Interior of a connected set is connected
$begingroup$
I know interior of connected set in a metric space need not be connected. Simplest example would be to take two tangent closed disk in Euclidean plane.
I am trying to construct a counterexample in $mathbb{R}$? I mean I am trying to find a connected set in $mathbb{R}$ such that interior is not connected.
Thanks for reading, and helping out.
general-topology
asked Jan 9 at 5:51
StammeringMathematicianStammeringMathematician
2,7211324
$endgroup$
add a comment |
$begingroup$
I know interior of connected set in a metric space need not be connected. Simplest example would be to take two tangent closed disk in Euclidean plane.
I am trying to construct a counterexample in $mathbb{R}$? I mean I am trying to find a connected set in $mathbb{R}$ such that interior is not connected.
Thanks for reading, and helping out.
general-topology
asked Jan 9 at 5:51
StammeringMathematicianStammeringMathematician
2,7211324
$endgroup$
2
$begingroup$
In $mathbb{R}$ this is not possible because the connected sets are the intervals.
$endgroup$
– Ian
Jan 9 at 5:54
2
$begingroup$
The only connected set on this case are intervals. Does that help?
$endgroup$
– Michael Hoppe
Jan 9 at 5:55
add a comment |
$begingroup$
I know interior of connected set in a metric space need not be connected. Simplest example would be to take two tangent closed disk in Euclidean plane.
I am trying to construct a counterexample in $mathbb{R}$? I mean I am trying to find a connected set in $mathbb{R}$ such that interior is not connected.
Thanks for reading, and helping out.
general-topology
asked Jan 9 at 5:51
StammeringMathematicianStammeringMathematician
2,7211324
$endgroup$
I know interior of connected set in a metric space need not be connected. Simplest example would be to take two tangent closed disk in Euclidean plane.
I am trying to construct a counterexample in $mathbb{R}$? I mean I am trying to find a connected set in $mathbb{R}$ such that interior is not connected.
Thanks for reading, and helping out.
general-topology
general-topology
asked Jan 9 at 5:51
StammeringMathematicianStammeringMathematician
2,7211324
asked Jan 9 at 5:51
StammeringMathematicianStammeringMathematician
2,7211324
asked Jan 9 at 5:51
StammeringMathematicianStammeringMathematician
2,7211324
asked Jan 9 at 5:51
StammeringMathematicianStammeringMathematician
2,7211324
asked Jan 9 at 5:51
StammeringMathematicianStammeringMathematician
2,7211324
2,7211324
2
$begingroup$
In $mathbb{R}$ this is not possible because the connected sets are the intervals.
$endgroup$
– Ian
Jan 9 at 5:54
2
$begingroup$
The only connected set on this case are intervals. Does that help?
$endgroup$
– Michael Hoppe
Jan 9 at 5:55
add a comment |
2
$begingroup$
In $mathbb{R}$ this is not possible because the connected sets are the intervals.
$endgroup$
– Ian
Jan 9 at 5:54
2
$begingroup$
The only connected set on this case are intervals. Does that help?
$endgroup$
– Michael Hoppe
Jan 9 at 5:55
2
2
$begingroup$
In $mathbb{R}$ this is not possible because the connected sets are the intervals.
$endgroup$
– Ian
Jan 9 at 5:54
$begingroup$
In $mathbb{R}$ this is not possible because the connected sets are the intervals.
$endgroup$
– Ian
Jan 9 at 5:54
2
2
$begingroup$
The only connected set on this case are intervals. Does that help?
$endgroup$
– Michael Hoppe
Jan 9 at 5:55
$begingroup$
The only connected set on this case are intervals. Does that help?
$endgroup$
– Michael Hoppe
Jan 9 at 5:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In $mathbb{R}$ a set is connected if and only if it is path connected. It follows that the only connected subsets are the intervals $Isubseteq mathbb{R}$. The interior of a closed interval $[a,b]$ is $(a,b)$, which is again connected. The other cases are the same.
So, every connected subset of $mathbb{R}$ has connected interior.
answered Jan 9 at 5:57
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
$begingroup$
What if the connected set if ${a}$? Is the empty set connected? Some people define "connected" as "having exactly one component".
$endgroup$
– bof
Jan 9 at 8:02
$begingroup$
I suppose that is an edge case : $X={a}$ is connected because the only way to write it as a disjoint union of its subsets is $X=Xsqcup varnothing$. The empty-set is connected for the same reason : $varnothing = Asqcup B $ for $A,Bsubseteq varnothing$ implies $A=B=varnothing$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 9 at 16:40
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In $mathbb{R}$ a set is connected if and only if it is path connected. It follows that the only connected subsets are the intervals $Isubseteq mathbb{R}$. The interior of a closed interval $[a,b]$ is $(a,b)$, which is again connected. The other cases are the same.
So, every connected subset of $mathbb{R}$ has connected interior.
answered Jan 9 at 5:57
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
$begingroup$
What if the connected set if ${a}$? Is the empty set connected? Some people define "connected" as "having exactly one component".
$endgroup$
– bof
Jan 9 at 8:02
$begingroup$
I suppose that is an edge case : $X={a}$ is connected because the only way to write it as a disjoint union of its subsets is $X=Xsqcup varnothing$. The empty-set is connected for the same reason : $varnothing = Asqcup B $ for $A,Bsubseteq varnothing$ implies $A=B=varnothing$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 9 at 16:40
add a comment |
$begingroup$
In $mathbb{R}$ a set is connected if and only if it is path connected. It follows that the only connected subsets are the intervals $Isubseteq mathbb{R}$. The interior of a closed interval $[a,b]$ is $(a,b)$, which is again connected. The other cases are the same.
So, every connected subset of $mathbb{R}$ has connected interior.
answered Jan 9 at 5:57
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
$begingroup$
What if the connected set if ${a}$? Is the empty set connected? Some people define "connected" as "having exactly one component".
$endgroup$
– bof
Jan 9 at 8:02
$begingroup$
I suppose that is an edge case : $X={a}$ is connected because the only way to write it as a disjoint union of its subsets is $X=Xsqcup varnothing$. The empty-set is connected for the same reason : $varnothing = Asqcup B $ for $A,Bsubseteq varnothing$ implies $A=B=varnothing$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 9 at 16:40
add a comment |
$begingroup$
In $mathbb{R}$ a set is connected if and only if it is path connected. It follows that the only connected subsets are the intervals $Isubseteq mathbb{R}$. The interior of a closed interval $[a,b]$ is $(a,b)$, which is again connected. The other cases are the same.
So, every connected subset of $mathbb{R}$ has connected interior.
answered Jan 9 at 5:57
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
In $mathbb{R}$ a set is connected if and only if it is path connected. It follows that the only connected subsets are the intervals $Isubseteq mathbb{R}$. The interior of a closed interval $[a,b]$ is $(a,b)$, which is again connected. The other cases are the same.
So, every connected subset of $mathbb{R}$ has connected interior.
answered Jan 9 at 5:57
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 9 at 5:57
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 9 at 5:57
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 9 at 5:57
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
10.6k41741
$begingroup$
What if the connected set if ${a}$? Is the empty set connected? Some people define "connected" as "having exactly one component".
$endgroup$
– bof
Jan 9 at 8:02
$begingroup$
I suppose that is an edge case : $X={a}$ is connected because the only way to write it as a disjoint union of its subsets is $X=Xsqcup varnothing$. The empty-set is connected for the same reason : $varnothing = Asqcup B $ for $A,Bsubseteq varnothing$ implies $A=B=varnothing$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 9 at 16:40
add a comment |
$begingroup$
What if the connected set if ${a}$? Is the empty set connected? Some people define "connected" as "having exactly one component".
$endgroup$
– bof
Jan 9 at 8:02
$begingroup$
I suppose that is an edge case : $X={a}$ is connected because the only way to write it as a disjoint union of its subsets is $X=Xsqcup varnothing$. The empty-set is connected for the same reason : $varnothing = Asqcup B $ for $A,Bsubseteq varnothing$ implies $A=B=varnothing$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 9 at 16:40
$begingroup$
What if the connected set if ${a}$? Is the empty set connected? Some people define "connected" as "having exactly one component".
$endgroup$
– bof
Jan 9 at 8:02
$begingroup$
What if the connected set if ${a}$? Is the empty set connected? Some people define "connected" as "having exactly one component".
$endgroup$
– bof
Jan 9 at 8:02
$begingroup$
I suppose that is an edge case : $X={a}$ is connected because the only way to write it as a disjoint union of its subsets is $X=Xsqcup varnothing$. The empty-set is connected for the same reason : $varnothing = Asqcup B $ for $A,Bsubseteq varnothing$ implies $A=B=varnothing$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 9 at 16:40
$begingroup$
I suppose that is an edge case : $X={a}$ is connected because the only way to write it as a disjoint union of its subsets is $X=Xsqcup varnothing$. The empty-set is connected for the same reason : $varnothing = Asqcup B $ for $A,Bsubseteq varnothing$ implies $A=B=varnothing$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 9 at 16:40
add a comment |
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2
$begingroup$
In $mathbb{R}$ this is not possible because the connected sets are the intervals.
$endgroup$
– Ian
Jan 9 at 5:54
2
$begingroup$
The only connected set on this case are intervals. Does that help?
$endgroup$
– Michael Hoppe
Jan 9 at 5:55