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







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  • 1




    $begingroup$
    Hint: $K$ and $L$ must be closed sets and $K cap L subseteq K$.
    $endgroup$
    – Randall
    Dec 5 '18 at 4:25
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: $K$ and $L$ must be closed sets and $K cap L subseteq K$.
    $endgroup$
    – Randall
    Dec 5 '18 at 4:25














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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share|cite|improve this question








edited Dec 12 '18 at 20:33









Davide Giraudo

127k16151265




127k16151265










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














  • 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










1 Answer
1






active

oldest

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0












$begingroup$


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.






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$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
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0












$begingroup$


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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$


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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$


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.






share|cite|improve this answer









$endgroup$




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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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