Spaces in which all closed sets are regular 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 $A=text{cl}(text{int}(A))$
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 4 hours ago
Eric Wofsey
191k14216349
asked 5 hours ago
Carlos JiménezCarlos Jiménez
2,4441621
$endgroup$
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 $A=text{cl}(text{int}(A))$
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 4 hours ago
Eric Wofsey
191k14216349
asked 5 hours ago
Carlos JiménezCarlos Jiménez
2,4441621
$endgroup$
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 $A=text{cl}(text{int}(A))$
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 4 hours ago
Eric Wofsey
191k14216349
asked 5 hours ago
Carlos JiménezCarlos Jiménez
2,4441621
$endgroup$
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 $A=text{cl}(text{int}(A))$
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 4 hours ago
Eric Wofsey
191k14216349
asked 5 hours ago
Carlos JiménezCarlos Jiménez
2,4441621
edited 4 hours ago
Eric Wofsey
191k14216349
asked 5 hours ago
Carlos JiménezCarlos Jiménez
2,4441621
edited 4 hours ago
Eric Wofsey
191k14216349
edited 4 hours ago
Eric Wofsey
191k14216349
edited 4 hours ago
Eric Wofsey
191k14216349
191k14216349
asked 5 hours ago
Carlos JiménezCarlos Jiménez
2,4441621
asked 5 hours ago
Carlos JiménezCarlos Jiménez
2,4441621
asked 5 hours ago
Carlos JiménezCarlos Jiménez
2,4441621
2,4441621
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 $yinoverline{{x}}$. Since $overline{{y}}$ 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 $overline{{y}}$. Since $yinoverline{{x}}$, we have $Usubseteq overline{{x}}$ as well and so $xin U$. Thus $xin overline{U}=overline{{y}}$ and so $overline{{x}}=overline{{y}}$. We see then that $U$ is the interior of $overline{{x}}$ and every element of $overline{{x}}$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overline{{x}}$). Thus $U=overline{{x}}$, so $overline{{x}}$ 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 4 hours 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 5 hours ago
Moishe KohanMoishe Kohan
48.4k344110
$endgroup$
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
5 hours 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
4 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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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 $yinoverline{{x}}$. Since $overline{{y}}$ 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 $overline{{y}}$. Since $yinoverline{{x}}$, we have $Usubseteq overline{{x}}$ as well and so $xin U$. Thus $xin overline{U}=overline{{y}}$ and so $overline{{x}}=overline{{y}}$. We see then that $U$ is the interior of $overline{{x}}$ and every element of $overline{{x}}$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overline{{x}}$). Thus $U=overline{{x}}$, so $overline{{x}}$ 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 4 hours 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 $yinoverline{{x}}$. Since $overline{{y}}$ 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 $overline{{y}}$. Since $yinoverline{{x}}$, we have $Usubseteq overline{{x}}$ as well and so $xin U$. Thus $xin overline{U}=overline{{y}}$ and so $overline{{x}}=overline{{y}}$. We see then that $U$ is the interior of $overline{{x}}$ and every element of $overline{{x}}$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overline{{x}}$). Thus $U=overline{{x}}$, so $overline{{x}}$ 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 4 hours 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 $yinoverline{{x}}$. Since $overline{{y}}$ 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 $overline{{y}}$. Since $yinoverline{{x}}$, we have $Usubseteq overline{{x}}$ as well and so $xin U$. Thus $xin overline{U}=overline{{y}}$ and so $overline{{x}}=overline{{y}}$. We see then that $U$ is the interior of $overline{{x}}$ and every element of $overline{{x}}$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overline{{x}}$). Thus $U=overline{{x}}$, so $overline{{x}}$ 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 4 hours 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 $yinoverline{{x}}$. Since $overline{{y}}$ 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 $overline{{y}}$. Since $yinoverline{{x}}$, we have $Usubseteq overline{{x}}$ as well and so $xin U$. Thus $xin overline{U}=overline{{y}}$ and so $overline{{x}}=overline{{y}}$. We see then that $U$ is the interior of $overline{{x}}$ and every element of $overline{{x}}$ is in $U$ (since $yin U$ and $y$ was originally an arbitrary element of $overline{{x}}$). Thus $U=overline{{x}}$, so $overline{{x}}$ 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 4 hours ago
Eric WofseyEric Wofsey
191k14216349
answered 4 hours ago
Eric WofseyEric Wofsey
191k14216349
answered 4 hours ago
Eric WofseyEric Wofsey
191k14216349
answered 4 hours 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 5 hours ago
Moishe KohanMoishe Kohan
48.4k344110
$endgroup$
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
5 hours 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
4 hours 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 5 hours ago
Moishe KohanMoishe Kohan
48.4k344110
$endgroup$
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
5 hours 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
4 hours 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 5 hours 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 5 hours ago
Moishe KohanMoishe Kohan
48.4k344110
answered 5 hours ago
Moishe KohanMoishe Kohan
48.4k344110
answered 5 hours ago
Moishe KohanMoishe Kohan
48.4k344110
answered 5 hours 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
5 hours 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
4 hours ago
add a comment |
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
5 hours 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
4 hours ago
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
5 hours ago
$begingroup$
Is there a non trivial example? I don't mind the separation axiom.
$endgroup$
– Carlos Jiménez
5 hours 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
4 hours 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
4 hours ago
add a comment |
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