Vietoris Topology
$begingroup$
let $ X $ be a topological space and $ operatorname{Exp}(X) $ is the set of all closed non-empty subsets of $X$ .
If $ U , V_{1}, V_{2}ldots V_{n}$ are the non-empty open subset in $ X$, define:
$$ langle U , V_{1}, V_{2}, ldots V_{n} rangle = { F in operatorname{Exp}(X)mid Fsubseteq U, forall 1 leq i leq n : Fcap V_{i} neq emptyset}$$
families $ B $includes all sets of the form $langle U , V_{1}, V_{2}, ldots, V_{n} rangle $ is the basis for a topology for $ Exp ( X ) $
This topology is called the Vietoris topology.
My question:
if $ X$ is a $T_{1}$ space, then is the Vietoris topology $ T_{1}$?
general-topology
edited Dec 31 '18 at 7:18
Martin Sleziak
45k10123277
asked Feb 24 '17 at 11:25
M.OM.O
613
$endgroup$
add a comment |
$begingroup$
let $ X $ be a topological space and $ operatorname{Exp}(X) $ is the set of all closed non-empty subsets of $X$ .
If $ U , V_{1}, V_{2}ldots V_{n}$ are the non-empty open subset in $ X$, define:
$$ langle U , V_{1}, V_{2}, ldots V_{n} rangle = { F in operatorname{Exp}(X)mid Fsubseteq U, forall 1 leq i leq n : Fcap V_{i} neq emptyset}$$
families $ B $includes all sets of the form $langle U , V_{1}, V_{2}, ldots, V_{n} rangle $ is the basis for a topology for $ Exp ( X ) $
This topology is called the Vietoris topology.
My question:
if $ X$ is a $T_{1}$ space, then is the Vietoris topology $ T_{1}$?
general-topology
edited Dec 31 '18 at 7:18
Martin Sleziak
45k10123277
asked Feb 24 '17 at 11:25
M.OM.O
613
$endgroup$
add a comment |
$begingroup$
let $ X $ be a topological space and $ operatorname{Exp}(X) $ is the set of all closed non-empty subsets of $X$ .
If $ U , V_{1}, V_{2}ldots V_{n}$ are the non-empty open subset in $ X$, define:
$$ langle U , V_{1}, V_{2}, ldots V_{n} rangle = { F in operatorname{Exp}(X)mid Fsubseteq U, forall 1 leq i leq n : Fcap V_{i} neq emptyset}$$
families $ B $includes all sets of the form $langle U , V_{1}, V_{2}, ldots, V_{n} rangle $ is the basis for a topology for $ Exp ( X ) $
This topology is called the Vietoris topology.
My question:
if $ X$ is a $T_{1}$ space, then is the Vietoris topology $ T_{1}$?
general-topology
edited Dec 31 '18 at 7:18
Martin Sleziak
45k10123277
asked Feb 24 '17 at 11:25
M.OM.O
613
$endgroup$
let $ X $ be a topological space and $ operatorname{Exp}(X) $ is the set of all closed non-empty subsets of $X$ .
If $ U , V_{1}, V_{2}ldots V_{n}$ are the non-empty open subset in $ X$, define:
$$ langle U , V_{1}, V_{2}, ldots V_{n} rangle = { F in operatorname{Exp}(X)mid Fsubseteq U, forall 1 leq i leq n : Fcap V_{i} neq emptyset}$$
families $ B $includes all sets of the form $langle U , V_{1}, V_{2}, ldots, V_{n} rangle $ is the basis for a topology for $ Exp ( X ) $
This topology is called the Vietoris topology.
My question:
if $ X$ is a $T_{1}$ space, then is the Vietoris topology $ T_{1}$?
general-topology
general-topology
edited Dec 31 '18 at 7:18
Martin Sleziak
45k10123277
asked Feb 24 '17 at 11:25
M.OM.O
613
edited Dec 31 '18 at 7:18
Martin Sleziak
45k10123277
asked Feb 24 '17 at 11:25
M.OM.O
613
edited Dec 31 '18 at 7:18
Martin Sleziak
45k10123277
edited Dec 31 '18 at 7:18
Martin Sleziak
45k10123277
edited Dec 31 '18 at 7:18
Martin Sleziak
45k10123277
45k10123277
asked Feb 24 '17 at 11:25
M.OM.O
613
asked Feb 24 '17 at 11:25
M.OM.O
613
asked Feb 24 '17 at 11:25
M.OM.O
613
613
add a comment |
add a comment |
1 Answer
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$begingroup$
This topology is called the Vietoris topology on $operatorname{Exp}(X)$, also denoted (as I usually do) by $H(X)$ or $2^X$, the hyperspace of $X$.
I normally use a standard subbase for this, for every non-empty open set $U subset X$,
define $[U] = {F in H(X): F cap U neq emptyset}$ and $langle U rangle = {F in H(X): F subseteq U}$. Then $langle U, V_1, ldots, V_nrangle = langle U rangle cap cap_{i=1}^n [V_i]$, so your base $B$ is the base generated by this subbase. Also note that $[U] = langle X, Urangle$, so is indeed open.
The subbase is especially handy when proving compactness of $H(X)$, using the Alexander subbase lemma, BTW.
As to the $T_1$ question: yes.
Suppose $A neq B$ are two different points in $H(X)$.
So we can assume that $exists x in A, x notin B$ (by symmetry, or we rename our sets).
Then $Xsetminus B$ is open, $A in [Xsetminus B] $ (as witnessed by $x$), $B notin [X setminus B]$ by definition, and $X setminus {x}$ is open as $X$ is $T_1$ and $B in langle Xsetminus {x}rangle $ and $A notin langle Xsetminus {x}rangle $, as witnesses by $x$ again.
So for two points we have open sets that contain one and not the other. So $H(X)$ is $T_1$.
answered Feb 24 '17 at 17:09
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
add a comment |
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1 Answer
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$begingroup$
This topology is called the Vietoris topology on $operatorname{Exp}(X)$, also denoted (as I usually do) by $H(X)$ or $2^X$, the hyperspace of $X$.
I normally use a standard subbase for this, for every non-empty open set $U subset X$,
define $[U] = {F in H(X): F cap U neq emptyset}$ and $langle U rangle = {F in H(X): F subseteq U}$. Then $langle U, V_1, ldots, V_nrangle = langle U rangle cap cap_{i=1}^n [V_i]$, so your base $B$ is the base generated by this subbase. Also note that $[U] = langle X, Urangle$, so is indeed open.
The subbase is especially handy when proving compactness of $H(X)$, using the Alexander subbase lemma, BTW.
As to the $T_1$ question: yes.
Suppose $A neq B$ are two different points in $H(X)$.
So we can assume that $exists x in A, x notin B$ (by symmetry, or we rename our sets).
Then $Xsetminus B$ is open, $A in [Xsetminus B] $ (as witnessed by $x$), $B notin [X setminus B]$ by definition, and $X setminus {x}$ is open as $X$ is $T_1$ and $B in langle Xsetminus {x}rangle $ and $A notin langle Xsetminus {x}rangle $, as witnesses by $x$ again.
So for two points we have open sets that contain one and not the other. So $H(X)$ is $T_1$.
answered Feb 24 '17 at 17:09
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
add a comment |
$begingroup$
This topology is called the Vietoris topology on $operatorname{Exp}(X)$, also denoted (as I usually do) by $H(X)$ or $2^X$, the hyperspace of $X$.
I normally use a standard subbase for this, for every non-empty open set $U subset X$,
define $[U] = {F in H(X): F cap U neq emptyset}$ and $langle U rangle = {F in H(X): F subseteq U}$. Then $langle U, V_1, ldots, V_nrangle = langle U rangle cap cap_{i=1}^n [V_i]$, so your base $B$ is the base generated by this subbase. Also note that $[U] = langle X, Urangle$, so is indeed open.
The subbase is especially handy when proving compactness of $H(X)$, using the Alexander subbase lemma, BTW.
As to the $T_1$ question: yes.
Suppose $A neq B$ are two different points in $H(X)$.
So we can assume that $exists x in A, x notin B$ (by symmetry, or we rename our sets).
Then $Xsetminus B$ is open, $A in [Xsetminus B] $ (as witnessed by $x$), $B notin [X setminus B]$ by definition, and $X setminus {x}$ is open as $X$ is $T_1$ and $B in langle Xsetminus {x}rangle $ and $A notin langle Xsetminus {x}rangle $, as witnesses by $x$ again.
So for two points we have open sets that contain one and not the other. So $H(X)$ is $T_1$.
answered Feb 24 '17 at 17:09
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
add a comment |
$begingroup$
This topology is called the Vietoris topology on $operatorname{Exp}(X)$, also denoted (as I usually do) by $H(X)$ or $2^X$, the hyperspace of $X$.
I normally use a standard subbase for this, for every non-empty open set $U subset X$,
define $[U] = {F in H(X): F cap U neq emptyset}$ and $langle U rangle = {F in H(X): F subseteq U}$. Then $langle U, V_1, ldots, V_nrangle = langle U rangle cap cap_{i=1}^n [V_i]$, so your base $B$ is the base generated by this subbase. Also note that $[U] = langle X, Urangle$, so is indeed open.
The subbase is especially handy when proving compactness of $H(X)$, using the Alexander subbase lemma, BTW.
As to the $T_1$ question: yes.
Suppose $A neq B$ are two different points in $H(X)$.
So we can assume that $exists x in A, x notin B$ (by symmetry, or we rename our sets).
Then $Xsetminus B$ is open, $A in [Xsetminus B] $ (as witnessed by $x$), $B notin [X setminus B]$ by definition, and $X setminus {x}$ is open as $X$ is $T_1$ and $B in langle Xsetminus {x}rangle $ and $A notin langle Xsetminus {x}rangle $, as witnesses by $x$ again.
So for two points we have open sets that contain one and not the other. So $H(X)$ is $T_1$.
answered Feb 24 '17 at 17:09
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
This topology is called the Vietoris topology on $operatorname{Exp}(X)$, also denoted (as I usually do) by $H(X)$ or $2^X$, the hyperspace of $X$.
I normally use a standard subbase for this, for every non-empty open set $U subset X$,
define $[U] = {F in H(X): F cap U neq emptyset}$ and $langle U rangle = {F in H(X): F subseteq U}$. Then $langle U, V_1, ldots, V_nrangle = langle U rangle cap cap_{i=1}^n [V_i]$, so your base $B$ is the base generated by this subbase. Also note that $[U] = langle X, Urangle$, so is indeed open.
The subbase is especially handy when proving compactness of $H(X)$, using the Alexander subbase lemma, BTW.
As to the $T_1$ question: yes.
Suppose $A neq B$ are two different points in $H(X)$.
So we can assume that $exists x in A, x notin B$ (by symmetry, or we rename our sets).
Then $Xsetminus B$ is open, $A in [Xsetminus B] $ (as witnessed by $x$), $B notin [X setminus B]$ by definition, and $X setminus {x}$ is open as $X$ is $T_1$ and $B in langle Xsetminus {x}rangle $ and $A notin langle Xsetminus {x}rangle $, as witnesses by $x$ again.
So for two points we have open sets that contain one and not the other. So $H(X)$ is $T_1$.
answered Feb 24 '17 at 17:09
Henno BrandsmaHenno Brandsma
117k349127
edited Oct 26 '18 at 13:30
answered Feb 24 '17 at 17:09
Henno BrandsmaHenno Brandsma
117k349127
answered Feb 24 '17 at 17:09
Henno BrandsmaHenno Brandsma
117k349127
answered Feb 24 '17 at 17:09
Henno BrandsmaHenno Brandsma
117k349127
117k349127
add a comment |
add a comment |
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