Separating points in a compact Hausdorff space.
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Suppose that $X$ is a compact Hausdorff space and let ${x_1,ldots,x_n}$ be a finite collection of points in $X$. It it possible to find open neighbourhoods $U_i$ of $x_i$ such that such that $x_j notin overline{U_i}$ for all $j ne i$ and $overline{U_i} cap overline{U_j} = varnothing$ for all $i ne j$? If not what extra separation axioms are needed?
I know for example that $X$ being compact Hausdorff implies that $X$ is normal, but that doesn't seem strong enough.
general-topology separation-axioms
asked Jan 21 at 23:30
ZorngoZorngo
689412
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add a comment |
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Suppose that $X$ is a compact Hausdorff space and let ${x_1,ldots,x_n}$ be a finite collection of points in $X$. It it possible to find open neighbourhoods $U_i$ of $x_i$ such that such that $x_j notin overline{U_i}$ for all $j ne i$ and $overline{U_i} cap overline{U_j} = varnothing$ for all $i ne j$? If not what extra separation axioms are needed?
I know for example that $X$ being compact Hausdorff implies that $X$ is normal, but that doesn't seem strong enough.
general-topology separation-axioms
asked Jan 21 at 23:30
ZorngoZorngo
689412
$endgroup$
add a comment |
$begingroup$
Suppose that $X$ is a compact Hausdorff space and let ${x_1,ldots,x_n}$ be a finite collection of points in $X$. It it possible to find open neighbourhoods $U_i$ of $x_i$ such that such that $x_j notin overline{U_i}$ for all $j ne i$ and $overline{U_i} cap overline{U_j} = varnothing$ for all $i ne j$? If not what extra separation axioms are needed?
I know for example that $X$ being compact Hausdorff implies that $X$ is normal, but that doesn't seem strong enough.
general-topology separation-axioms
asked Jan 21 at 23:30
ZorngoZorngo
689412
$endgroup$
Suppose that $X$ is a compact Hausdorff space and let ${x_1,ldots,x_n}$ be a finite collection of points in $X$. It it possible to find open neighbourhoods $U_i$ of $x_i$ such that such that $x_j notin overline{U_i}$ for all $j ne i$ and $overline{U_i} cap overline{U_j} = varnothing$ for all $i ne j$? If not what extra separation axioms are needed?
I know for example that $X$ being compact Hausdorff implies that $X$ is normal, but that doesn't seem strong enough.
general-topology separation-axioms
general-topology separation-axioms
asked Jan 21 at 23:30
ZorngoZorngo
689412
asked Jan 21 at 23:30
ZorngoZorngo
689412
asked Jan 21 at 23:30
ZorngoZorngo
689412
asked Jan 21 at 23:30
ZorngoZorngo
689412
asked Jan 21 at 23:30
ZorngoZorngo
689412
689412
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3 Answers
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Hint: use the fact that compact Hausdorff spaces are normal and regular.
answered Jan 21 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
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Thanks to Kavi for the hint! It made me realise what I was missing. Here's a rough outline of the answer for completion.
Use regularity of $X$ to obtain disjoint open sets $U$ and $V$ such that $x_1 in U$ and ${x_2, ldots, x_n} subseteq V$. Then $U subseteq V^c$ so that $overline{U} subseteq V^c$. In particular, $overline{U} cap V = varnothing$. Now use normality of $X$ to obtain disjoint open neighbourhoods $W$ of $overline{U}$ and $Y$ of ${x_2, ldots, x_n}$. Then $Y subseteq W^c$ so that $overline Y subseteq W^c subseteq overline{U}^c$. Hence, $overline{Y} cap overline{U} = varnothing$. A similar process can be undertaken to separate off the other points.
answered Jan 21 at 23:52
ZorngoZorngo
689412
$endgroup$
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normality is overkill.
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– Henno Brandsma
Jan 22 at 6:38
add a comment |
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You can use Hausdorffness to separate the $x_i$ into pairwise disjoint $U_i$ such that $x_i in U_i$, $i=1,ldots,n$ (this follows by a simple induction on $n$ that this can be done in any Hausdorff space) and then regularity to find $x_i in V_i subseteq U_i$, $V_i$ open, with $overline{V_i} subseteq U_i$. No normality is needed, just Hausdorffness and regularity. And compact Hausdorff spaces are regular.
answered Jan 22 at 5:35
Henno BrandsmaHenno Brandsma
111k348118
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3 Answers
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oldest
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3 Answers
3
active
oldest
votes
active
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active
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$begingroup$
Hint: use the fact that compact Hausdorff spaces are normal and regular.
answered Jan 21 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
add a comment |
$begingroup$
Hint: use the fact that compact Hausdorff spaces are normal and regular.
answered Jan 21 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
add a comment |
$begingroup$
Hint: use the fact that compact Hausdorff spaces are normal and regular.
answered Jan 21 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
Hint: use the fact that compact Hausdorff spaces are normal and regular.
answered Jan 21 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 21 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 21 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 21 at 23:32
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
62.7k42262
add a comment |
add a comment |
$begingroup$
Thanks to Kavi for the hint! It made me realise what I was missing. Here's a rough outline of the answer for completion.
Use regularity of $X$ to obtain disjoint open sets $U$ and $V$ such that $x_1 in U$ and ${x_2, ldots, x_n} subseteq V$. Then $U subseteq V^c$ so that $overline{U} subseteq V^c$. In particular, $overline{U} cap V = varnothing$. Now use normality of $X$ to obtain disjoint open neighbourhoods $W$ of $overline{U}$ and $Y$ of ${x_2, ldots, x_n}$. Then $Y subseteq W^c$ so that $overline Y subseteq W^c subseteq overline{U}^c$. Hence, $overline{Y} cap overline{U} = varnothing$. A similar process can be undertaken to separate off the other points.
answered Jan 21 at 23:52
ZorngoZorngo
689412
$endgroup$
$begingroup$
normality is overkill.
$endgroup$
– Henno Brandsma
Jan 22 at 6:38
add a comment |
$begingroup$
Thanks to Kavi for the hint! It made me realise what I was missing. Here's a rough outline of the answer for completion.
Use regularity of $X$ to obtain disjoint open sets $U$ and $V$ such that $x_1 in U$ and ${x_2, ldots, x_n} subseteq V$. Then $U subseteq V^c$ so that $overline{U} subseteq V^c$. In particular, $overline{U} cap V = varnothing$. Now use normality of $X$ to obtain disjoint open neighbourhoods $W$ of $overline{U}$ and $Y$ of ${x_2, ldots, x_n}$. Then $Y subseteq W^c$ so that $overline Y subseteq W^c subseteq overline{U}^c$. Hence, $overline{Y} cap overline{U} = varnothing$. A similar process can be undertaken to separate off the other points.
answered Jan 21 at 23:52
ZorngoZorngo
689412
$endgroup$
$begingroup$
normality is overkill.
$endgroup$
– Henno Brandsma
Jan 22 at 6:38
add a comment |
$begingroup$
Thanks to Kavi for the hint! It made me realise what I was missing. Here's a rough outline of the answer for completion.
Use regularity of $X$ to obtain disjoint open sets $U$ and $V$ such that $x_1 in U$ and ${x_2, ldots, x_n} subseteq V$. Then $U subseteq V^c$ so that $overline{U} subseteq V^c$. In particular, $overline{U} cap V = varnothing$. Now use normality of $X$ to obtain disjoint open neighbourhoods $W$ of $overline{U}$ and $Y$ of ${x_2, ldots, x_n}$. Then $Y subseteq W^c$ so that $overline Y subseteq W^c subseteq overline{U}^c$. Hence, $overline{Y} cap overline{U} = varnothing$. A similar process can be undertaken to separate off the other points.
answered Jan 21 at 23:52
ZorngoZorngo
689412
$endgroup$
Thanks to Kavi for the hint! It made me realise what I was missing. Here's a rough outline of the answer for completion.
Use regularity of $X$ to obtain disjoint open sets $U$ and $V$ such that $x_1 in U$ and ${x_2, ldots, x_n} subseteq V$. Then $U subseteq V^c$ so that $overline{U} subseteq V^c$. In particular, $overline{U} cap V = varnothing$. Now use normality of $X$ to obtain disjoint open neighbourhoods $W$ of $overline{U}$ and $Y$ of ${x_2, ldots, x_n}$. Then $Y subseteq W^c$ so that $overline Y subseteq W^c subseteq overline{U}^c$. Hence, $overline{Y} cap overline{U} = varnothing$. A similar process can be undertaken to separate off the other points.
answered Jan 21 at 23:52
ZorngoZorngo
689412
answered Jan 21 at 23:52
ZorngoZorngo
689412
answered Jan 21 at 23:52
ZorngoZorngo
689412
answered Jan 21 at 23:52
ZorngoZorngo
689412
689412
$begingroup$
normality is overkill.
$endgroup$
– Henno Brandsma
Jan 22 at 6:38
add a comment |
$begingroup$
normality is overkill.
$endgroup$
– Henno Brandsma
Jan 22 at 6:38
$begingroup$
normality is overkill.
$endgroup$
– Henno Brandsma
Jan 22 at 6:38
$begingroup$
normality is overkill.
$endgroup$
– Henno Brandsma
Jan 22 at 6:38
add a comment |
$begingroup$
You can use Hausdorffness to separate the $x_i$ into pairwise disjoint $U_i$ such that $x_i in U_i$, $i=1,ldots,n$ (this follows by a simple induction on $n$ that this can be done in any Hausdorff space) and then regularity to find $x_i in V_i subseteq U_i$, $V_i$ open, with $overline{V_i} subseteq U_i$. No normality is needed, just Hausdorffness and regularity. And compact Hausdorff spaces are regular.
answered Jan 22 at 5:35
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
add a comment |
$begingroup$
You can use Hausdorffness to separate the $x_i$ into pairwise disjoint $U_i$ such that $x_i in U_i$, $i=1,ldots,n$ (this follows by a simple induction on $n$ that this can be done in any Hausdorff space) and then regularity to find $x_i in V_i subseteq U_i$, $V_i$ open, with $overline{V_i} subseteq U_i$. No normality is needed, just Hausdorffness and regularity. And compact Hausdorff spaces are regular.
answered Jan 22 at 5:35
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
add a comment |
$begingroup$
You can use Hausdorffness to separate the $x_i$ into pairwise disjoint $U_i$ such that $x_i in U_i$, $i=1,ldots,n$ (this follows by a simple induction on $n$ that this can be done in any Hausdorff space) and then regularity to find $x_i in V_i subseteq U_i$, $V_i$ open, with $overline{V_i} subseteq U_i$. No normality is needed, just Hausdorffness and regularity. And compact Hausdorff spaces are regular.
answered Jan 22 at 5:35
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
You can use Hausdorffness to separate the $x_i$ into pairwise disjoint $U_i$ such that $x_i in U_i$, $i=1,ldots,n$ (this follows by a simple induction on $n$ that this can be done in any Hausdorff space) and then regularity to find $x_i in V_i subseteq U_i$, $V_i$ open, with $overline{V_i} subseteq U_i$. No normality is needed, just Hausdorffness and regularity. And compact Hausdorff spaces are regular.
answered Jan 22 at 5:35
Henno BrandsmaHenno Brandsma
111k348118
edited Jan 22 at 6:40
answered Jan 22 at 5:35
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 22 at 5:35
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 22 at 5:35
Henno BrandsmaHenno Brandsma
111k348118
111k348118
add a comment |
add a comment |
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