Spaces in which all closed sets are regular closed The Next CEO of Stack OverflowA space is regular if each closed set $Z$ is the intersection of all open sets containing $Z$?Examples of topologies in which all open sets are regular?All zero dimensional spaces are completely regular.All finite Baire measures are Closed-regular?On the small first countable regular spacesShow that if $X$ and $Y$ are regular, then so is the product space $Xtimes Y$.Regular spaces and Hausdorff spaceHow to Show that Points and Closed Sets Can be Separated by Closed Sets in a T3 (Regular) SpaceSaturated sets and topological spacesOn regular closed sets which are not open-closed
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Spaces in which all closed sets are regular closed
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Spaces in which all closed sets are regular closed
The Next CEO of Stack OverflowA space is regular if each closed set $Z$ is the intersection of all open sets containing $Z$?Examples of topologies in which all open sets are regular?All zero dimensional spaces are completely regular.All finite Baire measures are Closed-regular?On the small first countable regular spacesShow that if $X$ and $Y$ are regular, then so is the product space $Xtimes Y$.Regular spaces and Hausdorff spaceHow to Show that Points and Closed Sets Can be Separated by Closed Sets in a T3 (Regular) SpaceSaturated sets and topological spacesOn regular closed sets which are not open-closed
$begingroup$
I was reading about the regular closed sets. The definition is
Let $X$ be a topological space and $Asubseteq X$. We say that $A$ is a regular closed if $A=textcl(textint(A))$
Then, one question came to my mind: is there a topological space $X$ such that $X$ isn't a discrete space and for that every closed subset of $X$ is a regular closed set?
Obviusly, if $X$ is discrete then every closed set is a regular closed, but, if $X$ isn't discrete, what happens? That example exists?
Thanks in advance.
general-topology examples-counterexamples
edited 21 mins ago
Eric Wofsey
191k14216349
asked 54 mins ago
Carlos JiménezCarlos Jiménez
2,4341621
$endgroup$
add a comment |
$begingroup$
I was reading about the regular closed sets. The definition is
Let $X$ be a topological space and $Asubseteq X$. We say that $A$ is a regular closed if $A=textcl(textint(A))$
Then, one question came to my mind: is there a topological space $X$ such that $X$ isn't a discrete space and for that every closed subset of $X$ is a regular closed set?
Obviusly, if $X$ is discrete then every closed set is a regular closed, but, if $X$ isn't discrete, what happens? That example exists?
Thanks in advance.
general-topology examples-counterexamples
edited 21 mins ago
Eric Wofsey
191k14216349
asked 54 mins ago
Carlos JiménezCarlos Jiménez
2,4341621
$endgroup$
add a comment |
$begingroup$
I was reading about the regular closed sets. The definition is
Let $X$ be a topological space and $Asubseteq X$. We say that $A$ is a regular closed if $A=textcl(textint(A))$
Then, one question came to my mind: is there a topological space $X$ such that $X$ isn't a discrete space and for that every closed subset of $X$ is a regular closed set?
Obviusly, if $X$ is discrete then every closed set is a regular closed, but, if $X$ isn't discrete, what happens? That example exists?
Thanks in advance.
general-topology examples-counterexamples
edited 21 mins ago
Eric Wofsey
191k14216349
asked 54 mins ago
Carlos JiménezCarlos Jiménez
2,4341621
$endgroup$
I was reading about the regular closed sets. The definition is
Let $X$ be a topological space and $Asubseteq X$. We say that $A$ is a regular closed if $A=textcl(textint(A))$
Then, one question came to my mind: is there a topological space $X$ such that $X$ isn't a discrete space and for that every closed subset of $X$ is a regular closed set?
Obviusly, if $X$ is discrete then every closed set is a regular closed, but, if $X$ isn't discrete, what happens? That example exists?
Thanks in advance.
general-topology examples-counterexamples
general-topology examples-counterexamples
edited 21 mins ago
Eric Wofsey
191k14216349
asked 54 mins ago
Carlos JiménezCarlos Jiménez
2,4341621
edited 21 mins ago
Eric Wofsey
191k14216349
asked 54 mins ago
Carlos JiménezCarlos Jiménez
2,4341621
edited 21 mins ago
Eric Wofsey
191k14216349
edited 21 mins ago
Eric Wofsey
191k14216349
edited 21 mins ago
Eric Wofsey
191k14216349
191k14216349
asked 54 mins ago
Carlos JiménezCarlos Jiménez
2,4341621
asked 54 mins ago
Carlos JiménezCarlos Jiménez
2,4341621
asked 54 mins ago
Carlos JiménezCarlos Jiménez
2,4341621
2,4341621
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Given a partition $P$ on a set $X$, we can define a topology whose open sets are unions of elements of $P$. In this topology, open sets and closed sets are the same, so all closed sets are regular closed. (If $P$ is the finest partition this is the discrete topology; if $P$ is the coarsest topology it is the indiscrete topology. Such topologies can also be characterized as the topologies in which closed sets and open sets coincide, or topologies whose $T_0$ quotient is discrete.)
I claim, though, that these are the only examples. Indeed, suppose $X$ is a topological space in which all closed sets are regular closed. Suppose $x,yin X$ are such that $yinoverlinex$. Since $overliney$ is regular closed, it is the closure of its interior $U$ which is in particular nonempty, and we must have $yin U$ since $y$ is dense in $overliney$. Since $yinoverlinex$, we have $Usubseteq overlinex$ as well and so $xin U$. Thus $xin overlineU=overliney$ and so $overlinex=overliney$. We see then that $U$ is the interior of $overlinex$ and every element of $overlinex$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overlinex$). Thus $U=overlinex$, so $overlinex$ is open.
So, we have shown that the closure of each singleton in $X$ is a clopen set and is equal to the closure of any of its elements. It follows easily that the collection of closures of singletons is a partition of $X$, and a subset of $X$ is open iff it is a union of elements of this partition.
answered 28 mins ago
Eric WofseyEric Wofsey
191k14216349
$endgroup$
add a comment |
$begingroup$
You can take any set $X$ with trivial topology. Then every closed subset in $X$ is trivially regular.
But if $X$ is $T_1$ and every closed subset is regular then $X$ is discrete.
answered 50 mins ago
Moishe KohanMoishe Kohan
48.4k344110
$endgroup$
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
48 mins ago
$begingroup$
@CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
$endgroup$
– Moishe Kohan
45 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given a partition $P$ on a set $X$, we can define a topology whose open sets are unions of elements of $P$. In this topology, open sets and closed sets are the same, so all closed sets are regular closed. (If $P$ is the finest partition this is the discrete topology; if $P$ is the coarsest topology it is the indiscrete topology. Such topologies can also be characterized as the topologies in which closed sets and open sets coincide, or topologies whose $T_0$ quotient is discrete.)
I claim, though, that these are the only examples. Indeed, suppose $X$ is a topological space in which all closed sets are regular closed. Suppose $x,yin X$ are such that $yinoverlinex$. Since $overliney$ is regular closed, it is the closure of its interior $U$ which is in particular nonempty, and we must have $yin U$ since $y$ is dense in $overliney$. Since $yinoverlinex$, we have $Usubseteq overlinex$ as well and so $xin U$. Thus $xin overlineU=overliney$ and so $overlinex=overliney$. We see then that $U$ is the interior of $overlinex$ and every element of $overlinex$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overlinex$). Thus $U=overlinex$, so $overlinex$ is open.
So, we have shown that the closure of each singleton in $X$ is a clopen set and is equal to the closure of any of its elements. It follows easily that the collection of closures of singletons is a partition of $X$, and a subset of $X$ is open iff it is a union of elements of this partition.
answered 28 mins ago
Eric WofseyEric Wofsey
191k14216349
$endgroup$
add a comment |
$begingroup$
Given a partition $P$ on a set $X$, we can define a topology whose open sets are unions of elements of $P$. In this topology, open sets and closed sets are the same, so all closed sets are regular closed. (If $P$ is the finest partition this is the discrete topology; if $P$ is the coarsest topology it is the indiscrete topology. Such topologies can also be characterized as the topologies in which closed sets and open sets coincide, or topologies whose $T_0$ quotient is discrete.)
I claim, though, that these are the only examples. Indeed, suppose $X$ is a topological space in which all closed sets are regular closed. Suppose $x,yin X$ are such that $yinoverlinex$. Since $overliney$ is regular closed, it is the closure of its interior $U$ which is in particular nonempty, and we must have $yin U$ since $y$ is dense in $overliney$. Since $yinoverlinex$, we have $Usubseteq overlinex$ as well and so $xin U$. Thus $xin overlineU=overliney$ and so $overlinex=overliney$. We see then that $U$ is the interior of $overlinex$ and every element of $overlinex$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overlinex$). Thus $U=overlinex$, so $overlinex$ is open.
So, we have shown that the closure of each singleton in $X$ is a clopen set and is equal to the closure of any of its elements. It follows easily that the collection of closures of singletons is a partition of $X$, and a subset of $X$ is open iff it is a union of elements of this partition.
answered 28 mins ago
Eric WofseyEric Wofsey
191k14216349
$endgroup$
add a comment |
$begingroup$
Given a partition $P$ on a set $X$, we can define a topology whose open sets are unions of elements of $P$. In this topology, open sets and closed sets are the same, so all closed sets are regular closed. (If $P$ is the finest partition this is the discrete topology; if $P$ is the coarsest topology it is the indiscrete topology. Such topologies can also be characterized as the topologies in which closed sets and open sets coincide, or topologies whose $T_0$ quotient is discrete.)
I claim, though, that these are the only examples. Indeed, suppose $X$ is a topological space in which all closed sets are regular closed. Suppose $x,yin X$ are such that $yinoverlinex$. Since $overliney$ is regular closed, it is the closure of its interior $U$ which is in particular nonempty, and we must have $yin U$ since $y$ is dense in $overliney$. Since $yinoverlinex$, we have $Usubseteq overlinex$ as well and so $xin U$. Thus $xin overlineU=overliney$ and so $overlinex=overliney$. We see then that $U$ is the interior of $overlinex$ and every element of $overlinex$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overlinex$). Thus $U=overlinex$, so $overlinex$ is open.
So, we have shown that the closure of each singleton in $X$ is a clopen set and is equal to the closure of any of its elements. It follows easily that the collection of closures of singletons is a partition of $X$, and a subset of $X$ is open iff it is a union of elements of this partition.
answered 28 mins ago
Eric WofseyEric Wofsey
191k14216349
$endgroup$
Given a partition $P$ on a set $X$, we can define a topology whose open sets are unions of elements of $P$. In this topology, open sets and closed sets are the same, so all closed sets are regular closed. (If $P$ is the finest partition this is the discrete topology; if $P$ is the coarsest topology it is the indiscrete topology. Such topologies can also be characterized as the topologies in which closed sets and open sets coincide, or topologies whose $T_0$ quotient is discrete.)
I claim, though, that these are the only examples. Indeed, suppose $X$ is a topological space in which all closed sets are regular closed. Suppose $x,yin X$ are such that $yinoverlinex$. Since $overliney$ is regular closed, it is the closure of its interior $U$ which is in particular nonempty, and we must have $yin U$ since $y$ is dense in $overliney$. Since $yinoverlinex$, we have $Usubseteq overlinex$ as well and so $xin U$. Thus $xin overlineU=overliney$ and so $overlinex=overliney$. We see then that $U$ is the interior of $overlinex$ and every element of $overlinex$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overlinex$). Thus $U=overlinex$, so $overlinex$ is open.
So, we have shown that the closure of each singleton in $X$ is a clopen set and is equal to the closure of any of its elements. It follows easily that the collection of closures of singletons is a partition of $X$, and a subset of $X$ is open iff it is a union of elements of this partition.
answered 28 mins ago
Eric WofseyEric Wofsey
191k14216349
answered 28 mins ago
Eric WofseyEric Wofsey
191k14216349
answered 28 mins ago
Eric WofseyEric Wofsey
191k14216349
answered 28 mins ago
Eric WofseyEric Wofsey
191k14216349
191k14216349
add a comment |
add a comment |
$begingroup$
You can take any set $X$ with trivial topology. Then every closed subset in $X$ is trivially regular.
But if $X$ is $T_1$ and every closed subset is regular then $X$ is discrete.
answered 50 mins ago
Moishe KohanMoishe Kohan
48.4k344110
$endgroup$
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
48 mins ago
$begingroup$
@CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
$endgroup$
– Moishe Kohan
45 mins ago
add a comment |
$begingroup$
You can take any set $X$ with trivial topology. Then every closed subset in $X$ is trivially regular.
But if $X$ is $T_1$ and every closed subset is regular then $X$ is discrete.
answered 50 mins ago
Moishe KohanMoishe Kohan
48.4k344110
$endgroup$
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
48 mins ago
$begingroup$
@CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
$endgroup$
– Moishe Kohan
45 mins ago
add a comment |
$begingroup$
You can take any set $X$ with trivial topology. Then every closed subset in $X$ is trivially regular.
But if $X$ is $T_1$ and every closed subset is regular then $X$ is discrete.
answered 50 mins ago
Moishe KohanMoishe Kohan
48.4k344110
$endgroup$
You can take any set $X$ with trivial topology. Then every closed subset in $X$ is trivially regular.
But if $X$ is $T_1$ and every closed subset is regular then $X$ is discrete.
answered 50 mins ago
Moishe KohanMoishe Kohan
48.4k344110
answered 50 mins ago
Moishe KohanMoishe Kohan
48.4k344110
answered 50 mins ago
Moishe KohanMoishe Kohan
48.4k344110
answered 50 mins ago
Moishe KohanMoishe Kohan
48.4k344110
48.4k344110
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
48 mins ago
$begingroup$
@CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
$endgroup$
– Moishe Kohan
45 mins ago
add a comment |
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
48 mins ago
$begingroup$
@CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
$endgroup$
– Moishe Kohan
45 mins ago
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
48 mins ago
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
48 mins ago
$begingroup$
@CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
$endgroup$
– Moishe Kohan
45 mins ago
$begingroup$
@CarlosJiménez: A less trivial example would be a space $X=X_1sqcup X_2$ where both $X_1, X_2$ are open and have trivial topology. As I said, $T_1$ implies discreteness in your setting.
$endgroup$
– Moishe Kohan
45 mins ago
add a comment |
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