I am not sure how to finish the proof of showing that the intersection of finitely many open subsets is open. The 2019 Stack Overflow Developer Survey Results Are InShow that the infinite intersection of nested non-empty closed subsets of a compact space is not emptyHelp Me Understand: Proof that Finite Intersection of Open Sets is OpenProve the existence of disjoint open subsetsWould the following proof be wrong? (About the intersections of compact subsets)Proof on Cross - Product of Open SubsetsIs this proof that intervals are connected correct?Finding a special subsequence of any Cauchy sequenceSequence of nested closed balls has not empty intersection, implies that a metric space is completeIs this a valid proof that if an element $x$, is within the closure of a set $A$ (in a metric space) then $exists (a_n) in A : a_n rightarrow x$All unbounded metric spaces are not compact. (Proof Verification)

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I am not sure how to finish the proof of showing that the intersection of finitely many open subsets is open.



The 2019 Stack Overflow Developer Survey Results Are InShow that the infinite intersection of nested non-empty closed subsets of a compact space is not emptyHelp Me Understand: Proof that Finite Intersection of Open Sets is OpenProve the existence of disjoint open subsetsWould the following proof be wrong? (About the intersections of compact subsets)Proof on Cross - Product of Open SubsetsIs this proof that intervals are connected correct?Finding a special subsequence of any Cauchy sequenceSequence of nested closed balls has not empty intersection, implies that a metric space is completeIs this a valid proof that if an element $x$, is within the closure of a set $A$ (in a metric space) then $exists (a_n) in A : a_n rightarrow x$All unbounded metric spaces are not compact. (Proof Verification)










2












$begingroup$


I am trying to prove the following:




Let $(X,d)$ be a metric space. An intersection of finitely many open subsets of $X$ is open.




The following is my attempt:



Let $x in bigcap_i=1^n U_i$ be arbitrary, where $U_i$'s are open subsets in $X$, for each $i=1,2,3,ldots,n-1,n.$



Since $x in bigcap_i=1^n U_i$, $x in U_i$ for all $i=1,2,3, ldots,n-1,n.$



Because all $U_i$'s are open, for each $i,$ there exists $epsilon_i>0$ such that $B_epsilon_i(x) subseteq U_i.$



Let $epsilon = minepsilon_1, epsilon_2, epsilon_3, ldots , epsilon_n-1, epsilon_n.$



Then $B_epsilon(x)subseteq B_epsilon_i(x)$ for all $i=1,2,3, ldots,n-1,n.$




Question:



I am not quite sure of how to finish the proof, by showing that $B_epsilon(x) subseteq bigcap_i=1^n U_i.$




My reasoning for the ending of the proof:



Intuitively, it makes sense that $B_epsilon(x) subseteq bigcap_i=1^n U_i$, since the open ball centred at $x$, with radius $epsilon$ contains all points that are only in all of the sets $B_epsilon_1(x),,B_epsilon_2(x),,B_epsilon_3(x),ldots,,B_epsilon_n-1(x)$ and $,B_epsilon_n(x).$



Would it be correct in writing
beginequation*
B_epsilon(x) = bigcap_i=1^n B_epsilon_i(x)?
endequation*



Then would it be the case that $$B_epsilon(x) =bigcap_i=1^n B_epsilon_i(x) subseteq bigcap_i=1^nU_i,$$



therefore $bigcap_i=1^nU_i$ is open?










share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    I am trying to prove the following:




    Let $(X,d)$ be a metric space. An intersection of finitely many open subsets of $X$ is open.




    The following is my attempt:



    Let $x in bigcap_i=1^n U_i$ be arbitrary, where $U_i$'s are open subsets in $X$, for each $i=1,2,3,ldots,n-1,n.$



    Since $x in bigcap_i=1^n U_i$, $x in U_i$ for all $i=1,2,3, ldots,n-1,n.$



    Because all $U_i$'s are open, for each $i,$ there exists $epsilon_i>0$ such that $B_epsilon_i(x) subseteq U_i.$



    Let $epsilon = minepsilon_1, epsilon_2, epsilon_3, ldots , epsilon_n-1, epsilon_n.$



    Then $B_epsilon(x)subseteq B_epsilon_i(x)$ for all $i=1,2,3, ldots,n-1,n.$




    Question:



    I am not quite sure of how to finish the proof, by showing that $B_epsilon(x) subseteq bigcap_i=1^n U_i.$




    My reasoning for the ending of the proof:



    Intuitively, it makes sense that $B_epsilon(x) subseteq bigcap_i=1^n U_i$, since the open ball centred at $x$, with radius $epsilon$ contains all points that are only in all of the sets $B_epsilon_1(x),,B_epsilon_2(x),,B_epsilon_3(x),ldots,,B_epsilon_n-1(x)$ and $,B_epsilon_n(x).$



    Would it be correct in writing
    beginequation*
    B_epsilon(x) = bigcap_i=1^n B_epsilon_i(x)?
    endequation*



    Then would it be the case that $$B_epsilon(x) =bigcap_i=1^n B_epsilon_i(x) subseteq bigcap_i=1^nU_i,$$



    therefore $bigcap_i=1^nU_i$ is open?










    share|cite|improve this question









    $endgroup$














      2












      2








      2





      $begingroup$


      I am trying to prove the following:




      Let $(X,d)$ be a metric space. An intersection of finitely many open subsets of $X$ is open.




      The following is my attempt:



      Let $x in bigcap_i=1^n U_i$ be arbitrary, where $U_i$'s are open subsets in $X$, for each $i=1,2,3,ldots,n-1,n.$



      Since $x in bigcap_i=1^n U_i$, $x in U_i$ for all $i=1,2,3, ldots,n-1,n.$



      Because all $U_i$'s are open, for each $i,$ there exists $epsilon_i>0$ such that $B_epsilon_i(x) subseteq U_i.$



      Let $epsilon = minepsilon_1, epsilon_2, epsilon_3, ldots , epsilon_n-1, epsilon_n.$



      Then $B_epsilon(x)subseteq B_epsilon_i(x)$ for all $i=1,2,3, ldots,n-1,n.$




      Question:



      I am not quite sure of how to finish the proof, by showing that $B_epsilon(x) subseteq bigcap_i=1^n U_i.$




      My reasoning for the ending of the proof:



      Intuitively, it makes sense that $B_epsilon(x) subseteq bigcap_i=1^n U_i$, since the open ball centred at $x$, with radius $epsilon$ contains all points that are only in all of the sets $B_epsilon_1(x),,B_epsilon_2(x),,B_epsilon_3(x),ldots,,B_epsilon_n-1(x)$ and $,B_epsilon_n(x).$



      Would it be correct in writing
      beginequation*
      B_epsilon(x) = bigcap_i=1^n B_epsilon_i(x)?
      endequation*



      Then would it be the case that $$B_epsilon(x) =bigcap_i=1^n B_epsilon_i(x) subseteq bigcap_i=1^nU_i,$$



      therefore $bigcap_i=1^nU_i$ is open?










      share|cite|improve this question









      $endgroup$




      I am trying to prove the following:




      Let $(X,d)$ be a metric space. An intersection of finitely many open subsets of $X$ is open.




      The following is my attempt:



      Let $x in bigcap_i=1^n U_i$ be arbitrary, where $U_i$'s are open subsets in $X$, for each $i=1,2,3,ldots,n-1,n.$



      Since $x in bigcap_i=1^n U_i$, $x in U_i$ for all $i=1,2,3, ldots,n-1,n.$



      Because all $U_i$'s are open, for each $i,$ there exists $epsilon_i>0$ such that $B_epsilon_i(x) subseteq U_i.$



      Let $epsilon = minepsilon_1, epsilon_2, epsilon_3, ldots , epsilon_n-1, epsilon_n.$



      Then $B_epsilon(x)subseteq B_epsilon_i(x)$ for all $i=1,2,3, ldots,n-1,n.$




      Question:



      I am not quite sure of how to finish the proof, by showing that $B_epsilon(x) subseteq bigcap_i=1^n U_i.$




      My reasoning for the ending of the proof:



      Intuitively, it makes sense that $B_epsilon(x) subseteq bigcap_i=1^n U_i$, since the open ball centred at $x$, with radius $epsilon$ contains all points that are only in all of the sets $B_epsilon_1(x),,B_epsilon_2(x),,B_epsilon_3(x),ldots,,B_epsilon_n-1(x)$ and $,B_epsilon_n(x).$



      Would it be correct in writing
      beginequation*
      B_epsilon(x) = bigcap_i=1^n B_epsilon_i(x)?
      endequation*



      Then would it be the case that $$B_epsilon(x) =bigcap_i=1^n B_epsilon_i(x) subseteq bigcap_i=1^nU_i,$$



      therefore $bigcap_i=1^nU_i$ is open?







      general-topology elementary-set-theory metric-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 13 hours ago









      GurjinderGurjinder

      594518




      594518




















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          The proof consists in exhibiting an open ball that is included in the intersection (around every point).



          For every set you know that you can find a suitable ball. Then the ball with minimum radius is the intersection of all these balls.



          Hence the minimal ball is included in every ball, hence in every set, thus in their intersection.




          The proof will not work with infinitely many sets, because then the minimum radius may fail to be positive.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            So what I have wrote is correct?
            $endgroup$
            – Gurjinder
            12 hours ago










          • $begingroup$
            @Gurjinder: is my answer unclear ?
            $endgroup$
            – Yves Daoust
            12 hours ago










          • $begingroup$
            Nah, it seems good. Just a question about the statement ‘Then the ball with minimum radius is the intersection of all these balls.’ This means $B_epsilon(x) = bigcap_i=1^n B_epsilon_i(x)$, as I have wrote? And as a side not, could I not just write $B_epsilon(x) subseteq bigcap_i=1^n B_epsilon_i(x)$ and conclude the proof the same way? Or when writing a mathematical proof, should you ‘precise’ as possible about a statement. For example, I don’t think it would be wrong to write $3 leq 3$, but would it be better to write $3 =3$?
            $endgroup$
            – Gurjinder
            12 hours ago











          • $begingroup$
            @Gurjinder: $3le3$ is a true statement.
            $endgroup$
            – Yves Daoust
            12 hours ago










          • $begingroup$
            Yeah, okay cool. Thanks for your assistance.
            $endgroup$
            – Gurjinder
            12 hours ago


















          1












          $begingroup$

          $B_epsilon(x)subseteq B_epsilon_i(x) $$ $ for all $i=1,2,3, ldots,n-1,n.$
          $B_epsilon_i(x) subseteq U_i $ for all $i=1,2,3, ldots,n-1,n.$
          $B_epsilon(x)subseteq U_i $ for all $i=1,2,3, ldots,n-1,n.$






          share|cite|improve this answer









          $endgroup$




















            0












            $begingroup$

            For the last portion, the only weakness is to prove that all the points in a ball is a subset of the points of a larger radius, when the center is the same point $p$. And also, the intersection of a finite number of such balls is the smallest one.



            Though this step can be considered trivial, we can't deduce the intersection of the balls by their openness, since it is yet to be proved.



            For two balls, if that doesn't hold, there exits a point $ain B_epsilon_1(p)$, but $a notin B_epsilon_2(p)$, where $epsilon_2 > epsilon_1$. We have $epsilon_1 > |a-p| ge epsilon_2$, which contradicts the assumption. Then it can be applied to finitely many.



            It seems the manipulation of open balls is more fundamental than openness or closeness.






            share|cite|improve this answer











            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              The proof consists in exhibiting an open ball that is included in the intersection (around every point).



              For every set you know that you can find a suitable ball. Then the ball with minimum radius is the intersection of all these balls.



              Hence the minimal ball is included in every ball, hence in every set, thus in their intersection.




              The proof will not work with infinitely many sets, because then the minimum radius may fail to be positive.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                So what I have wrote is correct?
                $endgroup$
                – Gurjinder
                12 hours ago










              • $begingroup$
                @Gurjinder: is my answer unclear ?
                $endgroup$
                – Yves Daoust
                12 hours ago










              • $begingroup$
                Nah, it seems good. Just a question about the statement ‘Then the ball with minimum radius is the intersection of all these balls.’ This means $B_epsilon(x) = bigcap_i=1^n B_epsilon_i(x)$, as I have wrote? And as a side not, could I not just write $B_epsilon(x) subseteq bigcap_i=1^n B_epsilon_i(x)$ and conclude the proof the same way? Or when writing a mathematical proof, should you ‘precise’ as possible about a statement. For example, I don’t think it would be wrong to write $3 leq 3$, but would it be better to write $3 =3$?
                $endgroup$
                – Gurjinder
                12 hours ago











              • $begingroup$
                @Gurjinder: $3le3$ is a true statement.
                $endgroup$
                – Yves Daoust
                12 hours ago










              • $begingroup$
                Yeah, okay cool. Thanks for your assistance.
                $endgroup$
                – Gurjinder
                12 hours ago















              3












              $begingroup$

              The proof consists in exhibiting an open ball that is included in the intersection (around every point).



              For every set you know that you can find a suitable ball. Then the ball with minimum radius is the intersection of all these balls.



              Hence the minimal ball is included in every ball, hence in every set, thus in their intersection.




              The proof will not work with infinitely many sets, because then the minimum radius may fail to be positive.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                So what I have wrote is correct?
                $endgroup$
                – Gurjinder
                12 hours ago










              • $begingroup$
                @Gurjinder: is my answer unclear ?
                $endgroup$
                – Yves Daoust
                12 hours ago










              • $begingroup$
                Nah, it seems good. Just a question about the statement ‘Then the ball with minimum radius is the intersection of all these balls.’ This means $B_epsilon(x) = bigcap_i=1^n B_epsilon_i(x)$, as I have wrote? And as a side not, could I not just write $B_epsilon(x) subseteq bigcap_i=1^n B_epsilon_i(x)$ and conclude the proof the same way? Or when writing a mathematical proof, should you ‘precise’ as possible about a statement. For example, I don’t think it would be wrong to write $3 leq 3$, but would it be better to write $3 =3$?
                $endgroup$
                – Gurjinder
                12 hours ago











              • $begingroup$
                @Gurjinder: $3le3$ is a true statement.
                $endgroup$
                – Yves Daoust
                12 hours ago










              • $begingroup$
                Yeah, okay cool. Thanks for your assistance.
                $endgroup$
                – Gurjinder
                12 hours ago













              3












              3








              3





              $begingroup$

              The proof consists in exhibiting an open ball that is included in the intersection (around every point).



              For every set you know that you can find a suitable ball. Then the ball with minimum radius is the intersection of all these balls.



              Hence the minimal ball is included in every ball, hence in every set, thus in their intersection.




              The proof will not work with infinitely many sets, because then the minimum radius may fail to be positive.






              share|cite|improve this answer









              $endgroup$



              The proof consists in exhibiting an open ball that is included in the intersection (around every point).



              For every set you know that you can find a suitable ball. Then the ball with minimum radius is the intersection of all these balls.



              Hence the minimal ball is included in every ball, hence in every set, thus in their intersection.




              The proof will not work with infinitely many sets, because then the minimum radius may fail to be positive.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 12 hours ago









              Yves DaoustYves Daoust

              133k676231




              133k676231











              • $begingroup$
                So what I have wrote is correct?
                $endgroup$
                – Gurjinder
                12 hours ago










              • $begingroup$
                @Gurjinder: is my answer unclear ?
                $endgroup$
                – Yves Daoust
                12 hours ago










              • $begingroup$
                Nah, it seems good. Just a question about the statement ‘Then the ball with minimum radius is the intersection of all these balls.’ This means $B_epsilon(x) = bigcap_i=1^n B_epsilon_i(x)$, as I have wrote? And as a side not, could I not just write $B_epsilon(x) subseteq bigcap_i=1^n B_epsilon_i(x)$ and conclude the proof the same way? Or when writing a mathematical proof, should you ‘precise’ as possible about a statement. For example, I don’t think it would be wrong to write $3 leq 3$, but would it be better to write $3 =3$?
                $endgroup$
                – Gurjinder
                12 hours ago











              • $begingroup$
                @Gurjinder: $3le3$ is a true statement.
                $endgroup$
                – Yves Daoust
                12 hours ago










              • $begingroup$
                Yeah, okay cool. Thanks for your assistance.
                $endgroup$
                – Gurjinder
                12 hours ago
















              • $begingroup$
                So what I have wrote is correct?
                $endgroup$
                – Gurjinder
                12 hours ago










              • $begingroup$
                @Gurjinder: is my answer unclear ?
                $endgroup$
                – Yves Daoust
                12 hours ago










              • $begingroup$
                Nah, it seems good. Just a question about the statement ‘Then the ball with minimum radius is the intersection of all these balls.’ This means $B_epsilon(x) = bigcap_i=1^n B_epsilon_i(x)$, as I have wrote? And as a side not, could I not just write $B_epsilon(x) subseteq bigcap_i=1^n B_epsilon_i(x)$ and conclude the proof the same way? Or when writing a mathematical proof, should you ‘precise’ as possible about a statement. For example, I don’t think it would be wrong to write $3 leq 3$, but would it be better to write $3 =3$?
                $endgroup$
                – Gurjinder
                12 hours ago











              • $begingroup$
                @Gurjinder: $3le3$ is a true statement.
                $endgroup$
                – Yves Daoust
                12 hours ago










              • $begingroup$
                Yeah, okay cool. Thanks for your assistance.
                $endgroup$
                – Gurjinder
                12 hours ago















              $begingroup$
              So what I have wrote is correct?
              $endgroup$
              – Gurjinder
              12 hours ago




              $begingroup$
              So what I have wrote is correct?
              $endgroup$
              – Gurjinder
              12 hours ago












              $begingroup$
              @Gurjinder: is my answer unclear ?
              $endgroup$
              – Yves Daoust
              12 hours ago




              $begingroup$
              @Gurjinder: is my answer unclear ?
              $endgroup$
              – Yves Daoust
              12 hours ago












              $begingroup$
              Nah, it seems good. Just a question about the statement ‘Then the ball with minimum radius is the intersection of all these balls.’ This means $B_epsilon(x) = bigcap_i=1^n B_epsilon_i(x)$, as I have wrote? And as a side not, could I not just write $B_epsilon(x) subseteq bigcap_i=1^n B_epsilon_i(x)$ and conclude the proof the same way? Or when writing a mathematical proof, should you ‘precise’ as possible about a statement. For example, I don’t think it would be wrong to write $3 leq 3$, but would it be better to write $3 =3$?
              $endgroup$
              – Gurjinder
              12 hours ago





              $begingroup$
              Nah, it seems good. Just a question about the statement ‘Then the ball with minimum radius is the intersection of all these balls.’ This means $B_epsilon(x) = bigcap_i=1^n B_epsilon_i(x)$, as I have wrote? And as a side not, could I not just write $B_epsilon(x) subseteq bigcap_i=1^n B_epsilon_i(x)$ and conclude the proof the same way? Or when writing a mathematical proof, should you ‘precise’ as possible about a statement. For example, I don’t think it would be wrong to write $3 leq 3$, but would it be better to write $3 =3$?
              $endgroup$
              – Gurjinder
              12 hours ago













              $begingroup$
              @Gurjinder: $3le3$ is a true statement.
              $endgroup$
              – Yves Daoust
              12 hours ago




              $begingroup$
              @Gurjinder: $3le3$ is a true statement.
              $endgroup$
              – Yves Daoust
              12 hours ago












              $begingroup$
              Yeah, okay cool. Thanks for your assistance.
              $endgroup$
              – Gurjinder
              12 hours ago




              $begingroup$
              Yeah, okay cool. Thanks for your assistance.
              $endgroup$
              – Gurjinder
              12 hours ago











              1












              $begingroup$

              $B_epsilon(x)subseteq B_epsilon_i(x) $$ $ for all $i=1,2,3, ldots,n-1,n.$
              $B_epsilon_i(x) subseteq U_i $ for all $i=1,2,3, ldots,n-1,n.$
              $B_epsilon(x)subseteq U_i $ for all $i=1,2,3, ldots,n-1,n.$






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                $B_epsilon(x)subseteq B_epsilon_i(x) $$ $ for all $i=1,2,3, ldots,n-1,n.$
                $B_epsilon_i(x) subseteq U_i $ for all $i=1,2,3, ldots,n-1,n.$
                $B_epsilon(x)subseteq U_i $ for all $i=1,2,3, ldots,n-1,n.$






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  $B_epsilon(x)subseteq B_epsilon_i(x) $$ $ for all $i=1,2,3, ldots,n-1,n.$
                  $B_epsilon_i(x) subseteq U_i $ for all $i=1,2,3, ldots,n-1,n.$
                  $B_epsilon(x)subseteq U_i $ for all $i=1,2,3, ldots,n-1,n.$






                  share|cite|improve this answer









                  $endgroup$



                  $B_epsilon(x)subseteq B_epsilon_i(x) $$ $ for all $i=1,2,3, ldots,n-1,n.$
                  $B_epsilon_i(x) subseteq U_i $ for all $i=1,2,3, ldots,n-1,n.$
                  $B_epsilon(x)subseteq U_i $ for all $i=1,2,3, ldots,n-1,n.$







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                  answered 13 hours ago









                  William ElliotWilliam Elliot

                  9,1562820




                  9,1562820





















                      0












                      $begingroup$

                      For the last portion, the only weakness is to prove that all the points in a ball is a subset of the points of a larger radius, when the center is the same point $p$. And also, the intersection of a finite number of such balls is the smallest one.



                      Though this step can be considered trivial, we can't deduce the intersection of the balls by their openness, since it is yet to be proved.



                      For two balls, if that doesn't hold, there exits a point $ain B_epsilon_1(p)$, but $a notin B_epsilon_2(p)$, where $epsilon_2 > epsilon_1$. We have $epsilon_1 > |a-p| ge epsilon_2$, which contradicts the assumption. Then it can be applied to finitely many.



                      It seems the manipulation of open balls is more fundamental than openness or closeness.






                      share|cite|improve this answer











                      $endgroup$

















                        0












                        $begingroup$

                        For the last portion, the only weakness is to prove that all the points in a ball is a subset of the points of a larger radius, when the center is the same point $p$. And also, the intersection of a finite number of such balls is the smallest one.



                        Though this step can be considered trivial, we can't deduce the intersection of the balls by their openness, since it is yet to be proved.



                        For two balls, if that doesn't hold, there exits a point $ain B_epsilon_1(p)$, but $a notin B_epsilon_2(p)$, where $epsilon_2 > epsilon_1$. We have $epsilon_1 > |a-p| ge epsilon_2$, which contradicts the assumption. Then it can be applied to finitely many.



                        It seems the manipulation of open balls is more fundamental than openness or closeness.






                        share|cite|improve this answer











                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          For the last portion, the only weakness is to prove that all the points in a ball is a subset of the points of a larger radius, when the center is the same point $p$. And also, the intersection of a finite number of such balls is the smallest one.



                          Though this step can be considered trivial, we can't deduce the intersection of the balls by their openness, since it is yet to be proved.



                          For two balls, if that doesn't hold, there exits a point $ain B_epsilon_1(p)$, but $a notin B_epsilon_2(p)$, where $epsilon_2 > epsilon_1$. We have $epsilon_1 > |a-p| ge epsilon_2$, which contradicts the assumption. Then it can be applied to finitely many.



                          It seems the manipulation of open balls is more fundamental than openness or closeness.






                          share|cite|improve this answer











                          $endgroup$



                          For the last portion, the only weakness is to prove that all the points in a ball is a subset of the points of a larger radius, when the center is the same point $p$. And also, the intersection of a finite number of such balls is the smallest one.



                          Though this step can be considered trivial, we can't deduce the intersection of the balls by their openness, since it is yet to be proved.



                          For two balls, if that doesn't hold, there exits a point $ain B_epsilon_1(p)$, but $a notin B_epsilon_2(p)$, where $epsilon_2 > epsilon_1$. We have $epsilon_1 > |a-p| ge epsilon_2$, which contradicts the assumption. Then it can be applied to finitely many.



                          It seems the manipulation of open balls is more fundamental than openness or closeness.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 11 hours ago

























                          answered 12 hours ago









                          Wenkuei P'eiWenkuei P'ei

                          216




                          216



























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