intersection of all neighborhoods of a point in zariski topology.

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Consider the affine space $A^n$ with the Zariski topology,
V a variety (with the induced topology) and let $Pin V$ a point. Let B be the set of all neighborhoods of the point P in V. Is it true that $P=bigcaplimits_U_iin B U_i$?



Obviously this is not true on a general topological space (e.g consider the trivial topology where the only open sets are $emptyset$ and the whole space), so the fact that we have Zariski topology is important.



Apart from that, I' m not sure on how to proceed. Intuition tells me that this is indeed true, but given the fact that open sets here are very big in size (they 're dense, so no two open sets can have an empty intersection), I 'm not confident of the result.
Any hint would be welcome.




general-topology algebraic-geometry




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    up vote
    4
    down vote

    favorite












    Consider the affine space $A^n$ with the Zariski topology,
    V a variety (with the induced topology) and let $Pin V$ a point. Let B be the set of all neighborhoods of the point P in V. Is it true that $P=bigcaplimits_U_iin B U_i$?



    Obviously this is not true on a general topological space (e.g consider the trivial topology where the only open sets are $emptyset$ and the whole space), so the fact that we have Zariski topology is important.



    Apart from that, I' m not sure on how to proceed. Intuition tells me that this is indeed true, but given the fact that open sets here are very big in size (they 're dense, so no two open sets can have an empty intersection), I 'm not confident of the result.
    Any hint would be welcome.




    general-topology algebraic-geometry




    share|cite|improve this question






















      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      Consider the affine space $A^n$ with the Zariski topology,
      V a variety (with the induced topology) and let $Pin V$ a point. Let B be the set of all neighborhoods of the point P in V. Is it true that $P=bigcaplimits_U_iin B U_i$?



      Obviously this is not true on a general topological space (e.g consider the trivial topology where the only open sets are $emptyset$ and the whole space), so the fact that we have Zariski topology is important.



      Apart from that, I' m not sure on how to proceed. Intuition tells me that this is indeed true, but given the fact that open sets here are very big in size (they 're dense, so no two open sets can have an empty intersection), I 'm not confident of the result.
      Any hint would be welcome.




      general-topology algebraic-geometry




      share|cite|improve this question












      Consider the affine space $A^n$ with the Zariski topology,
      V a variety (with the induced topology) and let $Pin V$ a point. Let B be the set of all neighborhoods of the point P in V. Is it true that $P=bigcaplimits_U_iin B U_i$?



      Obviously this is not true on a general topological space (e.g consider the trivial topology where the only open sets are $emptyset$ and the whole space), so the fact that we have Zariski topology is important.



      Apart from that, I' m not sure on how to proceed. Intuition tells me that this is indeed true, but given the fact that open sets here are very big in size (they 're dense, so no two open sets can have an empty intersection), I 'm not confident of the result.
      Any hint would be welcome.





      general-topology algebraic-geometry





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      asked Aug 9 at 17:55









      Foivos

      35929




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          $(a_1,...,a_n)=V(X-a_1,...,X-a_n)$, this implies that $U(a_1,...,a_n)=A^n-(a_1,..,a_n)$ is open.



          $(a_1,..,a_n)=cap U(b_1,...,b_n), (b_1,..,b_n)neq (a_1,..,a_n)$.






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            Follows from the fact that $Bbb A^n$ is $T_1$ and that subspace of $T_1$ space is $T_1$.






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              2 Answers
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              2 Answers
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              up vote
              4
              down vote



              accepted










              $(a_1,...,a_n)=V(X-a_1,...,X-a_n)$, this implies that $U(a_1,...,a_n)=A^n-(a_1,..,a_n)$ is open.



              $(a_1,..,a_n)=cap U(b_1,...,b_n), (b_1,..,b_n)neq (a_1,..,a_n)$.






              share|cite|improve this answer
























                up vote
                4
                down vote



                accepted










                $(a_1,...,a_n)=V(X-a_1,...,X-a_n)$, this implies that $U(a_1,...,a_n)=A^n-(a_1,..,a_n)$ is open.



                $(a_1,..,a_n)=cap U(b_1,...,b_n), (b_1,..,b_n)neq (a_1,..,a_n)$.






                share|cite|improve this answer






















                  up vote
                  4
                  down vote



                  accepted







                  up vote
                  4
                  down vote



                  accepted






                  $(a_1,...,a_n)=V(X-a_1,...,X-a_n)$, this implies that $U(a_1,...,a_n)=A^n-(a_1,..,a_n)$ is open.



                  $(a_1,..,a_n)=cap U(b_1,...,b_n), (b_1,..,b_n)neq (a_1,..,a_n)$.






                  share|cite|improve this answer












                  $(a_1,...,a_n)=V(X-a_1,...,X-a_n)$, this implies that $U(a_1,...,a_n)=A^n-(a_1,..,a_n)$ is open.



                  $(a_1,..,a_n)=cap U(b_1,...,b_n), (b_1,..,b_n)neq (a_1,..,a_n)$.







                  share|cite|improve this answer












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                  answered Aug 9 at 18:01









                  Tsemo Aristide

                  51.5k11243




                  51.5k11243




















                      up vote
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                      Follows from the fact that $Bbb A^n$ is $T_1$ and that subspace of $T_1$ space is $T_1$.






                      share|cite|improve this answer
























                        up vote
                        6
                        down vote













                        Follows from the fact that $Bbb A^n$ is $T_1$ and that subspace of $T_1$ space is $T_1$.






                        share|cite|improve this answer






















                          up vote
                          6
                          down vote










                          up vote
                          6
                          down vote









                          Follows from the fact that $Bbb A^n$ is $T_1$ and that subspace of $T_1$ space is $T_1$.






                          share|cite|improve this answer












                          Follows from the fact that $Bbb A^n$ is $T_1$ and that subspace of $T_1$ space is $T_1$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Aug 9 at 18:00









                          Kenny Lau

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