Let (X, d) be a metric space and let A be a subset of X and O be an open subset of X. Prove that $O cap...












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  • The intersection of a closure with an open set

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I don't know how to use the fact that O is open. Its interior must be itself, so I guess I should take the complement of the sets in question. But I'm not sure how to proceed.






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    This question already has an answer here:




    • The intersection of a closure with an open set

      3 answers




    I don't know how to use the fact that O is open. Its interior must be itself, so I guess I should take the complement of the sets in question. But I'm not sure how to proceed.






    real-analysis







    share|cite|improve this question













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      This question already has an answer here:




      • The intersection of a closure with an open set

        3 answers




      I don't know how to use the fact that O is open. Its interior must be itself, so I guess I should take the complement of the sets in question. But I'm not sure how to proceed.






      real-analysis







      share|cite|improve this question














      This question already has an answer here:




      • The intersection of a closure with an open set

        3 answers




      I don't know how to use the fact that O is open. Its interior must be itself, so I guess I should take the complement of the sets in question. But I'm not sure how to proceed.





      This question already has an answer here:




      • The intersection of a closure with an open set

        3 answers







      real-analysis




      real-analysis






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      asked Nov 26 at 9:58









      davidh

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          Suppose $x in O cap overline{A}$. This means that $x in O$ and $x in overline{A}$. Suppose $B$ is any open ball containing $x$, then $O cap B$ is also open (this is where we use $O$ is open) and contains $x$, and as $x in overline{A}$, we know that $(B cap O) cap A neq emptyset$. But this means that $B cap (O cap A) neq emptyset$ as well and as $B$ was arbitrary, $x in overline{O cap A}$, as required to prove the inclusion.






          share|cite|improve this answer























          • I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
            – davidh
            Nov 26 at 10:07










          • couldn't x be in closure of A but not in A?
            – davidh
            Nov 26 at 10:07










          • $x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
            – Henno Brandsma
            Nov 26 at 10:09










          • I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
            – davidh
            Nov 26 at 10:13












          • @davidh yes we use both directions.
            – Henno Brandsma
            Nov 26 at 10:14


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          Suppose $x in O cap overline{A}$. This means that $x in O$ and $x in overline{A}$. Suppose $B$ is any open ball containing $x$, then $O cap B$ is also open (this is where we use $O$ is open) and contains $x$, and as $x in overline{A}$, we know that $(B cap O) cap A neq emptyset$. But this means that $B cap (O cap A) neq emptyset$ as well and as $B$ was arbitrary, $x in overline{O cap A}$, as required to prove the inclusion.






          share|cite|improve this answer























          • I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
            – davidh
            Nov 26 at 10:07










          • couldn't x be in closure of A but not in A?
            – davidh
            Nov 26 at 10:07










          • $x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
            – Henno Brandsma
            Nov 26 at 10:09










          • I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
            – davidh
            Nov 26 at 10:13












          • @davidh yes we use both directions.
            – Henno Brandsma
            Nov 26 at 10:14
















          1














          Suppose $x in O cap overline{A}$. This means that $x in O$ and $x in overline{A}$. Suppose $B$ is any open ball containing $x$, then $O cap B$ is also open (this is where we use $O$ is open) and contains $x$, and as $x in overline{A}$, we know that $(B cap O) cap A neq emptyset$. But this means that $B cap (O cap A) neq emptyset$ as well and as $B$ was arbitrary, $x in overline{O cap A}$, as required to prove the inclusion.






          share|cite|improve this answer























          • I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
            – davidh
            Nov 26 at 10:07










          • couldn't x be in closure of A but not in A?
            – davidh
            Nov 26 at 10:07










          • $x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
            – Henno Brandsma
            Nov 26 at 10:09










          • I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
            – davidh
            Nov 26 at 10:13












          • @davidh yes we use both directions.
            – Henno Brandsma
            Nov 26 at 10:14














          1












          1








          1






          Suppose $x in O cap overline{A}$. This means that $x in O$ and $x in overline{A}$. Suppose $B$ is any open ball containing $x$, then $O cap B$ is also open (this is where we use $O$ is open) and contains $x$, and as $x in overline{A}$, we know that $(B cap O) cap A neq emptyset$. But this means that $B cap (O cap A) neq emptyset$ as well and as $B$ was arbitrary, $x in overline{O cap A}$, as required to prove the inclusion.






          share|cite|improve this answer














          Suppose $x in O cap overline{A}$. This means that $x in O$ and $x in overline{A}$. Suppose $B$ is any open ball containing $x$, then $O cap B$ is also open (this is where we use $O$ is open) and contains $x$, and as $x in overline{A}$, we know that $(B cap O) cap A neq emptyset$. But this means that $B cap (O cap A) neq emptyset$ as well and as $B$ was arbitrary, $x in overline{O cap A}$, as required to prove the inclusion.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 26 at 10:18

























          answered Nov 26 at 10:03









          Henno Brandsma

          104k346113




          104k346113












          • I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
            – davidh
            Nov 26 at 10:07










          • couldn't x be in closure of A but not in A?
            – davidh
            Nov 26 at 10:07










          • $x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
            – Henno Brandsma
            Nov 26 at 10:09










          • I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
            – davidh
            Nov 26 at 10:13












          • @davidh yes we use both directions.
            – Henno Brandsma
            Nov 26 at 10:14


















          • I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
            – davidh
            Nov 26 at 10:07










          • couldn't x be in closure of A but not in A?
            – davidh
            Nov 26 at 10:07










          • $x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
            – Henno Brandsma
            Nov 26 at 10:09










          • I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
            – davidh
            Nov 26 at 10:13












          • @davidh yes we use both directions.
            – Henno Brandsma
            Nov 26 at 10:14
















          I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
          – davidh
          Nov 26 at 10:07




          I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
          – davidh
          Nov 26 at 10:07












          couldn't x be in closure of A but not in A?
          – davidh
          Nov 26 at 10:07




          couldn't x be in closure of A but not in A?
          – davidh
          Nov 26 at 10:07












          $x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
          – Henno Brandsma
          Nov 26 at 10:09




          $x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
          – Henno Brandsma
          Nov 26 at 10:09












          I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
          – davidh
          Nov 26 at 10:13






          I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
          – davidh
          Nov 26 at 10:13














          @davidh yes we use both directions.
          – Henno Brandsma
          Nov 26 at 10:14




          @davidh yes we use both directions.
          – Henno Brandsma
          Nov 26 at 10:14



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