Mixing
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
asked Aug 9 at 17:55
Foivos
35929
add a comment |Â
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
asked Aug 9 at 17:55
Foivos
35929
add a comment |Â
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
asked Aug 9 at 17:55
Foivos
35929
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
asked Aug 9 at 17:55
Foivos
35929
asked Aug 9 at 17:55
Foivos
35929
asked Aug 9 at 17:55
Foivos
35929
asked Aug 9 at 17:55
Foivos
35929
35929
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
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.
$(a_1,..,a_n)=cap U(b_1,...,b_n), (b_1,..,b_n)neq (a_1,..,a_n)$.
answered Aug 9 at 18:01
Tsemo Aristide
51.9k11244
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up vote
6
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Follows from the fact that $Bbb A^n$ is $T_1$ and that subspace of $T_1$ space is $T_1$.
answered Aug 9 at 18:00
Kenny Lau
19k2157
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
$(a_1,..,a_n)=cap U(b_1,...,b_n), (b_1,..,b_n)neq (a_1,..,a_n)$.
answered Aug 9 at 18:01
Tsemo Aristide
51.9k11244
add a comment |Â
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.
$(a_1,..,a_n)=cap U(b_1,...,b_n), (b_1,..,b_n)neq (a_1,..,a_n)$.
answered Aug 9 at 18:01
Tsemo Aristide
51.9k11244
add a comment |Â
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.
$(a_1,..,a_n)=cap U(b_1,...,b_n), (b_1,..,b_n)neq (a_1,..,a_n)$.
answered Aug 9 at 18:01
Tsemo Aristide
51.9k11244
$(a_1,...,a_n)=V(X-a_1,...,X-a_n)$, this implies that $U(a_1,...,a_n)=A^n-(a_1,..,a_n)$ is open.
$(a_1,..,a_n)=cap U(b_1,...,b_n), (b_1,..,b_n)neq (a_1,..,a_n)$.
answered Aug 9 at 18:01
Tsemo Aristide
51.9k11244
answered Aug 9 at 18:01
Tsemo Aristide
51.9k11244
answered Aug 9 at 18:01
Tsemo Aristide
51.9k11244
answered Aug 9 at 18:01
Tsemo Aristide
51.9k11244
51.9k11244
add a comment |Â
add a comment |Â
up vote
6
down vote
Follows from the fact that $Bbb A^n$ is $T_1$ and that subspace of $T_1$ space is $T_1$.
answered Aug 9 at 18:00
Kenny Lau
19k2157
add a comment |Â
up vote
6
down vote
Follows from the fact that $Bbb A^n$ is $T_1$ and that subspace of $T_1$ space is $T_1$.
answered Aug 9 at 18:00
Kenny Lau
19k2157
add a comment |Â
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 $T_1$.
answered Aug 9 at 18:00
Kenny Lau
19k2157
Follows from the fact that $Bbb A^n$ is $T_1$ and that subspace of $T_1$ space is $T_1$.
answered Aug 9 at 18:00
Kenny Lau
19k2157
answered Aug 9 at 18:00
Kenny Lau
19k2157
answered Aug 9 at 18:00
Kenny Lau
19k2157
answered Aug 9 at 18:00
Kenny Lau
19k2157
19k2157
add a comment |Â
add a comment |Â
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