an open cover of $K$(and $L$). Since $K$ is compact,$mathcal{B}$ has a finite subcover,say,${A_1,A_2,ldots,A_n,{(K cap L)}^C}$. So ${A_1,A_2,ldots,A_n}$ is a finite subcover of $mathcal{A}$ that covers $K cap L$. Hence, $K cap L$ is compact.
answered Dec 5 '18 at 5:03
Thomas ShelbyThomas Shelby
3,7192525
$endgroup$
$begingroup$
Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
$endgroup$
– user198044
Dec 5 '18 at 6:19
1
$begingroup$
@JackBauer Even if $(Kcap L)^C $ is not included in the finite subcover, still the union of that subcover and $(Kcap L)^C $ will cover $K$, right? So basically I was trying to incorporate the two cases into one by taking the above mentioned subcover. If there is any mistake , please let me know.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 9:04
1
$begingroup$
I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
$endgroup$
– user198044
Dec 12 '18 at 20:11
add a comment |
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$begingroup$
Theorem:Any compact subset of a Hausdorff space is closed.
You can find a proof here.
Let $mathcal{A}$ be an open cover of $K cap L$. Since $K$ and $L$ are closed sets ,$K cap L$ is also closed. So ${(K cap L)}^C$ is open. Then $mathcal{B}=mathcal{A}cup {(K cap L)}^C$ is an open cover of $K$(and $L$). Since $K$ is compact,$mathcal{B}$ has a finite subcover,say,${A_1,A_2,ldots,A_n,{(K cap L)}^C}$. So ${A_1,A_2,ldots,A_n}$ is a finite subcover of $mathcal{A}$ that covers $K cap L$. Hence, $K cap L$ is compact.
answered Dec 5 '18 at 5:03
Thomas ShelbyThomas Shelby
3,7192525
$endgroup$
$begingroup$
Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
$endgroup$
– user198044
Dec 5 '18 at 6:19
1
$begingroup$
@JackBauer Even if $(Kcap L)^C $ is not included in the finite subcover, still the union of that subcover and $(Kcap L)^C $ will cover $K$, right? So basically I was trying to incorporate the two cases into one by taking the above mentioned subcover. If there is any mistake , please let me know.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 9:04
1
$begingroup$
I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
$endgroup$
– user198044
Dec 12 '18 at 20:11
add a comment |
$begingroup$
Theorem:Any compact subset of a Hausdorff space is closed.
You can find a proof here.
Let $mathcal{A}$ be an open cover of $K cap L$. Since $K$ and $L$ are closed sets ,$K cap L$ is also closed. So ${(K cap L)}^C$ is open. Then $mathcal{B}=mathcal{A}cup {(K cap L)}^C$ is an open cover of $K$(and $L$). Since $K$ is compact,$mathcal{B}$ has a finite subcover,say,${A_1,A_2,ldots,A_n,{(K cap L)}^C}$. So ${A_1,A_2,ldots,A_n}$ is a finite subcover of $mathcal{A}$ that covers $K cap L$. Hence, $K cap L$ is compact.
answered Dec 5 '18 at 5:03
Thomas ShelbyThomas Shelby
3,7192525
$endgroup$
$begingroup$
Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
$endgroup$
– user198044
Dec 5 '18 at 6:19
1
$begingroup$
@JackBauer Even if $(Kcap L)^C $ is not included in the finite subcover, still the union of that subcover and $(Kcap L)^C $ will cover $K$, right? So basically I was trying to incorporate the two cases into one by taking the above mentioned subcover. If there is any mistake , please let me know.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 9:04
1
$begingroup$
I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
$endgroup$
– user198044
Dec 12 '18 at 20:11
add a comment |
$begingroup$
Theorem:Any compact subset of a Hausdorff space is closed.
You can find a proof here.
Let $mathcal{A}$ be an open cover of $K cap L$. Since $K$ and $L$ are closed sets ,$K cap L$ is also closed. So ${(K cap L)}^C$ is open. Then $mathcal{B}=mathcal{A}cup {(K cap L)}^C$ is an open cover of $K$(and $L$). Since $K$ is compact,$mathcal{B}$ has a finite subcover,say,${A_1,A_2,ldots,A_n,{(K cap L)}^C}$. So ${A_1,A_2,ldots,A_n}$ is a finite subcover of $mathcal{A}$ that covers $K cap L$. Hence, $K cap L$ is compact.
answered Dec 5 '18 at 5:03
Thomas ShelbyThomas Shelby
3,7192525
$endgroup$
Theorem:Any compact subset of a Hausdorff space is closed.
You can find a proof here.
Let $mathcal{A}$ be an open cover of $K cap L$. Since $K$ and $L$ are closed sets ,$K cap L$ is also closed. So ${(K cap L)}^C$ is open. Then $mathcal{B}=mathcal{A}cup {(K cap L)}^C$ is an open cover of $K$(and $L$). Since $K$ is compact,$mathcal{B}$ has a finite subcover,say,${A_1,A_2,ldots,A_n,{(K cap L)}^C}$. So ${A_1,A_2,ldots,A_n}$ is a finite subcover of $mathcal{A}$ that covers $K cap L$. Hence, $K cap L$ is compact.
answered Dec 5 '18 at 5:03
Thomas ShelbyThomas Shelby
3,7192525
answered Dec 5 '18 at 5:03
Thomas ShelbyThomas Shelby
3,7192525
answered Dec 5 '18 at 5:03
Thomas ShelbyThomas Shelby
3,7192525
answered Dec 5 '18 at 5:03
Thomas ShelbyThomas Shelby
3,7192525
3,7192525
$begingroup$
Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
$endgroup$
– user198044
Dec 5 '18 at 6:19
1
$begingroup$
@JackBauer Even if $(Kcap L)^C $ is not included in the finite subcover, still the union of that subcover and $(Kcap L)^C $ will cover $K$, right? So basically I was trying to incorporate the two cases into one by taking the above mentioned subcover. If there is any mistake , please let me know.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 9:04
1
$begingroup$
I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
$endgroup$
– user198044
Dec 12 '18 at 20:11
add a comment |
$begingroup$
Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
$endgroup$
– user198044
Dec 5 '18 at 6:19
1
$begingroup$
@JackBauer Even if $(Kcap L)^C $ is not included in the finite subcover, still the union of that subcover and $(Kcap L)^C $ will cover $K$, right? So basically I was trying to incorporate the two cases into one by taking the above mentioned subcover. If there is any mistake , please let me know.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 9:04
1
$begingroup$
I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
$endgroup$
– user198044
Dec 12 '18 at 20:11
$begingroup$
Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
$endgroup$
– user198044
Dec 5 '18 at 6:19
$begingroup$
Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
$endgroup$
– user198044
Dec 5 '18 at 6:19
1
1
$begingroup$
@JackBauer Even if $(Kcap L)^C $ is not included in the finite subcover, still the union of that subcover and $(Kcap L)^C $ will cover $K$, right? So basically I was trying to incorporate the two cases into one by taking the above mentioned subcover. If there is any mistake , please let me know.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 9:04
$begingroup$
@JackBauer Even if $(Kcap L)^C $ is not included in the finite subcover, still the union of that subcover and $(Kcap L)^C $ will cover $K$, right? So basically I was trying to incorporate the two cases into one by taking the above mentioned subcover. If there is any mistake , please let me know.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 9:04
1
1
$begingroup$
I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
$endgroup$
– user198044
Dec 12 '18 at 20:11
$begingroup$
I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
$endgroup$
– user198044
Dec 12 '18 at 20:11
add a comment |
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$begingroup$
Hint: $K$ and $L$ must be closed sets and $K cap L subseteq K$.
$endgroup$
– Randall
Dec 5 '18 at 4:25