proof compactness of sets2 questions regarding compactness and closedCharacterization of compactness in weak* topologyContinuous one-to-one mapping of a compact spaceCharacterizing $sigma$-compactness via closed setsMetric space and compactnessa problem about compactness and sequential compactness in metric spaceProof that boundedness of continuous Real Valued functions implies CompactnessShow compactnessChecking the compactness of setsCompactness in the compact complement topology

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proof compactness of sets


2 questions regarding compactness and closedCharacterization of compactness in weak* topologyContinuous one-to-one mapping of a compact spaceCharacterizing $sigma$-compactness via closed setsMetric space and compactnessa problem about compactness and sequential compactness in metric spaceProof that boundedness of continuous Real Valued functions implies CompactnessShow compactnessChecking the compactness of setsCompactness in the compact complement topology













0












$begingroup$


Let $$K ⊂ R^m$$ and $$L ⊂ R^n$$ be compact subsets.



Show:



a) The set $$ K × L: = (x, y): x ∈ K, y ∈ L ⊂ R^m+n$$ is also compact.



I have to show that this is bounded and closed.
But how do I do that?










share|cite|improve this question









New contributor




breakshooter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 2




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Lord Shark the Unknown
    Mar 11 at 21:15






  • 1




    $begingroup$
    Bounded should not be a problem in either case. Have you tried proving that either $Ktimes L$ or $K+L$ is closed? What would the structure of such a proof look like?
    $endgroup$
    – Greg Martin
    Mar 11 at 22:20










  • $begingroup$
    I don't know, can you help me?
    $endgroup$
    – breakshooter
    Mar 11 at 23:51















0












$begingroup$


Let $$K ⊂ R^m$$ and $$L ⊂ R^n$$ be compact subsets.



Show:



a) The set $$ K × L: = (x, y): x ∈ K, y ∈ L ⊂ R^m+n$$ is also compact.



I have to show that this is bounded and closed.
But how do I do that?










share|cite|improve this question









New contributor




breakshooter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 2




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Lord Shark the Unknown
    Mar 11 at 21:15






  • 1




    $begingroup$
    Bounded should not be a problem in either case. Have you tried proving that either $Ktimes L$ or $K+L$ is closed? What would the structure of such a proof look like?
    $endgroup$
    – Greg Martin
    Mar 11 at 22:20










  • $begingroup$
    I don't know, can you help me?
    $endgroup$
    – breakshooter
    Mar 11 at 23:51













0












0








0





$begingroup$


Let $$K ⊂ R^m$$ and $$L ⊂ R^n$$ be compact subsets.



Show:



a) The set $$ K × L: = (x, y): x ∈ K, y ∈ L ⊂ R^m+n$$ is also compact.



I have to show that this is bounded and closed.
But how do I do that?










share|cite|improve this question









New contributor




breakshooter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Let $$K ⊂ R^m$$ and $$L ⊂ R^n$$ be compact subsets.



Show:



a) The set $$ K × L: = (x, y): x ∈ K, y ∈ L ⊂ R^m+n$$ is also compact.



I have to show that this is bounded and closed.
But how do I do that?







compactness






share|cite|improve this question









New contributor




breakshooter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




breakshooter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Mar 11 at 23:50







breakshooter













New contributor




breakshooter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Mar 11 at 21:14









breakshooterbreakshooter

11




11




New contributor




breakshooter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





breakshooter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






breakshooter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 2




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Lord Shark the Unknown
    Mar 11 at 21:15






  • 1




    $begingroup$
    Bounded should not be a problem in either case. Have you tried proving that either $Ktimes L$ or $K+L$ is closed? What would the structure of such a proof look like?
    $endgroup$
    – Greg Martin
    Mar 11 at 22:20










  • $begingroup$
    I don't know, can you help me?
    $endgroup$
    – breakshooter
    Mar 11 at 23:51












  • 2




    $begingroup$
    Please see math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Lord Shark the Unknown
    Mar 11 at 21:15






  • 1




    $begingroup$
    Bounded should not be a problem in either case. Have you tried proving that either $Ktimes L$ or $K+L$ is closed? What would the structure of such a proof look like?
    $endgroup$
    – Greg Martin
    Mar 11 at 22:20










  • $begingroup$
    I don't know, can you help me?
    $endgroup$
    – breakshooter
    Mar 11 at 23:51







2




2




$begingroup$
Please see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Mar 11 at 21:15




$begingroup$
Please see math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Lord Shark the Unknown
Mar 11 at 21:15




1




1




$begingroup$
Bounded should not be a problem in either case. Have you tried proving that either $Ktimes L$ or $K+L$ is closed? What would the structure of such a proof look like?
$endgroup$
– Greg Martin
Mar 11 at 22:20




$begingroup$
Bounded should not be a problem in either case. Have you tried proving that either $Ktimes L$ or $K+L$ is closed? What would the structure of such a proof look like?
$endgroup$
– Greg Martin
Mar 11 at 22:20












$begingroup$
I don't know, can you help me?
$endgroup$
– breakshooter
Mar 11 at 23:51




$begingroup$
I don't know, can you help me?
$endgroup$
– breakshooter
Mar 11 at 23:51










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