Mixing
Every minimal Hausdorff space is H-closed
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A Hausdorff topological space $(X,mathcal T)$ is called H-closed or absolutely closed if it is closed in any Hausdorff space, which contains $X$ as a subspace.$newcommand{ol}[1]{overline{#1}}newcommand{mc}[1]{mathcal{#1}}$
A Hausdorff topological space $(X,mathcal T)$ is called minimal Hausdorff if there is no Hausdorff topology on $X$ that is strictly weaker than $mathcal T$. (I.e., it is minimal element of the set of all Hausdorff topologies on $X$ with partial ordering $subseteq$.)
How to show that every minimal Hausdorff space is H-closed?
Is there an easy counterexample for the other direction? (I guess that finding a counterexample is not that easy - otherwise Willard would very probably mention an example.)
This is part of Problem 17M from Willard: General Topology, p.127
In the preceding problems from that book I have already shown some results on H-closed and minimal Hausdorff spaces:
Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open filter in $X$ has a cluster point.
Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open cover $mc C$ of $X$ contains a finite subsystem $mc D$ such that $bigcup {ol D; Dinmc D}=X$, i.e., the closures of the sets from $mc D$ cover $X$.
Let $X$ be a Hausdorff space. $X$ is minimal Hausdorff $Leftrightarrow$ every open filter with unique cluster point converges.
Hint it Willard's book suggest to show that if $X$ Hausdorff and not H-closed, then it is not minimal. I've tried to use the characterization of H-closed spaces using filters. Which means that there is an open filter in $X$ which has no cluster point. But if I tried to use a construction similar to the one I used in the proof of the characterization of minimal Hausdorff spaces via open ultrafilters, the new topology was not necessarily strictly weaker than the original one.
general-topology examples-counterexamples separation-axioms
asked Apr 23 '12 at 5:17
Martin SleziakMartin Sleziak
44.9k10122277
$endgroup$
add a comment |
$begingroup$
A Hausdorff topological space $(X,mathcal T)$ is called H-closed or absolutely closed if it is closed in any Hausdorff space, which contains $X$ as a subspace.$newcommand{ol}[1]{overline{#1}}newcommand{mc}[1]{mathcal{#1}}$
A Hausdorff topological space $(X,mathcal T)$ is called minimal Hausdorff if there is no Hausdorff topology on $X$ that is strictly weaker than $mathcal T$. (I.e., it is minimal element of the set of all Hausdorff topologies on $X$ with partial ordering $subseteq$.)
How to show that every minimal Hausdorff space is H-closed?
Is there an easy counterexample for the other direction? (I guess that finding a counterexample is not that easy - otherwise Willard would very probably mention an example.)
This is part of Problem 17M from Willard: General Topology, p.127
In the preceding problems from that book I have already shown some results on H-closed and minimal Hausdorff spaces:
Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open filter in $X$ has a cluster point.
Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open cover $mc C$ of $X$ contains a finite subsystem $mc D$ such that $bigcup {ol D; Dinmc D}=X$, i.e., the closures of the sets from $mc D$ cover $X$.
Let $X$ be a Hausdorff space. $X$ is minimal Hausdorff $Leftrightarrow$ every open filter with unique cluster point converges.
Hint it Willard's book suggest to show that if $X$ Hausdorff and not H-closed, then it is not minimal. I've tried to use the characterization of H-closed spaces using filters. Which means that there is an open filter in $X$ which has no cluster point. But if I tried to use a construction similar to the one I used in the proof of the characterization of minimal Hausdorff spaces via open ultrafilters, the new topology was not necessarily strictly weaker than the original one.
general-topology examples-counterexamples separation-axioms
asked Apr 23 '12 at 5:17
Martin SleziakMartin Sleziak
44.9k10122277
$endgroup$
1
$begingroup$
There is some info and many references in the exercises to Engelking's topology book. According to exercise 3.12.5 e) a Hausdorff space is minimal Hausdorff (Engelking calls it H-minimal) if and only if it is H-closed and semiregular (the regular open sets are a basis of the topology). See also exercise 1.7.8.
$endgroup$
– t.b.
Apr 23 '12 at 7:56
$begingroup$
Thanks a lot @t.b. (Engelking is my favorite reference for general topology; I don't know why, but I did not think about checking that book until you reminded me.)
$endgroup$
– Martin Sleziak
Apr 25 '12 at 16:58
add a comment |
$begingroup$
A Hausdorff topological space $(X,mathcal T)$ is called H-closed or absolutely closed if it is closed in any Hausdorff space, which contains $X$ as a subspace.$newcommand{ol}[1]{overline{#1}}newcommand{mc}[1]{mathcal{#1}}$
A Hausdorff topological space $(X,mathcal T)$ is called minimal Hausdorff if there is no Hausdorff topology on $X$ that is strictly weaker than $mathcal T$. (I.e., it is minimal element of the set of all Hausdorff topologies on $X$ with partial ordering $subseteq$.)
How to show that every minimal Hausdorff space is H-closed?
Is there an easy counterexample for the other direction? (I guess that finding a counterexample is not that easy - otherwise Willard would very probably mention an example.)
This is part of Problem 17M from Willard: General Topology, p.127
In the preceding problems from that book I have already shown some results on H-closed and minimal Hausdorff spaces:
Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open filter in $X$ has a cluster point.
Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open cover $mc C$ of $X$ contains a finite subsystem $mc D$ such that $bigcup {ol D; Dinmc D}=X$, i.e., the closures of the sets from $mc D$ cover $X$.
Let $X$ be a Hausdorff space. $X$ is minimal Hausdorff $Leftrightarrow$ every open filter with unique cluster point converges.
Hint it Willard's book suggest to show that if $X$ Hausdorff and not H-closed, then it is not minimal. I've tried to use the characterization of H-closed spaces using filters. Which means that there is an open filter in $X$ which has no cluster point. But if I tried to use a construction similar to the one I used in the proof of the characterization of minimal Hausdorff spaces via open ultrafilters, the new topology was not necessarily strictly weaker than the original one.
general-topology examples-counterexamples separation-axioms
asked Apr 23 '12 at 5:17
Martin SleziakMartin Sleziak
44.9k10122277
$endgroup$
A Hausdorff topological space $(X,mathcal T)$ is called H-closed or absolutely closed if it is closed in any Hausdorff space, which contains $X$ as a subspace.$newcommand{ol}[1]{overline{#1}}newcommand{mc}[1]{mathcal{#1}}$
A Hausdorff topological space $(X,mathcal T)$ is called minimal Hausdorff if there is no Hausdorff topology on $X$ that is strictly weaker than $mathcal T$. (I.e., it is minimal element of the set of all Hausdorff topologies on $X$ with partial ordering $subseteq$.)
How to show that every minimal Hausdorff space is H-closed?
Is there an easy counterexample for the other direction? (I guess that finding a counterexample is not that easy - otherwise Willard would very probably mention an example.)
This is part of Problem 17M from Willard: General Topology, p.127
In the preceding problems from that book I have already shown some results on H-closed and minimal Hausdorff spaces:
Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open filter in $X$ has a cluster point.
Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open cover $mc C$ of $X$ contains a finite subsystem $mc D$ such that $bigcup {ol D; Dinmc D}=X$, i.e., the closures of the sets from $mc D$ cover $X$.
Let $X$ be a Hausdorff space. $X$ is minimal Hausdorff $Leftrightarrow$ every open filter with unique cluster point converges.
Hint it Willard's book suggest to show that if $X$ Hausdorff and not H-closed, then it is not minimal. I've tried to use the characterization of H-closed spaces using filters. Which means that there is an open filter in $X$ which has no cluster point. But if I tried to use a construction similar to the one I used in the proof of the characterization of minimal Hausdorff spaces via open ultrafilters, the new topology was not necessarily strictly weaker than the original one.
general-topology examples-counterexamples separation-axioms
general-topology examples-counterexamples separation-axioms
asked Apr 23 '12 at 5:17
Martin SleziakMartin Sleziak
44.9k10122277
asked Apr 23 '12 at 5:17
Martin SleziakMartin Sleziak
44.9k10122277
edited Jun 5 '17 at 5:34
Martin Sleziak
asked Apr 23 '12 at 5:17
Martin SleziakMartin Sleziak
44.9k10122277
asked Apr 23 '12 at 5:17
Martin SleziakMartin Sleziak
44.9k10122277
asked Apr 23 '12 at 5:17
Martin SleziakMartin Sleziak
44.9k10122277
44.9k10122277
1
$begingroup$
There is some info and many references in the exercises to Engelking's topology book. According to exercise 3.12.5 e) a Hausdorff space is minimal Hausdorff (Engelking calls it H-minimal) if and only if it is H-closed and semiregular (the regular open sets are a basis of the topology). See also exercise 1.7.8.
$endgroup$
– t.b.
Apr 23 '12 at 7:56
$begingroup$
Thanks a lot @t.b. (Engelking is my favorite reference for general topology; I don't know why, but I did not think about checking that book until you reminded me.)
$endgroup$
– Martin Sleziak
Apr 25 '12 at 16:58
add a comment |
1
$begingroup$
There is some info and many references in the exercises to Engelking's topology book. According to exercise 3.12.5 e) a Hausdorff space is minimal Hausdorff (Engelking calls it H-minimal) if and only if it is H-closed and semiregular (the regular open sets are a basis of the topology). See also exercise 1.7.8.
$endgroup$
– t.b.
Apr 23 '12 at 7:56
$begingroup$
Thanks a lot @t.b. (Engelking is my favorite reference for general topology; I don't know why, but I did not think about checking that book until you reminded me.)
$endgroup$
– Martin Sleziak
Apr 25 '12 at 16:58
1
1
$begingroup$
There is some info and many references in the exercises to Engelking's topology book. According to exercise 3.12.5 e) a Hausdorff space is minimal Hausdorff (Engelking calls it H-minimal) if and only if it is H-closed and semiregular (the regular open sets are a basis of the topology). See also exercise 1.7.8.
$endgroup$
– t.b.
Apr 23 '12 at 7:56
$begingroup$
There is some info and many references in the exercises to Engelking's topology book. According to exercise 3.12.5 e) a Hausdorff space is minimal Hausdorff (Engelking calls it H-minimal) if and only if it is H-closed and semiregular (the regular open sets are a basis of the topology). See also exercise 1.7.8.
$endgroup$
– t.b.
Apr 23 '12 at 7:56
$begingroup$
Thanks a lot @t.b. (Engelking is my favorite reference for general topology; I don't know why, but I did not think about checking that book until you reminded me.)
$endgroup$
– Martin Sleziak
Apr 25 '12 at 16:58
$begingroup$
Thanks a lot @t.b. (Engelking is my favorite reference for general topology; I don't know why, but I did not think about checking that book until you reminded me.)
$endgroup$
– Martin Sleziak
Apr 25 '12 at 16:58
add a comment |
1 Answer
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$begingroup$
I found the following proof in the paper
Horst Herrlich: $T_v$-Abgeschlossenheit und $T_v$-Minimalität, Mathematische Zeitschrift, Volume 88, Number 3, 285-294, DOI: 10.1007/BF01111687. The proof is given there in a greater generality; for $T_v$-minimal and $T_v$-closed spaces, where $vin{2,3,4}$.
Here is a translation of H. Herrlich's proof:$newcommand{mc}[1]{mathcal{#1}}$
Let $(X,mc T)$ be a space which is not H-closed. Then there exists a $T_2$-space $(X',mc T')$ such that $X'=Xcup{a}$ and $X$ is not closed in $(X',mc T')$. If we choose an arbitrary element $x_0in X$ then
$$mc T''={M; Minmc T; x_0in M Rightarrow Mcup{a}inmc T'}$$
is a $T_2$-topology on $X$ which is strictly weaker than $mc T$. Hence $(X,mc T)$ is not $T_2$-minimal.
Some minor details:
The topology $mc T''$ is Hausdorff: If we have $x_0ne y$, $yin X$ then there are $mc T'$-neighborhoods $U_xni x$, $V_1ni y$ which are disjoint. Similarly, we have $U_ani a$, $V_2ni y$, which are disjoint. Hence $U_xcup (U_acap X)$ and $V_1cap V_2$ are $mc T''$-neighborhoods separating the points $x$ and $y$. The points different from $x_0$ have the same neighborhoods as in $mc T$.
The fact that $mc T''$ is strictly weaker than $mc T$ follows from the fact, that ${a}$ is not isolated in $X'$ (equivalently, $X$ is not closed in $X'$). Since $(X,mc T')$ is Hausdorff, we have disjoint neighborhoods $U_{x_0}ni x$ and $U_ani a$, which separate $x_0$ and $a$ in this space. The set $U_{x_0}$ is not open in $(X,mc T'')$.
The following example is given in Willard's book, Problem 17M/4, as an example of an H-closed space which is not compact.
Let $newcommand{N}{mathbb N}N^*={0}cup{frac1n; ninmathbb N}$ with the topology inherited from real line. (I.e., a convergent sequence.) Then we take a topological product $NtimesN^*$, where $N$ has discrete topology. (I.e. this is just the topological sum of countably many integer sequences.) We adjoin a new point $q$ with the neighborhood basis consisting of the sets of the form $U_{n_0}(q)={(n,1/m)inNtimesN^*; nge n_0}cup{q}$. (I.e., $U_{n_0}(q)$ consists of isolated points in all but finitely many sequences.) Let us call this space $X$.
Note: A similar space is described as example 100 in Counterexamples in Topology
p.119-120. If you look only at the left half of the picture given in this book, it depicts a typical basic neighborhood of the point $q$.
A topological space is called semiregular, if
regular open sets form a base. A set $U$ is regular open if $U=operatorname{Int} overline U$.
It is known that a space $X$ is Hausdorff minimal if and only if it is H-closed and semiregular.
The space $X$ described above is an H-closed space, but it is not semiregular, since closure of each set $U_{n_0}(q)$ contains all points $(n,0)$ for $nge n_0$ in its interior. Hence no regular open set containing $p$ is contained in the basic set $U_{n_0}(q)$. Since this space is not semiregular it is not Hausdorff minimal. Thus this is an example of a topological space which is H-closed but not Hausdorff minimal.
$endgroup$
1
$begingroup$
I find it helpful to view $langle X,mathcal{T}''rangle$ as the quotient of $X'$ obtained by identifying $a$ and $x_0$.
$endgroup$
– Brian M. Scott
Apr 29 '12 at 4:12
add a comment |
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$begingroup$
I found the following proof in the paper
Horst Herrlich: $T_v$-Abgeschlossenheit und $T_v$-Minimalität, Mathematische Zeitschrift, Volume 88, Number 3, 285-294, DOI: 10.1007/BF01111687. The proof is given there in a greater generality; for $T_v$-minimal and $T_v$-closed spaces, where $vin{2,3,4}$.
Here is a translation of H. Herrlich's proof:$newcommand{mc}[1]{mathcal{#1}}$
Let $(X,mc T)$ be a space which is not H-closed. Then there exists a $T_2$-space $(X',mc T')$ such that $X'=Xcup{a}$ and $X$ is not closed in $(X',mc T')$. If we choose an arbitrary element $x_0in X$ then
$$mc T''={M; Minmc T; x_0in M Rightarrow Mcup{a}inmc T'}$$
is a $T_2$-topology on $X$ which is strictly weaker than $mc T$. Hence $(X,mc T)$ is not $T_2$-minimal.
Some minor details:
The topology $mc T''$ is Hausdorff: If we have $x_0ne y$, $yin X$ then there are $mc T'$-neighborhoods $U_xni x$, $V_1ni y$ which are disjoint. Similarly, we have $U_ani a$, $V_2ni y$, which are disjoint. Hence $U_xcup (U_acap X)$ and $V_1cap V_2$ are $mc T''$-neighborhoods separating the points $x$ and $y$. The points different from $x_0$ have the same neighborhoods as in $mc T$.
The fact that $mc T''$ is strictly weaker than $mc T$ follows from the fact, that ${a}$ is not isolated in $X'$ (equivalently, $X$ is not closed in $X'$). Since $(X,mc T')$ is Hausdorff, we have disjoint neighborhoods $U_{x_0}ni x$ and $U_ani a$, which separate $x_0$ and $a$ in this space. The set $U_{x_0}$ is not open in $(X,mc T'')$.
The following example is given in Willard's book, Problem 17M/4, as an example of an H-closed space which is not compact.
Let $newcommand{N}{mathbb N}N^*={0}cup{frac1n; ninmathbb N}$ with the topology inherited from real line. (I.e., a convergent sequence.) Then we take a topological product $NtimesN^*$, where $N$ has discrete topology. (I.e. this is just the topological sum of countably many integer sequences.) We adjoin a new point $q$ with the neighborhood basis consisting of the sets of the form $U_{n_0}(q)={(n,1/m)inNtimesN^*; nge n_0}cup{q}$. (I.e., $U_{n_0}(q)$ consists of isolated points in all but finitely many sequences.) Let us call this space $X$.
Note: A similar space is described as example 100 in Counterexamples in Topology
p.119-120. If you look only at the left half of the picture given in this book, it depicts a typical basic neighborhood of the point $q$.
A topological space is called semiregular, if
regular open sets form a base. A set $U$ is regular open if $U=operatorname{Int} overline U$.
It is known that a space $X$ is Hausdorff minimal if and only if it is H-closed and semiregular.
The space $X$ described above is an H-closed space, but it is not semiregular, since closure of each set $U_{n_0}(q)$ contains all points $(n,0)$ for $nge n_0$ in its interior. Hence no regular open set containing $p$ is contained in the basic set $U_{n_0}(q)$. Since this space is not semiregular it is not Hausdorff minimal. Thus this is an example of a topological space which is H-closed but not Hausdorff minimal.
$endgroup$
1
$begingroup$
I find it helpful to view $langle X,mathcal{T}''rangle$ as the quotient of $X'$ obtained by identifying $a$ and $x_0$.
$endgroup$
– Brian M. Scott
Apr 29 '12 at 4:12
add a comment |
$begingroup$
I found the following proof in the paper
Horst Herrlich: $T_v$-Abgeschlossenheit und $T_v$-Minimalität, Mathematische Zeitschrift, Volume 88, Number 3, 285-294, DOI: 10.1007/BF01111687. The proof is given there in a greater generality; for $T_v$-minimal and $T_v$-closed spaces, where $vin{2,3,4}$.
Here is a translation of H. Herrlich's proof:$newcommand{mc}[1]{mathcal{#1}}$
Let $(X,mc T)$ be a space which is not H-closed. Then there exists a $T_2$-space $(X',mc T')$ such that $X'=Xcup{a}$ and $X$ is not closed in $(X',mc T')$. If we choose an arbitrary element $x_0in X$ then
$$mc T''={M; Minmc T; x_0in M Rightarrow Mcup{a}inmc T'}$$
is a $T_2$-topology on $X$ which is strictly weaker than $mc T$. Hence $(X,mc T)$ is not $T_2$-minimal.
Some minor details:
The topology $mc T''$ is Hausdorff: If we have $x_0ne y$, $yin X$ then there are $mc T'$-neighborhoods $U_xni x$, $V_1ni y$ which are disjoint. Similarly, we have $U_ani a$, $V_2ni y$, which are disjoint. Hence $U_xcup (U_acap X)$ and $V_1cap V_2$ are $mc T''$-neighborhoods separating the points $x$ and $y$. The points different from $x_0$ have the same neighborhoods as in $mc T$.
The fact that $mc T''$ is strictly weaker than $mc T$ follows from the fact, that ${a}$ is not isolated in $X'$ (equivalently, $X$ is not closed in $X'$). Since $(X,mc T')$ is Hausdorff, we have disjoint neighborhoods $U_{x_0}ni x$ and $U_ani a$, which separate $x_0$ and $a$ in this space. The set $U_{x_0}$ is not open in $(X,mc T'')$.
The following example is given in Willard's book, Problem 17M/4, as an example of an H-closed space which is not compact.
Let $newcommand{N}{mathbb N}N^*={0}cup{frac1n; ninmathbb N}$ with the topology inherited from real line. (I.e., a convergent sequence.) Then we take a topological product $NtimesN^*$, where $N$ has discrete topology. (I.e. this is just the topological sum of countably many integer sequences.) We adjoin a new point $q$ with the neighborhood basis consisting of the sets of the form $U_{n_0}(q)={(n,1/m)inNtimesN^*; nge n_0}cup{q}$. (I.e., $U_{n_0}(q)$ consists of isolated points in all but finitely many sequences.) Let us call this space $X$.
Note: A similar space is described as example 100 in Counterexamples in Topology
p.119-120. If you look only at the left half of the picture given in this book, it depicts a typical basic neighborhood of the point $q$.
A topological space is called semiregular, if
regular open sets form a base. A set $U$ is regular open if $U=operatorname{Int} overline U$.
It is known that a space $X$ is Hausdorff minimal if and only if it is H-closed and semiregular.
The space $X$ described above is an H-closed space, but it is not semiregular, since closure of each set $U_{n_0}(q)$ contains all points $(n,0)$ for $nge n_0$ in its interior. Hence no regular open set containing $p$ is contained in the basic set $U_{n_0}(q)$. Since this space is not semiregular it is not Hausdorff minimal. Thus this is an example of a topological space which is H-closed but not Hausdorff minimal.
$endgroup$
1
$begingroup$
I find it helpful to view $langle X,mathcal{T}''rangle$ as the quotient of $X'$ obtained by identifying $a$ and $x_0$.
$endgroup$
– Brian M. Scott
Apr 29 '12 at 4:12
add a comment |
$begingroup$
I found the following proof in the paper
Horst Herrlich: $T_v$-Abgeschlossenheit und $T_v$-Minimalität, Mathematische Zeitschrift, Volume 88, Number 3, 285-294, DOI: 10.1007/BF01111687. The proof is given there in a greater generality; for $T_v$-minimal and $T_v$-closed spaces, where $vin{2,3,4}$.
Here is a translation of H. Herrlich's proof:$newcommand{mc}[1]{mathcal{#1}}$
Let $(X,mc T)$ be a space which is not H-closed. Then there exists a $T_2$-space $(X',mc T')$ such that $X'=Xcup{a}$ and $X$ is not closed in $(X',mc T')$. If we choose an arbitrary element $x_0in X$ then
$$mc T''={M; Minmc T; x_0in M Rightarrow Mcup{a}inmc T'}$$
is a $T_2$-topology on $X$ which is strictly weaker than $mc T$. Hence $(X,mc T)$ is not $T_2$-minimal.
Some minor details:
The topology $mc T''$ is Hausdorff: If we have $x_0ne y$, $yin X$ then there are $mc T'$-neighborhoods $U_xni x$, $V_1ni y$ which are disjoint. Similarly, we have $U_ani a$, $V_2ni y$, which are disjoint. Hence $U_xcup (U_acap X)$ and $V_1cap V_2$ are $mc T''$-neighborhoods separating the points $x$ and $y$. The points different from $x_0$ have the same neighborhoods as in $mc T$.
The fact that $mc T''$ is strictly weaker than $mc T$ follows from the fact, that ${a}$ is not isolated in $X'$ (equivalently, $X$ is not closed in $X'$). Since $(X,mc T')$ is Hausdorff, we have disjoint neighborhoods $U_{x_0}ni x$ and $U_ani a$, which separate $x_0$ and $a$ in this space. The set $U_{x_0}$ is not open in $(X,mc T'')$.
The following example is given in Willard's book, Problem 17M/4, as an example of an H-closed space which is not compact.
Let $newcommand{N}{mathbb N}N^*={0}cup{frac1n; ninmathbb N}$ with the topology inherited from real line. (I.e., a convergent sequence.) Then we take a topological product $NtimesN^*$, where $N$ has discrete topology. (I.e. this is just the topological sum of countably many integer sequences.) We adjoin a new point $q$ with the neighborhood basis consisting of the sets of the form $U_{n_0}(q)={(n,1/m)inNtimesN^*; nge n_0}cup{q}$. (I.e., $U_{n_0}(q)$ consists of isolated points in all but finitely many sequences.) Let us call this space $X$.
Note: A similar space is described as example 100 in Counterexamples in Topology
p.119-120. If you look only at the left half of the picture given in this book, it depicts a typical basic neighborhood of the point $q$.
A topological space is called semiregular, if
regular open sets form a base. A set $U$ is regular open if $U=operatorname{Int} overline U$.
It is known that a space $X$ is Hausdorff minimal if and only if it is H-closed and semiregular.
The space $X$ described above is an H-closed space, but it is not semiregular, since closure of each set $U_{n_0}(q)$ contains all points $(n,0)$ for $nge n_0$ in its interior. Hence no regular open set containing $p$ is contained in the basic set $U_{n_0}(q)$. Since this space is not semiregular it is not Hausdorff minimal. Thus this is an example of a topological space which is H-closed but not Hausdorff minimal.
$endgroup$
I found the following proof in the paper
Horst Herrlich: $T_v$-Abgeschlossenheit und $T_v$-Minimalität, Mathematische Zeitschrift, Volume 88, Number 3, 285-294, DOI: 10.1007/BF01111687. The proof is given there in a greater generality; for $T_v$-minimal and $T_v$-closed spaces, where $vin{2,3,4}$.
Here is a translation of H. Herrlich's proof:$newcommand{mc}[1]{mathcal{#1}}$
Let $(X,mc T)$ be a space which is not H-closed. Then there exists a $T_2$-space $(X',mc T')$ such that $X'=Xcup{a}$ and $X$ is not closed in $(X',mc T')$. If we choose an arbitrary element $x_0in X$ then
$$mc T''={M; Minmc T; x_0in M Rightarrow Mcup{a}inmc T'}$$
is a $T_2$-topology on $X$ which is strictly weaker than $mc T$. Hence $(X,mc T)$ is not $T_2$-minimal.
Some minor details:
The topology $mc T''$ is Hausdorff: If we have $x_0ne y$, $yin X$ then there are $mc T'$-neighborhoods $U_xni x$, $V_1ni y$ which are disjoint. Similarly, we have $U_ani a$, $V_2ni y$, which are disjoint. Hence $U_xcup (U_acap X)$ and $V_1cap V_2$ are $mc T''$-neighborhoods separating the points $x$ and $y$. The points different from $x_0$ have the same neighborhoods as in $mc T$.
The fact that $mc T''$ is strictly weaker than $mc T$ follows from the fact, that ${a}$ is not isolated in $X'$ (equivalently, $X$ is not closed in $X'$). Since $(X,mc T')$ is Hausdorff, we have disjoint neighborhoods $U_{x_0}ni x$ and $U_ani a$, which separate $x_0$ and $a$ in this space. The set $U_{x_0}$ is not open in $(X,mc T'')$.
The following example is given in Willard's book, Problem 17M/4, as an example of an H-closed space which is not compact.
Let $newcommand{N}{mathbb N}N^*={0}cup{frac1n; ninmathbb N}$ with the topology inherited from real line. (I.e., a convergent sequence.) Then we take a topological product $NtimesN^*$, where $N$ has discrete topology. (I.e. this is just the topological sum of countably many integer sequences.) We adjoin a new point $q$ with the neighborhood basis consisting of the sets of the form $U_{n_0}(q)={(n,1/m)inNtimesN^*; nge n_0}cup{q}$. (I.e., $U_{n_0}(q)$ consists of isolated points in all but finitely many sequences.) Let us call this space $X$.
Note: A similar space is described as example 100 in Counterexamples in Topology
p.119-120. If you look only at the left half of the picture given in this book, it depicts a typical basic neighborhood of the point $q$.
A topological space is called semiregular, if
regular open sets form a base. A set $U$ is regular open if $U=operatorname{Int} overline U$.
It is known that a space $X$ is Hausdorff minimal if and only if it is H-closed and semiregular.
The space $X$ described above is an H-closed space, but it is not semiregular, since closure of each set $U_{n_0}(q)$ contains all points $(n,0)$ for $nge n_0$ in its interior. Hence no regular open set containing $p$ is contained in the basic set $U_{n_0}(q)$. Since this space is not semiregular it is not Hausdorff minimal. Thus this is an example of a topological space which is H-closed but not Hausdorff minimal.
edited Jan 29 at 19:46
community wiki
7 revs
Martin Sleziak
1
$begingroup$
I find it helpful to view $langle X,mathcal{T}''rangle$ as the quotient of $X'$ obtained by identifying $a$ and $x_0$.
$endgroup$
– Brian M. Scott
Apr 29 '12 at 4:12
add a comment |
1
$begingroup$
I find it helpful to view $langle X,mathcal{T}''rangle$ as the quotient of $X'$ obtained by identifying $a$ and $x_0$.
$endgroup$
– Brian M. Scott
Apr 29 '12 at 4:12
1
1
$begingroup$
I find it helpful to view $langle X,mathcal{T}''rangle$ as the quotient of $X'$ obtained by identifying $a$ and $x_0$.
$endgroup$
– Brian M. Scott
Apr 29 '12 at 4:12
$begingroup$
I find it helpful to view $langle X,mathcal{T}''rangle$ as the quotient of $X'$ obtained by identifying $a$ and $x_0$.
$endgroup$
– Brian M. Scott
Apr 29 '12 at 4:12
add a comment |
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There is some info and many references in the exercises to Engelking's topology book. According to exercise 3.12.5 e) a Hausdorff space is minimal Hausdorff (Engelking calls it H-minimal) if and only if it is H-closed and semiregular (the regular open sets are a basis of the topology). See also exercise 1.7.8.
$endgroup$
– t.b.
Apr 23 '12 at 7:56
$begingroup$
Thanks a lot @t.b. (Engelking is my favorite reference for general topology; I don't know why, but I did not think about checking that book until you reminded me.)
$endgroup$
– Martin Sleziak
Apr 25 '12 at 16:58