Let $X$ be a Hausdorff and Separable topological space and $S$ a dense subset.
I am redoing a demonstration of Set-Theoretic Topology and I can not prove the following:
Let $X$ be a Hausdorff and Separable topological space and $S$ a dense subset. Prove that if $p$, $q$ $in X$ are distinct points, there exists $A subset S$ such that p belongs to the closure of A and q does not belong.
general-topology
asked Nov 22 at 22:06
Lucas
225
add a comment |
I am redoing a demonstration of Set-Theoretic Topology and I can not prove the following:
Let $X$ be a Hausdorff and Separable topological space and $S$ a dense subset. Prove that if $p$, $q$ $in X$ are distinct points, there exists $A subset S$ such that p belongs to the closure of A and q does not belong.
general-topology
asked Nov 22 at 22:06
Lucas
225
add a comment |
I am redoing a demonstration of Set-Theoretic Topology and I can not prove the following:
Let $X$ be a Hausdorff and Separable topological space and $S$ a dense subset. Prove that if $p$, $q$ $in X$ are distinct points, there exists $A subset S$ such that p belongs to the closure of A and q does not belong.
general-topology
asked Nov 22 at 22:06
Lucas
225
I am redoing a demonstration of Set-Theoretic Topology and I can not prove the following:
Let $X$ be a Hausdorff and Separable topological space and $S$ a dense subset. Prove that if $p$, $q$ $in X$ are distinct points, there exists $A subset S$ such that p belongs to the closure of A and q does not belong.
general-topology
general-topology
asked Nov 22 at 22:06
Lucas
225
asked Nov 22 at 22:06
Lucas
225
asked Nov 22 at 22:06
Lucas
225
asked Nov 22 at 22:06
Lucas
225
asked Nov 22 at 22:06
Lucas
225
225
add a comment |
add a comment |
1 Answer
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If $U$ is an open set containing $p$ and $V$ an open set containing $q$ such that $U cap V = emptyset$, then consider $A = S cap U$.
$qnotin overline{A}$ as $V$ is a neighbourhood of $q$ disjoint from $A$, while
supposing that $O$ is any open set containing $p$, so is $U cap O$ and we know that $S$ must intersect the non-empty open set $U cap O$ and so $O$ intersects $A$, showing that $p in overline{A}$.
If $X$ is separable, then we can take $S$ countable, and the map that sends $(p,q)$ to this subset shows that $|X|=|X|^2 le |mathscr{P}(S)| = mathfrak{c}$, so we get an upper bound on the szie of $X$. Note that on a $T_1$ space you can get no such bound, as the cofinite topology on any size $X$ is separable and $T_1$. So Hausdorffness is essential.
answered Nov 22 at 22:16
Henno Brandsma
104k346113
add a comment |
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1 Answer
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If $U$ is an open set containing $p$ and $V$ an open set containing $q$ such that $U cap V = emptyset$, then consider $A = S cap U$.
$qnotin overline{A}$ as $V$ is a neighbourhood of $q$ disjoint from $A$, while
supposing that $O$ is any open set containing $p$, so is $U cap O$ and we know that $S$ must intersect the non-empty open set $U cap O$ and so $O$ intersects $A$, showing that $p in overline{A}$.
If $X$ is separable, then we can take $S$ countable, and the map that sends $(p,q)$ to this subset shows that $|X|=|X|^2 le |mathscr{P}(S)| = mathfrak{c}$, so we get an upper bound on the szie of $X$. Note that on a $T_1$ space you can get no such bound, as the cofinite topology on any size $X$ is separable and $T_1$. So Hausdorffness is essential.
answered Nov 22 at 22:16
Henno Brandsma
104k346113
add a comment |
If $U$ is an open set containing $p$ and $V$ an open set containing $q$ such that $U cap V = emptyset$, then consider $A = S cap U$.
$qnotin overline{A}$ as $V$ is a neighbourhood of $q$ disjoint from $A$, while
supposing that $O$ is any open set containing $p$, so is $U cap O$ and we know that $S$ must intersect the non-empty open set $U cap O$ and so $O$ intersects $A$, showing that $p in overline{A}$.
If $X$ is separable, then we can take $S$ countable, and the map that sends $(p,q)$ to this subset shows that $|X|=|X|^2 le |mathscr{P}(S)| = mathfrak{c}$, so we get an upper bound on the szie of $X$. Note that on a $T_1$ space you can get no such bound, as the cofinite topology on any size $X$ is separable and $T_1$. So Hausdorffness is essential.
answered Nov 22 at 22:16
Henno Brandsma
104k346113
add a comment |
If $U$ is an open set containing $p$ and $V$ an open set containing $q$ such that $U cap V = emptyset$, then consider $A = S cap U$.
$qnotin overline{A}$ as $V$ is a neighbourhood of $q$ disjoint from $A$, while
supposing that $O$ is any open set containing $p$, so is $U cap O$ and we know that $S$ must intersect the non-empty open set $U cap O$ and so $O$ intersects $A$, showing that $p in overline{A}$.
If $X$ is separable, then we can take $S$ countable, and the map that sends $(p,q)$ to this subset shows that $|X|=|X|^2 le |mathscr{P}(S)| = mathfrak{c}$, so we get an upper bound on the szie of $X$. Note that on a $T_1$ space you can get no such bound, as the cofinite topology on any size $X$ is separable and $T_1$. So Hausdorffness is essential.
answered Nov 22 at 22:16
Henno Brandsma
104k346113
If $U$ is an open set containing $p$ and $V$ an open set containing $q$ such that $U cap V = emptyset$, then consider $A = S cap U$.
$qnotin overline{A}$ as $V$ is a neighbourhood of $q$ disjoint from $A$, while
supposing that $O$ is any open set containing $p$, so is $U cap O$ and we know that $S$ must intersect the non-empty open set $U cap O$ and so $O$ intersects $A$, showing that $p in overline{A}$.
If $X$ is separable, then we can take $S$ countable, and the map that sends $(p,q)$ to this subset shows that $|X|=|X|^2 le |mathscr{P}(S)| = mathfrak{c}$, so we get an upper bound on the szie of $X$. Note that on a $T_1$ space you can get no such bound, as the cofinite topology on any size $X$ is separable and $T_1$. So Hausdorffness is essential.
answered Nov 22 at 22:16
Henno Brandsma
104k346113
answered Nov 22 at 22:16
Henno Brandsma
104k346113
answered Nov 22 at 22:16
Henno Brandsma
104k346113
answered Nov 22 at 22:16
Henno Brandsma
104k346113
104k346113
add a comment |
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