Mixing
Shows by an example that a $gamma$-hyperconnected space in topology may not be hyperconnected.
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In a topological space $(X,T)$ a subset A of X is said to be preopen if $A subset int(cl(A))$ and $gamma$-open if $A cap B$ is preopen for every preopen set B in $X.$ $(X, T)$ is said to be hyperconnected if closure of any open set gives $X$ and $gamma$-hyperconnected if $gamma$-closure of each $gamma$-open set gives $X$.
general-topology
asked Jun 13 '17 at 10:26
Birojit DasBirojit Das
179
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show 17 more comments
$begingroup$
In a topological space $(X,T)$ a subset A of X is said to be preopen if $A subset int(cl(A))$ and $gamma$-open if $A cap B$ is preopen for every preopen set B in $X.$ $(X, T)$ is said to be hyperconnected if closure of any open set gives $X$ and $gamma$-hyperconnected if $gamma$-closure of each $gamma$-open set gives $X$.
general-topology
asked Jun 13 '17 at 10:26
Birojit DasBirojit Das
179
$endgroup$
$begingroup$
You should state your question the body. Also, how do you define $γ$-closure? Is it true that every open set is $γ$-open?
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– user87690
Jun 13 '17 at 15:29
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Thanks for your comment. Let me tell you, I tried to solve it further and I found an example.
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– Birojit Das
Jun 15 '17 at 6:45
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You shouldn't have found an example. Since every open set is $γ$-open, every $γ$-hyperconnected space is hyperconnected, isn't it?
$endgroup$
– user87690
Jun 15 '17 at 19:04
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You are talking about topological space. In bitopology I found that example.
$endgroup$
– Birojit Das
Jun 18 '17 at 2:33
$begingroup$
You didn't mention bitopology in the question. Also, if you have just two unrelated topologies, it is trivial. You should first pose the question correctly.
$endgroup$
– user87690
Jun 18 '17 at 11:25
|
show 17 more comments
$begingroup$
In a topological space $(X,T)$ a subset A of X is said to be preopen if $A subset int(cl(A))$ and $gamma$-open if $A cap B$ is preopen for every preopen set B in $X.$ $(X, T)$ is said to be hyperconnected if closure of any open set gives $X$ and $gamma$-hyperconnected if $gamma$-closure of each $gamma$-open set gives $X$.
general-topology
asked Jun 13 '17 at 10:26
Birojit DasBirojit Das
179
$endgroup$
In a topological space $(X,T)$ a subset A of X is said to be preopen if $A subset int(cl(A))$ and $gamma$-open if $A cap B$ is preopen for every preopen set B in $X.$ $(X, T)$ is said to be hyperconnected if closure of any open set gives $X$ and $gamma$-hyperconnected if $gamma$-closure of each $gamma$-open set gives $X$.
general-topology
general-topology
asked Jun 13 '17 at 10:26
Birojit DasBirojit Das
179
asked Jun 13 '17 at 10:26
Birojit DasBirojit Das
179
asked Jun 13 '17 at 10:26
Birojit DasBirojit Das
179
asked Jun 13 '17 at 10:26
Birojit DasBirojit Das
179
asked Jun 13 '17 at 10:26
Birojit DasBirojit Das
179
179
$begingroup$
You should state your question the body. Also, how do you define $γ$-closure? Is it true that every open set is $γ$-open?
$endgroup$
– user87690
Jun 13 '17 at 15:29
$begingroup$
Thanks for your comment. Let me tell you, I tried to solve it further and I found an example.
$endgroup$
– Birojit Das
Jun 15 '17 at 6:45
$begingroup$
You shouldn't have found an example. Since every open set is $γ$-open, every $γ$-hyperconnected space is hyperconnected, isn't it?
$endgroup$
– user87690
Jun 15 '17 at 19:04
$begingroup$
You are talking about topological space. In bitopology I found that example.
$endgroup$
– Birojit Das
Jun 18 '17 at 2:33
$begingroup$
You didn't mention bitopology in the question. Also, if you have just two unrelated topologies, it is trivial. You should first pose the question correctly.
$endgroup$
– user87690
Jun 18 '17 at 11:25
|
show 17 more comments
$begingroup$
You should state your question the body. Also, how do you define $γ$-closure? Is it true that every open set is $γ$-open?
$endgroup$
– user87690
Jun 13 '17 at 15:29
$begingroup$
Thanks for your comment. Let me tell you, I tried to solve it further and I found an example.
$endgroup$
– Birojit Das
Jun 15 '17 at 6:45
$begingroup$
You shouldn't have found an example. Since every open set is $γ$-open, every $γ$-hyperconnected space is hyperconnected, isn't it?
$endgroup$
– user87690
Jun 15 '17 at 19:04
$begingroup$
You are talking about topological space. In bitopology I found that example.
$endgroup$
– Birojit Das
Jun 18 '17 at 2:33
$begingroup$
You didn't mention bitopology in the question. Also, if you have just two unrelated topologies, it is trivial. You should first pose the question correctly.
$endgroup$
– user87690
Jun 18 '17 at 11:25
$begingroup$
You should state your question the body. Also, how do you define $γ$-closure? Is it true that every open set is $γ$-open?
$endgroup$
– user87690
Jun 13 '17 at 15:29
$begingroup$
You should state your question the body. Also, how do you define $γ$-closure? Is it true that every open set is $γ$-open?
$endgroup$
– user87690
Jun 13 '17 at 15:29
$begingroup$
Thanks for your comment. Let me tell you, I tried to solve it further and I found an example.
$endgroup$
– Birojit Das
Jun 15 '17 at 6:45
$begingroup$
Thanks for your comment. Let me tell you, I tried to solve it further and I found an example.
$endgroup$
– Birojit Das
Jun 15 '17 at 6:45
$begingroup$
You shouldn't have found an example. Since every open set is $γ$-open, every $γ$-hyperconnected space is hyperconnected, isn't it?
$endgroup$
– user87690
Jun 15 '17 at 19:04
$begingroup$
You shouldn't have found an example. Since every open set is $γ$-open, every $γ$-hyperconnected space is hyperconnected, isn't it?
$endgroup$
– user87690
Jun 15 '17 at 19:04
$begingroup$
You are talking about topological space. In bitopology I found that example.
$endgroup$
– Birojit Das
Jun 18 '17 at 2:33
$begingroup$
You are talking about topological space. In bitopology I found that example.
$endgroup$
– Birojit Das
Jun 18 '17 at 2:33
$begingroup$
You didn't mention bitopology in the question. Also, if you have just two unrelated topologies, it is trivial. You should first pose the question correctly.
$endgroup$
– user87690
Jun 18 '17 at 11:25
$begingroup$
You didn't mention bitopology in the question. Also, if you have just two unrelated topologies, it is trivial. You should first pose the question correctly.
$endgroup$
– user87690
Jun 18 '17 at 11:25
|
show 17 more comments
1 Answer
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$begingroup$
Let me summarize the picture. We start with some topology $T$ on a set $X$. The topology induces a finer family of all preopen sets, say $pT$. This in general does not have to be a topology. So we may fix this by introducing the collection of all $γ$-open sets – $γT$. It seems to me that $γT$ is a topology.
In general, we may say that a set $X$ is $(T_1, T_2)$-hyperconnected (where $T_1$ and $T_2$ are topologies on $X$) if for every nonempty sets $U ∈ T_1$, $V ∈ T_2$ we have $U ∩ V ≠ ∅$. This way,
$(X, T)$ is hyperconnected if it is $(T, T)$-hyperconnected,
$(X, T)$ is $γ$-hyperconnected if it is $(γT, γT)$-hyperconnected,
- and in the comments we have also considered the condition of being $(T, γT)$-hyperconnected.
Since we have $Τ ⊆ γT ⊆ pT$, we have also the implications $γ$-hyperconnected $implies$ $(T, γT)$-hyperconnected $implies$ hyperconnected.
If $T$ is indiscrete, then both $pT$ and $γT$ are discrete. This shows $(T, γT)$-hyperconnected $notRightarrow$ $γ$-hyperconnected.
It turns out that $X$ is hyperconnected if and only if it is $(T, γT)$-hyperconnected. For every nonempty preopen set $A$ there is an open set $U$ such that $A ⊆ U$ and $A$ is dense in $U$. So if $X$ is hyperconnected, for every nonempty open set $V$ we have $U ∩ V ≠ ∅$, and so $A ∩ V ≠ ∅$ since $A$ is dense in $U$. That means $X$ is even $(T, pT)$-hyperconnected.
So it is really like this: $γ$-hyperconnected $implies$ hyperconnected, and the indiscrete space is a counterexample for the other implication.
answered Jan 24 at 11:59
user87690user87690
6,5761825
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$begingroup$
Let me summarize the picture. We start with some topology $T$ on a set $X$. The topology induces a finer family of all preopen sets, say $pT$. This in general does not have to be a topology. So we may fix this by introducing the collection of all $γ$-open sets – $γT$. It seems to me that $γT$ is a topology.
In general, we may say that a set $X$ is $(T_1, T_2)$-hyperconnected (where $T_1$ and $T_2$ are topologies on $X$) if for every nonempty sets $U ∈ T_1$, $V ∈ T_2$ we have $U ∩ V ≠ ∅$. This way,
$(X, T)$ is hyperconnected if it is $(T, T)$-hyperconnected,
$(X, T)$ is $γ$-hyperconnected if it is $(γT, γT)$-hyperconnected,
- and in the comments we have also considered the condition of being $(T, γT)$-hyperconnected.
Since we have $Τ ⊆ γT ⊆ pT$, we have also the implications $γ$-hyperconnected $implies$ $(T, γT)$-hyperconnected $implies$ hyperconnected.
If $T$ is indiscrete, then both $pT$ and $γT$ are discrete. This shows $(T, γT)$-hyperconnected $notRightarrow$ $γ$-hyperconnected.
It turns out that $X$ is hyperconnected if and only if it is $(T, γT)$-hyperconnected. For every nonempty preopen set $A$ there is an open set $U$ such that $A ⊆ U$ and $A$ is dense in $U$. So if $X$ is hyperconnected, for every nonempty open set $V$ we have $U ∩ V ≠ ∅$, and so $A ∩ V ≠ ∅$ since $A$ is dense in $U$. That means $X$ is even $(T, pT)$-hyperconnected.
So it is really like this: $γ$-hyperconnected $implies$ hyperconnected, and the indiscrete space is a counterexample for the other implication.
answered Jan 24 at 11:59
user87690user87690
6,5761825
$endgroup$
add a comment |
$begingroup$
Let me summarize the picture. We start with some topology $T$ on a set $X$. The topology induces a finer family of all preopen sets, say $pT$. This in general does not have to be a topology. So we may fix this by introducing the collection of all $γ$-open sets – $γT$. It seems to me that $γT$ is a topology.
In general, we may say that a set $X$ is $(T_1, T_2)$-hyperconnected (where $T_1$ and $T_2$ are topologies on $X$) if for every nonempty sets $U ∈ T_1$, $V ∈ T_2$ we have $U ∩ V ≠ ∅$. This way,
$(X, T)$ is hyperconnected if it is $(T, T)$-hyperconnected,
$(X, T)$ is $γ$-hyperconnected if it is $(γT, γT)$-hyperconnected,
- and in the comments we have also considered the condition of being $(T, γT)$-hyperconnected.
Since we have $Τ ⊆ γT ⊆ pT$, we have also the implications $γ$-hyperconnected $implies$ $(T, γT)$-hyperconnected $implies$ hyperconnected.
If $T$ is indiscrete, then both $pT$ and $γT$ are discrete. This shows $(T, γT)$-hyperconnected $notRightarrow$ $γ$-hyperconnected.
It turns out that $X$ is hyperconnected if and only if it is $(T, γT)$-hyperconnected. For every nonempty preopen set $A$ there is an open set $U$ such that $A ⊆ U$ and $A$ is dense in $U$. So if $X$ is hyperconnected, for every nonempty open set $V$ we have $U ∩ V ≠ ∅$, and so $A ∩ V ≠ ∅$ since $A$ is dense in $U$. That means $X$ is even $(T, pT)$-hyperconnected.
So it is really like this: $γ$-hyperconnected $implies$ hyperconnected, and the indiscrete space is a counterexample for the other implication.
answered Jan 24 at 11:59
user87690user87690
6,5761825
$endgroup$
add a comment |
$begingroup$
Let me summarize the picture. We start with some topology $T$ on a set $X$. The topology induces a finer family of all preopen sets, say $pT$. This in general does not have to be a topology. So we may fix this by introducing the collection of all $γ$-open sets – $γT$. It seems to me that $γT$ is a topology.
In general, we may say that a set $X$ is $(T_1, T_2)$-hyperconnected (where $T_1$ and $T_2$ are topologies on $X$) if for every nonempty sets $U ∈ T_1$, $V ∈ T_2$ we have $U ∩ V ≠ ∅$. This way,
$(X, T)$ is hyperconnected if it is $(T, T)$-hyperconnected,
$(X, T)$ is $γ$-hyperconnected if it is $(γT, γT)$-hyperconnected,
- and in the comments we have also considered the condition of being $(T, γT)$-hyperconnected.
Since we have $Τ ⊆ γT ⊆ pT$, we have also the implications $γ$-hyperconnected $implies$ $(T, γT)$-hyperconnected $implies$ hyperconnected.
If $T$ is indiscrete, then both $pT$ and $γT$ are discrete. This shows $(T, γT)$-hyperconnected $notRightarrow$ $γ$-hyperconnected.
It turns out that $X$ is hyperconnected if and only if it is $(T, γT)$-hyperconnected. For every nonempty preopen set $A$ there is an open set $U$ such that $A ⊆ U$ and $A$ is dense in $U$. So if $X$ is hyperconnected, for every nonempty open set $V$ we have $U ∩ V ≠ ∅$, and so $A ∩ V ≠ ∅$ since $A$ is dense in $U$. That means $X$ is even $(T, pT)$-hyperconnected.
So it is really like this: $γ$-hyperconnected $implies$ hyperconnected, and the indiscrete space is a counterexample for the other implication.
answered Jan 24 at 11:59
user87690user87690
6,5761825
$endgroup$
Let me summarize the picture. We start with some topology $T$ on a set $X$. The topology induces a finer family of all preopen sets, say $pT$. This in general does not have to be a topology. So we may fix this by introducing the collection of all $γ$-open sets – $γT$. It seems to me that $γT$ is a topology.
In general, we may say that a set $X$ is $(T_1, T_2)$-hyperconnected (where $T_1$ and $T_2$ are topologies on $X$) if for every nonempty sets $U ∈ T_1$, $V ∈ T_2$ we have $U ∩ V ≠ ∅$. This way,
$(X, T)$ is hyperconnected if it is $(T, T)$-hyperconnected,
$(X, T)$ is $γ$-hyperconnected if it is $(γT, γT)$-hyperconnected,
- and in the comments we have also considered the condition of being $(T, γT)$-hyperconnected.
Since we have $Τ ⊆ γT ⊆ pT$, we have also the implications $γ$-hyperconnected $implies$ $(T, γT)$-hyperconnected $implies$ hyperconnected.
If $T$ is indiscrete, then both $pT$ and $γT$ are discrete. This shows $(T, γT)$-hyperconnected $notRightarrow$ $γ$-hyperconnected.
It turns out that $X$ is hyperconnected if and only if it is $(T, γT)$-hyperconnected. For every nonempty preopen set $A$ there is an open set $U$ such that $A ⊆ U$ and $A$ is dense in $U$. So if $X$ is hyperconnected, for every nonempty open set $V$ we have $U ∩ V ≠ ∅$, and so $A ∩ V ≠ ∅$ since $A$ is dense in $U$. That means $X$ is even $(T, pT)$-hyperconnected.
So it is really like this: $γ$-hyperconnected $implies$ hyperconnected, and the indiscrete space is a counterexample for the other implication.
answered Jan 24 at 11:59
user87690user87690
6,5761825
answered Jan 24 at 11:59
user87690user87690
6,5761825
answered Jan 24 at 11:59
user87690user87690
6,5761825
answered Jan 24 at 11:59
user87690user87690
6,5761825
6,5761825
add a comment |
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$begingroup$
You should state your question the body. Also, how do you define $γ$-closure? Is it true that every open set is $γ$-open?
$endgroup$
– user87690
Jun 13 '17 at 15:29
$begingroup$
Thanks for your comment. Let me tell you, I tried to solve it further and I found an example.
$endgroup$
– Birojit Das
Jun 15 '17 at 6:45
$begingroup$
You shouldn't have found an example. Since every open set is $γ$-open, every $γ$-hyperconnected space is hyperconnected, isn't it?
$endgroup$
– user87690
Jun 15 '17 at 19:04
$begingroup$
You are talking about topological space. In bitopology I found that example.
$endgroup$
– Birojit Das
Jun 18 '17 at 2:33
$begingroup$
You didn't mention bitopology in the question. Also, if you have just two unrelated topologies, it is trivial. You should first pose the question correctly.
$endgroup$
– user87690
Jun 18 '17 at 11:25