Is it true that the intersection of the closures of sets $A$ and $ B$ is equal to the closure of their...












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Is it true that the intersection of the closures of sets $A$ and $B$ is equal to the closure of their intersection?
$ cl(A)cap{cl(B)}=cl(Acap{B})$ ?










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closed as off-topic by Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen Dec 19 '18 at 13:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


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    $begingroup$


    Is it true that the intersection of the closures of sets $A$ and $B$ is equal to the closure of their intersection?
    $ cl(A)cap{cl(B)}=cl(Acap{B})$ ?










    share|cite|improve this question











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    closed as off-topic by Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen Dec 19 '18 at 13:27


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen

    If this question can be reworded to fit the rules in the help center, please edit the question.



















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      Is it true that the intersection of the closures of sets $A$ and $B$ is equal to the closure of their intersection?
      $ cl(A)cap{cl(B)}=cl(Acap{B})$ ?










      share|cite|improve this question











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      Is it true that the intersection of the closures of sets $A$ and $B$ is equal to the closure of their intersection?
      $ cl(A)cap{cl(B)}=cl(Acap{B})$ ?







      general-topology






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      edited Dec 19 '18 at 7:24









      dmtri

      1,7632521




      1,7632521










      asked Dec 19 '18 at 6:39









      Zhaniya ShaimakhanovaZhaniya Shaimakhanova

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      111




      closed as off-topic by Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen Dec 19 '18 at 13:27


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen Dec 19 '18 at 13:27


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          3 Answers
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          No, the rationals and the irrationals (in the reals) are disjoint so $operatorname{cl}(A cap B) = operatorname{cl}{emptyset}=emptyset$ while $operatorname{cl}(A) = operatorname{cl}(B) = mathbb{R}$






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            No. Take $A=(-1,0), B=(0,1)$. Note that $0$ is in the closure of both of these sets.






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              No, it's not true. Look at this page, or this question.



              A simple counterexample stolen from the question cited above is as follows:



              Take $A = (0,1)$ and $B = (1,2)$. Then we have
              $$operatorname{cl}(A) cap operatorname{cl}(B) = [0,1] cap [1,2] = lbrace 1 rbrace $$
              but
              $$operatorname{cl}(A cap B) = operatorname{cl}(emptyset) = emptyset$$






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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                4












                $begingroup$

                No, the rationals and the irrationals (in the reals) are disjoint so $operatorname{cl}(A cap B) = operatorname{cl}{emptyset}=emptyset$ while $operatorname{cl}(A) = operatorname{cl}(B) = mathbb{R}$






                share|cite|improve this answer









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                  4












                  $begingroup$

                  No, the rationals and the irrationals (in the reals) are disjoint so $operatorname{cl}(A cap B) = operatorname{cl}{emptyset}=emptyset$ while $operatorname{cl}(A) = operatorname{cl}(B) = mathbb{R}$






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    No, the rationals and the irrationals (in the reals) are disjoint so $operatorname{cl}(A cap B) = operatorname{cl}{emptyset}=emptyset$ while $operatorname{cl}(A) = operatorname{cl}(B) = mathbb{R}$






                    share|cite|improve this answer









                    $endgroup$



                    No, the rationals and the irrationals (in the reals) are disjoint so $operatorname{cl}(A cap B) = operatorname{cl}{emptyset}=emptyset$ while $operatorname{cl}(A) = operatorname{cl}(B) = mathbb{R}$







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                    answered Dec 19 '18 at 6:44









                    Henno BrandsmaHenno Brandsma

                    114k348124




                    114k348124























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                        No. Take $A=(-1,0), B=(0,1)$. Note that $0$ is in the closure of both of these sets.






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                          2












                          $begingroup$

                          No. Take $A=(-1,0), B=(0,1)$. Note that $0$ is in the closure of both of these sets.






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                            2












                            2








                            2





                            $begingroup$

                            No. Take $A=(-1,0), B=(0,1)$. Note that $0$ is in the closure of both of these sets.






                            share|cite|improve this answer









                            $endgroup$



                            No. Take $A=(-1,0), B=(0,1)$. Note that $0$ is in the closure of both of these sets.







                            share|cite|improve this answer












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                            share|cite|improve this answer










                            answered Dec 19 '18 at 6:41









                            Kavi Rama MurthyKavi Rama Murthy

                            71.5k53170




                            71.5k53170























                                1












                                $begingroup$

                                No, it's not true. Look at this page, or this question.



                                A simple counterexample stolen from the question cited above is as follows:



                                Take $A = (0,1)$ and $B = (1,2)$. Then we have
                                $$operatorname{cl}(A) cap operatorname{cl}(B) = [0,1] cap [1,2] = lbrace 1 rbrace $$
                                but
                                $$operatorname{cl}(A cap B) = operatorname{cl}(emptyset) = emptyset$$






                                share|cite|improve this answer











                                $endgroup$


















                                  1












                                  $begingroup$

                                  No, it's not true. Look at this page, or this question.



                                  A simple counterexample stolen from the question cited above is as follows:



                                  Take $A = (0,1)$ and $B = (1,2)$. Then we have
                                  $$operatorname{cl}(A) cap operatorname{cl}(B) = [0,1] cap [1,2] = lbrace 1 rbrace $$
                                  but
                                  $$operatorname{cl}(A cap B) = operatorname{cl}(emptyset) = emptyset$$






                                  share|cite|improve this answer











                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    No, it's not true. Look at this page, or this question.



                                    A simple counterexample stolen from the question cited above is as follows:



                                    Take $A = (0,1)$ and $B = (1,2)$. Then we have
                                    $$operatorname{cl}(A) cap operatorname{cl}(B) = [0,1] cap [1,2] = lbrace 1 rbrace $$
                                    but
                                    $$operatorname{cl}(A cap B) = operatorname{cl}(emptyset) = emptyset$$






                                    share|cite|improve this answer











                                    $endgroup$



                                    No, it's not true. Look at this page, or this question.



                                    A simple counterexample stolen from the question cited above is as follows:



                                    Take $A = (0,1)$ and $B = (1,2)$. Then we have
                                    $$operatorname{cl}(A) cap operatorname{cl}(B) = [0,1] cap [1,2] = lbrace 1 rbrace $$
                                    but
                                    $$operatorname{cl}(A cap B) = operatorname{cl}(emptyset) = emptyset$$







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Dec 20 '18 at 2:05

























                                    answered Dec 19 '18 at 6:44









                                    bubbabubba

                                    30.7k33188




                                    30.7k33188















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