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
















0














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




















  • 1




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














0












0








0







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















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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edited Dec 12 '18 at 20:33









Davide Giraudo

125k16150261




125k16150261










asked Dec 5 '18 at 4:09









general1597general1597

313




313








  • 1




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














  • 1




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








1




1




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




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










1 Answer
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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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  • Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
    – user198044
    Dec 5 '18 at 6:19








  • 1




    @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.
    – Thomas Shelby
    Dec 5 '18 at 9:04








  • 1




    I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
    – user198044
    Dec 12 '18 at 20:11











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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





















  • Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
    – user198044
    Dec 5 '18 at 6:19








  • 1




    @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.
    – Thomas Shelby
    Dec 5 '18 at 9:04








  • 1




    I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
    – user198044
    Dec 12 '18 at 20:11
















0















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





















  • Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
    – user198044
    Dec 5 '18 at 6:19








  • 1




    @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.
    – Thomas Shelby
    Dec 5 '18 at 9:04








  • 1




    I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
    – user198044
    Dec 12 '18 at 20:11














0












0








0







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













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

2,007219




2,007219












  • Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
    – user198044
    Dec 5 '18 at 6:19








  • 1




    @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.
    – Thomas Shelby
    Dec 5 '18 at 9:04








  • 1




    I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
    – user198044
    Dec 12 '18 at 20:11


















  • Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
    – user198044
    Dec 5 '18 at 6:19








  • 1




    @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.
    – Thomas Shelby
    Dec 5 '18 at 9:04








  • 1




    I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
    – user198044
    Dec 12 '18 at 20:11
















Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
– user198044
Dec 5 '18 at 6:19






Why don't you have 2 cases to account for whether or not $(Kcap L)^C$ is included in the finite subcover?
– user198044
Dec 5 '18 at 6:19






1




1




@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.
– Thomas Shelby
Dec 5 '18 at 9:04






@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.
– Thomas Shelby
Dec 5 '18 at 9:04






1




1




I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
– user198044
Dec 12 '18 at 20:11




I didn't analyze anymore, but if you did "incorporate the two cases", then that's brilliant.
– user198044
Dec 12 '18 at 20:11


















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