Proving the open disc is not compact












0















Prove that the open disc $D={(x,y):x^2+y^2<1}$ considered as a subspace of $mathbb{R}^2$.




I can find the following covering of the disc $B_{1-frac{1}{n}}(0,0):ninmathbb{N}$.



$Dsubseteqbigcup_limits{i=1}^{infty}B_{1-frac{1}{n}}(0,0)$.



This covering is infinite and there is no finite sub covering then $D$ is not compact.



Question:



Is this right? If not. What should I do to prove $D$ is not compact?



Thanks in advance!






real-analysis general-topology







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  • Your proof is correct.
    – drhab
    Nov 28 '18 at 16:07










  • Take this note to your hand. A disc is in that case not a subspace but a subset
    – Fakemistake
    Nov 28 '18 at 21:12










  • @Fakemistake It is not me that said that but the book.
    – Pedro Gomes
    Nov 28 '18 at 21:31










  • @Fakemistake The information that $D$ is a subspace of $mathbb{R}^2$ is essential because it tells us which topology on the set $D$ is considered.
    – Paul Frost
    Dec 1 '18 at 21:23
















0















Prove that the open disc $D={(x,y):x^2+y^2<1}$ considered as a subspace of $mathbb{R}^2$.




I can find the following covering of the disc $B_{1-frac{1}{n}}(0,0):ninmathbb{N}$.



$Dsubseteqbigcup_limits{i=1}^{infty}B_{1-frac{1}{n}}(0,0)$.



This covering is infinite and there is no finite sub covering then $D$ is not compact.



Question:



Is this right? If not. What should I do to prove $D$ is not compact?



Thanks in advance!






real-analysis general-topology







share|cite|improve this question






















  • Your proof is correct.
    – drhab
    Nov 28 '18 at 16:07










  • Take this note to your hand. A disc is in that case not a subspace but a subset
    – Fakemistake
    Nov 28 '18 at 21:12










  • @Fakemistake It is not me that said that but the book.
    – Pedro Gomes
    Nov 28 '18 at 21:31










  • @Fakemistake The information that $D$ is a subspace of $mathbb{R}^2$ is essential because it tells us which topology on the set $D$ is considered.
    – Paul Frost
    Dec 1 '18 at 21:23














0












0








0








Prove that the open disc $D={(x,y):x^2+y^2<1}$ considered as a subspace of $mathbb{R}^2$.




I can find the following covering of the disc $B_{1-frac{1}{n}}(0,0):ninmathbb{N}$.



$Dsubseteqbigcup_limits{i=1}^{infty}B_{1-frac{1}{n}}(0,0)$.



This covering is infinite and there is no finite sub covering then $D$ is not compact.



Question:



Is this right? If not. What should I do to prove $D$ is not compact?



Thanks in advance!






real-analysis general-topology







share|cite|improve this question














Prove that the open disc $D={(x,y):x^2+y^2<1}$ considered as a subspace of $mathbb{R}^2$.




I can find the following covering of the disc $B_{1-frac{1}{n}}(0,0):ninmathbb{N}$.



$Dsubseteqbigcup_limits{i=1}^{infty}B_{1-frac{1}{n}}(0,0)$.



This covering is infinite and there is no finite sub covering then $D$ is not compact.



Question:



Is this right? If not. What should I do to prove $D$ is not compact?



Thanks in advance!






real-analysis general-topology




real-analysis general-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 28 '18 at 16:04









Pedro Gomes

1,7262720




1,7262720












  • Your proof is correct.
    – drhab
    Nov 28 '18 at 16:07










  • Take this note to your hand. A disc is in that case not a subspace but a subset
    – Fakemistake
    Nov 28 '18 at 21:12










  • @Fakemistake It is not me that said that but the book.
    – Pedro Gomes
    Nov 28 '18 at 21:31










  • @Fakemistake The information that $D$ is a subspace of $mathbb{R}^2$ is essential because it tells us which topology on the set $D$ is considered.
    – Paul Frost
    Dec 1 '18 at 21:23


















  • Your proof is correct.
    – drhab
    Nov 28 '18 at 16:07










  • Take this note to your hand. A disc is in that case not a subspace but a subset
    – Fakemistake
    Nov 28 '18 at 21:12










  • @Fakemistake It is not me that said that but the book.
    – Pedro Gomes
    Nov 28 '18 at 21:31










  • @Fakemistake The information that $D$ is a subspace of $mathbb{R}^2$ is essential because it tells us which topology on the set $D$ is considered.
    – Paul Frost
    Dec 1 '18 at 21:23
















Your proof is correct.
– drhab
Nov 28 '18 at 16:07




Your proof is correct.
– drhab
Nov 28 '18 at 16:07












Take this note to your hand. A disc is in that case not a subspace but a subset
– Fakemistake
Nov 28 '18 at 21:12




Take this note to your hand. A disc is in that case not a subspace but a subset
– Fakemistake
Nov 28 '18 at 21:12












@Fakemistake It is not me that said that but the book.
– Pedro Gomes
Nov 28 '18 at 21:31




@Fakemistake It is not me that said that but the book.
– Pedro Gomes
Nov 28 '18 at 21:31












@Fakemistake The information that $D$ is a subspace of $mathbb{R}^2$ is essential because it tells us which topology on the set $D$ is considered.
– Paul Frost
Dec 1 '18 at 21:23




@Fakemistake The information that $D$ is a subspace of $mathbb{R}^2$ is essential because it tells us which topology on the set $D$ is considered.
– Paul Frost
Dec 1 '18 at 21:23










3 Answers
3






active

oldest

votes


















1














You are right. But you have to show, that each collection of finitely many of your discs do not cover the set $D $.






share|cite|improve this answer





















  • I am going to complete the proof in the present comment for you to check if this is right.
    – Pedro Gomes
    Nov 28 '18 at 16:14










  • If there was a finite subcover then $exists minmathbb{N}$ such that $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)supseteq D$ However if $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)=B_{1-frac{1}{m}}(0,0)$ If $xin B_{1-frac{1}{m}}(0,0)$ then $d(x,0)<1-frac{1}{m}<1$ which implies $Dsupseteq B_{1-frac{1}{m}}(0,0)$
    – Pedro Gomes
    Nov 28 '18 at 16:20










  • Now your proof is fine!
    – Fred
    Nov 28 '18 at 17:19










  • I disagree that this proof is fine. It should be shown that $Dnotsubset B_{1-frac{1}{m}}(0,0)$ but you showed that $B_{1-frac{1}{m}}(0,0)subseteq D$ which is not the same.
    – FWE
    Nov 29 '18 at 12:23





















2














In $mathbb{R}^2$, a compact set is closed and bounded...



Here, the set $D$ is not closed.






share|cite|improve this answer





























    1














    You should show that $$Dnotsubset B_{1-frac{1}{m}}(0,0)$$



    Therfore you should show that there is some $(x,y)in D$ where $(x,y)notin B_{1-frac{1}{m}}(0,0)$.



    For that you might just choose $(x,y)=(1-frac{1}{m},0)$.




    alternatively



    Use the theorem of Heine and Borel (sometimes referred to as Borel and Lebesgue) which says that in Euclidean space $mathbb{R}^n$ the compact subsets are identical to the ones that are closed and bounded.






    share|cite|improve this answer























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      You are right. But you have to show, that each collection of finitely many of your discs do not cover the set $D $.






      share|cite|improve this answer





















      • I am going to complete the proof in the present comment for you to check if this is right.
        – Pedro Gomes
        Nov 28 '18 at 16:14










      • If there was a finite subcover then $exists minmathbb{N}$ such that $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)supseteq D$ However if $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)=B_{1-frac{1}{m}}(0,0)$ If $xin B_{1-frac{1}{m}}(0,0)$ then $d(x,0)<1-frac{1}{m}<1$ which implies $Dsupseteq B_{1-frac{1}{m}}(0,0)$
        – Pedro Gomes
        Nov 28 '18 at 16:20










      • Now your proof is fine!
        – Fred
        Nov 28 '18 at 17:19










      • I disagree that this proof is fine. It should be shown that $Dnotsubset B_{1-frac{1}{m}}(0,0)$ but you showed that $B_{1-frac{1}{m}}(0,0)subseteq D$ which is not the same.
        – FWE
        Nov 29 '18 at 12:23


















      1














      You are right. But you have to show, that each collection of finitely many of your discs do not cover the set $D $.






      share|cite|improve this answer





















      • I am going to complete the proof in the present comment for you to check if this is right.
        – Pedro Gomes
        Nov 28 '18 at 16:14










      • If there was a finite subcover then $exists minmathbb{N}$ such that $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)supseteq D$ However if $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)=B_{1-frac{1}{m}}(0,0)$ If $xin B_{1-frac{1}{m}}(0,0)$ then $d(x,0)<1-frac{1}{m}<1$ which implies $Dsupseteq B_{1-frac{1}{m}}(0,0)$
        – Pedro Gomes
        Nov 28 '18 at 16:20










      • Now your proof is fine!
        – Fred
        Nov 28 '18 at 17:19










      • I disagree that this proof is fine. It should be shown that $Dnotsubset B_{1-frac{1}{m}}(0,0)$ but you showed that $B_{1-frac{1}{m}}(0,0)subseteq D$ which is not the same.
        – FWE
        Nov 29 '18 at 12:23
















      1












      1








      1






      You are right. But you have to show, that each collection of finitely many of your discs do not cover the set $D $.






      share|cite|improve this answer












      You are right. But you have to show, that each collection of finitely many of your discs do not cover the set $D $.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 28 '18 at 16:08









      Fred

      44.2k1845




      44.2k1845












      • I am going to complete the proof in the present comment for you to check if this is right.
        – Pedro Gomes
        Nov 28 '18 at 16:14










      • If there was a finite subcover then $exists minmathbb{N}$ such that $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)supseteq D$ However if $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)=B_{1-frac{1}{m}}(0,0)$ If $xin B_{1-frac{1}{m}}(0,0)$ then $d(x,0)<1-frac{1}{m}<1$ which implies $Dsupseteq B_{1-frac{1}{m}}(0,0)$
        – Pedro Gomes
        Nov 28 '18 at 16:20










      • Now your proof is fine!
        – Fred
        Nov 28 '18 at 17:19










      • I disagree that this proof is fine. It should be shown that $Dnotsubset B_{1-frac{1}{m}}(0,0)$ but you showed that $B_{1-frac{1}{m}}(0,0)subseteq D$ which is not the same.
        – FWE
        Nov 29 '18 at 12:23




















      • I am going to complete the proof in the present comment for you to check if this is right.
        – Pedro Gomes
        Nov 28 '18 at 16:14










      • If there was a finite subcover then $exists minmathbb{N}$ such that $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)supseteq D$ However if $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)=B_{1-frac{1}{m}}(0,0)$ If $xin B_{1-frac{1}{m}}(0,0)$ then $d(x,0)<1-frac{1}{m}<1$ which implies $Dsupseteq B_{1-frac{1}{m}}(0,0)$
        – Pedro Gomes
        Nov 28 '18 at 16:20










      • Now your proof is fine!
        – Fred
        Nov 28 '18 at 17:19










      • I disagree that this proof is fine. It should be shown that $Dnotsubset B_{1-frac{1}{m}}(0,0)$ but you showed that $B_{1-frac{1}{m}}(0,0)subseteq D$ which is not the same.
        – FWE
        Nov 29 '18 at 12:23


















      I am going to complete the proof in the present comment for you to check if this is right.
      – Pedro Gomes
      Nov 28 '18 at 16:14




      I am going to complete the proof in the present comment for you to check if this is right.
      – Pedro Gomes
      Nov 28 '18 at 16:14












      If there was a finite subcover then $exists minmathbb{N}$ such that $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)supseteq D$ However if $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)=B_{1-frac{1}{m}}(0,0)$ If $xin B_{1-frac{1}{m}}(0,0)$ then $d(x,0)<1-frac{1}{m}<1$ which implies $Dsupseteq B_{1-frac{1}{m}}(0,0)$
      – Pedro Gomes
      Nov 28 '18 at 16:20




      If there was a finite subcover then $exists minmathbb{N}$ such that $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)supseteq D$ However if $bigcup_{i=1}^{m}B_{1-frac{1}{n}}(0,0)=B_{1-frac{1}{m}}(0,0)$ If $xin B_{1-frac{1}{m}}(0,0)$ then $d(x,0)<1-frac{1}{m}<1$ which implies $Dsupseteq B_{1-frac{1}{m}}(0,0)$
      – Pedro Gomes
      Nov 28 '18 at 16:20












      Now your proof is fine!
      – Fred
      Nov 28 '18 at 17:19




      Now your proof is fine!
      – Fred
      Nov 28 '18 at 17:19












      I disagree that this proof is fine. It should be shown that $Dnotsubset B_{1-frac{1}{m}}(0,0)$ but you showed that $B_{1-frac{1}{m}}(0,0)subseteq D$ which is not the same.
      – FWE
      Nov 29 '18 at 12:23






      I disagree that this proof is fine. It should be shown that $Dnotsubset B_{1-frac{1}{m}}(0,0)$ but you showed that $B_{1-frac{1}{m}}(0,0)subseteq D$ which is not the same.
      – FWE
      Nov 29 '18 at 12:23













      2














      In $mathbb{R}^2$, a compact set is closed and bounded...



      Here, the set $D$ is not closed.






      share|cite|improve this answer


























        2














        In $mathbb{R}^2$, a compact set is closed and bounded...



        Here, the set $D$ is not closed.






        share|cite|improve this answer
























          2












          2








          2






          In $mathbb{R}^2$, a compact set is closed and bounded...



          Here, the set $D$ is not closed.






          share|cite|improve this answer












          In $mathbb{R}^2$, a compact set is closed and bounded...



          Here, the set $D$ is not closed.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 '18 at 16:47









          Goldy

          42414




          42414























              1














              You should show that $$Dnotsubset B_{1-frac{1}{m}}(0,0)$$



              Therfore you should show that there is some $(x,y)in D$ where $(x,y)notin B_{1-frac{1}{m}}(0,0)$.



              For that you might just choose $(x,y)=(1-frac{1}{m},0)$.




              alternatively



              Use the theorem of Heine and Borel (sometimes referred to as Borel and Lebesgue) which says that in Euclidean space $mathbb{R}^n$ the compact subsets are identical to the ones that are closed and bounded.






              share|cite|improve this answer




























                1














                You should show that $$Dnotsubset B_{1-frac{1}{m}}(0,0)$$



                Therfore you should show that there is some $(x,y)in D$ where $(x,y)notin B_{1-frac{1}{m}}(0,0)$.



                For that you might just choose $(x,y)=(1-frac{1}{m},0)$.




                alternatively



                Use the theorem of Heine and Borel (sometimes referred to as Borel and Lebesgue) which says that in Euclidean space $mathbb{R}^n$ the compact subsets are identical to the ones that are closed and bounded.






                share|cite|improve this answer


























                  1












                  1








                  1






                  You should show that $$Dnotsubset B_{1-frac{1}{m}}(0,0)$$



                  Therfore you should show that there is some $(x,y)in D$ where $(x,y)notin B_{1-frac{1}{m}}(0,0)$.



                  For that you might just choose $(x,y)=(1-frac{1}{m},0)$.




                  alternatively



                  Use the theorem of Heine and Borel (sometimes referred to as Borel and Lebesgue) which says that in Euclidean space $mathbb{R}^n$ the compact subsets are identical to the ones that are closed and bounded.






                  share|cite|improve this answer














                  You should show that $$Dnotsubset B_{1-frac{1}{m}}(0,0)$$



                  Therfore you should show that there is some $(x,y)in D$ where $(x,y)notin B_{1-frac{1}{m}}(0,0)$.



                  For that you might just choose $(x,y)=(1-frac{1}{m},0)$.




                  alternatively



                  Use the theorem of Heine and Borel (sometimes referred to as Borel and Lebesgue) which says that in Euclidean space $mathbb{R}^n$ the compact subsets are identical to the ones that are closed and bounded.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 1 '18 at 19:38

























                  answered Nov 28 '18 at 16:42









                  FWE

                  1,039616




                  1,039616






























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