Connected topologies on $mathbb{R}$ strictly between the usual topology and the lower-limit topology
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It is well-known that the usual order/metric topology on $mathbb{R}$ is connected, and the lower-limit topology is not connected (it is even totally disconnected). We also know that the lower-limit topology is strictly finer than the usual topology.
Are there connected topologies on $mathbb{R}$ strictly between these two? (That is, is there is a connected topology on $mathbb{R}$ which is strictly finer than the usual topology, but coarser than the lower limit topology?)
I know that given any lower-limit basic open set $[a,b)$ (for $a < b$) the topology generated by the subbase consisting of $[a,b)$ and all of the usual open sets is not connected (because $[a,+infty) = [a,b) cup ( frac{a+b}{2} , + infty )$ and $mathbb{R} setminus [a,+infty) = (-infty , a )$ are both open in this topology). But perhaps there are more complicated lower-limit-open sets that can be added to yield a connected topology.
Definitions
A topological space $X$ is connected if the only subsets of $X$ that are clopen (closed and open) are $emptyset$ and $X$.
The lower-limit topology on $mathbb{R}$ is the topology generated by the base ${ [a,b) : a,b in mathbb{R} , a < b }$.
general-topology sorgenfrey-line
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add a comment |
$begingroup$
It is well-known that the usual order/metric topology on $mathbb{R}$ is connected, and the lower-limit topology is not connected (it is even totally disconnected). We also know that the lower-limit topology is strictly finer than the usual topology.
Are there connected topologies on $mathbb{R}$ strictly between these two? (That is, is there is a connected topology on $mathbb{R}$ which is strictly finer than the usual topology, but coarser than the lower limit topology?)
I know that given any lower-limit basic open set $[a,b)$ (for $a < b$) the topology generated by the subbase consisting of $[a,b)$ and all of the usual open sets is not connected (because $[a,+infty) = [a,b) cup ( frac{a+b}{2} , + infty )$ and $mathbb{R} setminus [a,+infty) = (-infty , a )$ are both open in this topology). But perhaps there are more complicated lower-limit-open sets that can be added to yield a connected topology.
Definitions
A topological space $X$ is connected if the only subsets of $X$ that are clopen (closed and open) are $emptyset$ and $X$.
The lower-limit topology on $mathbb{R}$ is the topology generated by the base ${ [a,b) : a,b in mathbb{R} , a < b }$.
general-topology sorgenfrey-line
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You are right that it doesn't work for $[a,b)$. But what if you add $U=(-infty, -1)cup[0,infty)$ to the standard topology? I'm writing this is as a comment because I'm not 100% sure if it works but it looks so.
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– freakish
Dec 6 '18 at 9:38
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@freakish Essentially the same problem. If your $U$ is open in the new topology, then so is $U cap ( frac{-1}{2} , +infty ) = [0,+infty)$, and clearly $mathbb{R} setminus [0,+infty) = ( - infty , 0 )$ is also open.
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– stochastic randomness
Dec 6 '18 at 9:42
$begingroup$
Ah yes, you're right.
$endgroup$
– freakish
Dec 6 '18 at 9:45
add a comment |
$begingroup$
It is well-known that the usual order/metric topology on $mathbb{R}$ is connected, and the lower-limit topology is not connected (it is even totally disconnected). We also know that the lower-limit topology is strictly finer than the usual topology.
Are there connected topologies on $mathbb{R}$ strictly between these two? (That is, is there is a connected topology on $mathbb{R}$ which is strictly finer than the usual topology, but coarser than the lower limit topology?)
I know that given any lower-limit basic open set $[a,b)$ (for $a < b$) the topology generated by the subbase consisting of $[a,b)$ and all of the usual open sets is not connected (because $[a,+infty) = [a,b) cup ( frac{a+b}{2} , + infty )$ and $mathbb{R} setminus [a,+infty) = (-infty , a )$ are both open in this topology). But perhaps there are more complicated lower-limit-open sets that can be added to yield a connected topology.
Definitions
A topological space $X$ is connected if the only subsets of $X$ that are clopen (closed and open) are $emptyset$ and $X$.
The lower-limit topology on $mathbb{R}$ is the topology generated by the base ${ [a,b) : a,b in mathbb{R} , a < b }$.
general-topology sorgenfrey-line
$endgroup$
It is well-known that the usual order/metric topology on $mathbb{R}$ is connected, and the lower-limit topology is not connected (it is even totally disconnected). We also know that the lower-limit topology is strictly finer than the usual topology.
Are there connected topologies on $mathbb{R}$ strictly between these two? (That is, is there is a connected topology on $mathbb{R}$ which is strictly finer than the usual topology, but coarser than the lower limit topology?)
I know that given any lower-limit basic open set $[a,b)$ (for $a < b$) the topology generated by the subbase consisting of $[a,b)$ and all of the usual open sets is not connected (because $[a,+infty) = [a,b) cup ( frac{a+b}{2} , + infty )$ and $mathbb{R} setminus [a,+infty) = (-infty , a )$ are both open in this topology). But perhaps there are more complicated lower-limit-open sets that can be added to yield a connected topology.
Definitions
A topological space $X$ is connected if the only subsets of $X$ that are clopen (closed and open) are $emptyset$ and $X$.
The lower-limit topology on $mathbb{R}$ is the topology generated by the base ${ [a,b) : a,b in mathbb{R} , a < b }$.
general-topology sorgenfrey-line
general-topology sorgenfrey-line
asked Dec 6 '18 at 9:29
stochastic randomnessstochastic randomness
41017
41017
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You are right that it doesn't work for $[a,b)$. But what if you add $U=(-infty, -1)cup[0,infty)$ to the standard topology? I'm writing this is as a comment because I'm not 100% sure if it works but it looks so.
$endgroup$
– freakish
Dec 6 '18 at 9:38
$begingroup$
@freakish Essentially the same problem. If your $U$ is open in the new topology, then so is $U cap ( frac{-1}{2} , +infty ) = [0,+infty)$, and clearly $mathbb{R} setminus [0,+infty) = ( - infty , 0 )$ is also open.
$endgroup$
– stochastic randomness
Dec 6 '18 at 9:42
$begingroup$
Ah yes, you're right.
$endgroup$
– freakish
Dec 6 '18 at 9:45
add a comment |
$begingroup$
You are right that it doesn't work for $[a,b)$. But what if you add $U=(-infty, -1)cup[0,infty)$ to the standard topology? I'm writing this is as a comment because I'm not 100% sure if it works but it looks so.
$endgroup$
– freakish
Dec 6 '18 at 9:38
$begingroup$
@freakish Essentially the same problem. If your $U$ is open in the new topology, then so is $U cap ( frac{-1}{2} , +infty ) = [0,+infty)$, and clearly $mathbb{R} setminus [0,+infty) = ( - infty , 0 )$ is also open.
$endgroup$
– stochastic randomness
Dec 6 '18 at 9:42
$begingroup$
Ah yes, you're right.
$endgroup$
– freakish
Dec 6 '18 at 9:45
$begingroup$
You are right that it doesn't work for $[a,b)$. But what if you add $U=(-infty, -1)cup[0,infty)$ to the standard topology? I'm writing this is as a comment because I'm not 100% sure if it works but it looks so.
$endgroup$
– freakish
Dec 6 '18 at 9:38
$begingroup$
You are right that it doesn't work for $[a,b)$. But what if you add $U=(-infty, -1)cup[0,infty)$ to the standard topology? I'm writing this is as a comment because I'm not 100% sure if it works but it looks so.
$endgroup$
– freakish
Dec 6 '18 at 9:38
$begingroup$
@freakish Essentially the same problem. If your $U$ is open in the new topology, then so is $U cap ( frac{-1}{2} , +infty ) = [0,+infty)$, and clearly $mathbb{R} setminus [0,+infty) = ( - infty , 0 )$ is also open.
$endgroup$
– stochastic randomness
Dec 6 '18 at 9:42
$begingroup$
@freakish Essentially the same problem. If your $U$ is open in the new topology, then so is $U cap ( frac{-1}{2} , +infty ) = [0,+infty)$, and clearly $mathbb{R} setminus [0,+infty) = ( - infty , 0 )$ is also open.
$endgroup$
– stochastic randomness
Dec 6 '18 at 9:42
$begingroup$
Ah yes, you're right.
$endgroup$
– freakish
Dec 6 '18 at 9:45
$begingroup$
Ah yes, you're right.
$endgroup$
– freakish
Dec 6 '18 at 9:45
add a comment |
1 Answer
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Are there connected topologies on $mathbb{R}$ strictly between these two?
Yes. For instance, let $sigma$ be a topology on $Bbb R$ generated by its standard topology $tau$ and a set $S=Bbb Rsetminus{-frac 1n:ninBbb N}$. The space $(Bbb R,sigma)$ is connected because $operatorname{int}_tau A=operatorname{int}_sigma A$ for each closed subset $A$ of $(Bbb R,sigma)$.
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add a comment |
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$begingroup$
Are there connected topologies on $mathbb{R}$ strictly between these two?
Yes. For instance, let $sigma$ be a topology on $Bbb R$ generated by its standard topology $tau$ and a set $S=Bbb Rsetminus{-frac 1n:ninBbb N}$. The space $(Bbb R,sigma)$ is connected because $operatorname{int}_tau A=operatorname{int}_sigma A$ for each closed subset $A$ of $(Bbb R,sigma)$.
$endgroup$
add a comment |
$begingroup$
Are there connected topologies on $mathbb{R}$ strictly between these two?
Yes. For instance, let $sigma$ be a topology on $Bbb R$ generated by its standard topology $tau$ and a set $S=Bbb Rsetminus{-frac 1n:ninBbb N}$. The space $(Bbb R,sigma)$ is connected because $operatorname{int}_tau A=operatorname{int}_sigma A$ for each closed subset $A$ of $(Bbb R,sigma)$.
$endgroup$
add a comment |
$begingroup$
Are there connected topologies on $mathbb{R}$ strictly between these two?
Yes. For instance, let $sigma$ be a topology on $Bbb R$ generated by its standard topology $tau$ and a set $S=Bbb Rsetminus{-frac 1n:ninBbb N}$. The space $(Bbb R,sigma)$ is connected because $operatorname{int}_tau A=operatorname{int}_sigma A$ for each closed subset $A$ of $(Bbb R,sigma)$.
$endgroup$
Are there connected topologies on $mathbb{R}$ strictly between these two?
Yes. For instance, let $sigma$ be a topology on $Bbb R$ generated by its standard topology $tau$ and a set $S=Bbb Rsetminus{-frac 1n:ninBbb N}$. The space $(Bbb R,sigma)$ is connected because $operatorname{int}_tau A=operatorname{int}_sigma A$ for each closed subset $A$ of $(Bbb R,sigma)$.
answered Dec 7 '18 at 21:52
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Alex RavskyAlex Ravsky
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$begingroup$
You are right that it doesn't work for $[a,b)$. But what if you add $U=(-infty, -1)cup[0,infty)$ to the standard topology? I'm writing this is as a comment because I'm not 100% sure if it works but it looks so.
$endgroup$
– freakish
Dec 6 '18 at 9:38
$begingroup$
@freakish Essentially the same problem. If your $U$ is open in the new topology, then so is $U cap ( frac{-1}{2} , +infty ) = [0,+infty)$, and clearly $mathbb{R} setminus [0,+infty) = ( - infty , 0 )$ is also open.
$endgroup$
– stochastic randomness
Dec 6 '18 at 9:42
$begingroup$
Ah yes, you're right.
$endgroup$
– freakish
Dec 6 '18 at 9:45