Intervals are connected and the only connected sets in $mathbb{R}$
$begingroup$
As the topic, prove that Intervals are connected and only connected in $mathbb{R}$. I know what is the definition of connected set. But not sure how to prove that.
general-topology
asked Nov 17 '12 at 5:54
MathematicsMathematics
2,20322146
$endgroup$
add a comment |
$begingroup$
As the topic, prove that Intervals are connected and only connected in $mathbb{R}$. I know what is the definition of connected set. But not sure how to prove that.
general-topology
asked Nov 17 '12 at 5:54
MathematicsMathematics
2,20322146
$endgroup$
1
$begingroup$
Maybe some of the discussion here is helpful? Set $A$ interval in $mathbb{R}implies$ connected
$endgroup$
– yunone
Nov 17 '12 at 5:56
1
$begingroup$
In Principles of Mathematical Analysis by Rudin, a proof for the same is given.
$endgroup$
– Gautam Shenoy
Nov 17 '12 at 5:58
add a comment |
$begingroup$
As the topic, prove that Intervals are connected and only connected in $mathbb{R}$. I know what is the definition of connected set. But not sure how to prove that.
general-topology
asked Nov 17 '12 at 5:54
MathematicsMathematics
2,20322146
$endgroup$
As the topic, prove that Intervals are connected and only connected in $mathbb{R}$. I know what is the definition of connected set. But not sure how to prove that.
general-topology
general-topology
asked Nov 17 '12 at 5:54
MathematicsMathematics
2,20322146
asked Nov 17 '12 at 5:54
MathematicsMathematics
2,20322146
asked Nov 17 '12 at 5:54
MathematicsMathematics
2,20322146
asked Nov 17 '12 at 5:54
MathematicsMathematics
2,20322146
asked Nov 17 '12 at 5:54
MathematicsMathematics
2,20322146
2,20322146
1
$begingroup$
Maybe some of the discussion here is helpful? Set $A$ interval in $mathbb{R}implies$ connected
$endgroup$
– yunone
Nov 17 '12 at 5:56
1
$begingroup$
In Principles of Mathematical Analysis by Rudin, a proof for the same is given.
$endgroup$
– Gautam Shenoy
Nov 17 '12 at 5:58
add a comment |
1
$begingroup$
Maybe some of the discussion here is helpful? Set $A$ interval in $mathbb{R}implies$ connected
$endgroup$
– yunone
Nov 17 '12 at 5:56
1
$begingroup$
In Principles of Mathematical Analysis by Rudin, a proof for the same is given.
$endgroup$
– Gautam Shenoy
Nov 17 '12 at 5:58
1
1
$begingroup$
Maybe some of the discussion here is helpful? Set $A$ interval in $mathbb{R}implies$ connected
$endgroup$
– yunone
Nov 17 '12 at 5:56
$begingroup$
Maybe some of the discussion here is helpful? Set $A$ interval in $mathbb{R}implies$ connected
$endgroup$
– yunone
Nov 17 '12 at 5:56
1
1
$begingroup$
In Principles of Mathematical Analysis by Rudin, a proof for the same is given.
$endgroup$
– Gautam Shenoy
Nov 17 '12 at 5:58
$begingroup$
In Principles of Mathematical Analysis by Rudin, a proof for the same is given.
$endgroup$
– Gautam Shenoy
Nov 17 '12 at 5:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$newcommand{cl}{operatorname{cl}}$HINTS: Suppose that $AsubseteqBbb R$ is not an interval; then there are points $a,bin A$ and $xinBbb Rsetminus A$ such that $a<x<b$. Use the sets $Acap(leftarrow,x)$ and $Acap(x,to)$ to show that $A$ is not connected.
The other direction is a bit harder. Suppose that $A$ is not connected. Then there is an open set $U$ in $Bbb R$ such that $Acap Unevarnothingne Asetminus U$ and $Acap U=
Acapcl U$; why? Fix $ain Acap U$ and $bin Asetminus U$ and show that $[a,b]nsubseteq A$, so that $A$ cannot be an interval.
answered Nov 17 '12 at 6:04
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
$begingroup$
Actually is it true that if two sets are disjoint, no matter it is open or not it is disconnected. As by the definition, it seems to requires that 2 sets are open
$endgroup$
– Mathematics
Nov 17 '12 at 6:21
$begingroup$
@Mathematics: $[0,1)$ and $[1,2]$ are disjoint, but their union is $[0,2]$, which is connected. It is important that $Acap U$ and $Asetminus U$ are both relatively open subsets of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:23
$begingroup$
$Acup Uneemptysetne Asetminus Uimplies Acup U ne Asetminus U$?I am a little bit confused with the notation
$endgroup$
– Mathematics
Nov 17 '12 at 6:37
$begingroup$
@Mathematics: No, it doesn’t. But it’s true that $Acup Une Asetminus U$, simply because $Unevarnothing$. I think that you’re making this more complicated than it really is; all I’m saying is that if $A$ is not connected, there must be an open $U$ in $Bbb R$ such that $Acap U$ and $Asetminuscl U$ are non-empty relatively open subsets of $A$ whose union is $A$ $-$ in other words, a disconnection of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:40
add a comment |
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1 Answer
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$begingroup$
$newcommand{cl}{operatorname{cl}}$HINTS: Suppose that $AsubseteqBbb R$ is not an interval; then there are points $a,bin A$ and $xinBbb Rsetminus A$ such that $a<x<b$. Use the sets $Acap(leftarrow,x)$ and $Acap(x,to)$ to show that $A$ is not connected.
The other direction is a bit harder. Suppose that $A$ is not connected. Then there is an open set $U$ in $Bbb R$ such that $Acap Unevarnothingne Asetminus U$ and $Acap U=
Acapcl U$; why? Fix $ain Acap U$ and $bin Asetminus U$ and show that $[a,b]nsubseteq A$, so that $A$ cannot be an interval.
answered Nov 17 '12 at 6:04
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
$begingroup$
Actually is it true that if two sets are disjoint, no matter it is open or not it is disconnected. As by the definition, it seems to requires that 2 sets are open
$endgroup$
– Mathematics
Nov 17 '12 at 6:21
$begingroup$
@Mathematics: $[0,1)$ and $[1,2]$ are disjoint, but their union is $[0,2]$, which is connected. It is important that $Acap U$ and $Asetminus U$ are both relatively open subsets of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:23
$begingroup$
$Acup Uneemptysetne Asetminus Uimplies Acup U ne Asetminus U$?I am a little bit confused with the notation
$endgroup$
– Mathematics
Nov 17 '12 at 6:37
$begingroup$
@Mathematics: No, it doesn’t. But it’s true that $Acup Une Asetminus U$, simply because $Unevarnothing$. I think that you’re making this more complicated than it really is; all I’m saying is that if $A$ is not connected, there must be an open $U$ in $Bbb R$ such that $Acap U$ and $Asetminuscl U$ are non-empty relatively open subsets of $A$ whose union is $A$ $-$ in other words, a disconnection of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:40
add a comment |
$begingroup$
$newcommand{cl}{operatorname{cl}}$HINTS: Suppose that $AsubseteqBbb R$ is not an interval; then there are points $a,bin A$ and $xinBbb Rsetminus A$ such that $a<x<b$. Use the sets $Acap(leftarrow,x)$ and $Acap(x,to)$ to show that $A$ is not connected.
The other direction is a bit harder. Suppose that $A$ is not connected. Then there is an open set $U$ in $Bbb R$ such that $Acap Unevarnothingne Asetminus U$ and $Acap U=
Acapcl U$; why? Fix $ain Acap U$ and $bin Asetminus U$ and show that $[a,b]nsubseteq A$, so that $A$ cannot be an interval.
answered Nov 17 '12 at 6:04
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
$begingroup$
Actually is it true that if two sets are disjoint, no matter it is open or not it is disconnected. As by the definition, it seems to requires that 2 sets are open
$endgroup$
– Mathematics
Nov 17 '12 at 6:21
$begingroup$
@Mathematics: $[0,1)$ and $[1,2]$ are disjoint, but their union is $[0,2]$, which is connected. It is important that $Acap U$ and $Asetminus U$ are both relatively open subsets of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:23
$begingroup$
$Acup Uneemptysetne Asetminus Uimplies Acup U ne Asetminus U$?I am a little bit confused with the notation
$endgroup$
– Mathematics
Nov 17 '12 at 6:37
$begingroup$
@Mathematics: No, it doesn’t. But it’s true that $Acup Une Asetminus U$, simply because $Unevarnothing$. I think that you’re making this more complicated than it really is; all I’m saying is that if $A$ is not connected, there must be an open $U$ in $Bbb R$ such that $Acap U$ and $Asetminuscl U$ are non-empty relatively open subsets of $A$ whose union is $A$ $-$ in other words, a disconnection of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:40
add a comment |
$begingroup$
$newcommand{cl}{operatorname{cl}}$HINTS: Suppose that $AsubseteqBbb R$ is not an interval; then there are points $a,bin A$ and $xinBbb Rsetminus A$ such that $a<x<b$. Use the sets $Acap(leftarrow,x)$ and $Acap(x,to)$ to show that $A$ is not connected.
The other direction is a bit harder. Suppose that $A$ is not connected. Then there is an open set $U$ in $Bbb R$ such that $Acap Unevarnothingne Asetminus U$ and $Acap U=
Acapcl U$; why? Fix $ain Acap U$ and $bin Asetminus U$ and show that $[a,b]nsubseteq A$, so that $A$ cannot be an interval.
answered Nov 17 '12 at 6:04
Brian M. ScottBrian M. Scott
461k40518920
$endgroup$
$newcommand{cl}{operatorname{cl}}$HINTS: Suppose that $AsubseteqBbb R$ is not an interval; then there are points $a,bin A$ and $xinBbb Rsetminus A$ such that $a<x<b$. Use the sets $Acap(leftarrow,x)$ and $Acap(x,to)$ to show that $A$ is not connected.
The other direction is a bit harder. Suppose that $A$ is not connected. Then there is an open set $U$ in $Bbb R$ such that $Acap Unevarnothingne Asetminus U$ and $Acap U=
Acapcl U$; why? Fix $ain Acap U$ and $bin Asetminus U$ and show that $[a,b]nsubseteq A$, so that $A$ cannot be an interval.
answered Nov 17 '12 at 6:04
Brian M. ScottBrian M. Scott
461k40518920
answered Nov 17 '12 at 6:04
Brian M. ScottBrian M. Scott
461k40518920
answered Nov 17 '12 at 6:04
Brian M. ScottBrian M. Scott
461k40518920
answered Nov 17 '12 at 6:04
Brian M. ScottBrian M. Scott
461k40518920
461k40518920
$begingroup$
Actually is it true that if two sets are disjoint, no matter it is open or not it is disconnected. As by the definition, it seems to requires that 2 sets are open
$endgroup$
– Mathematics
Nov 17 '12 at 6:21
$begingroup$
@Mathematics: $[0,1)$ and $[1,2]$ are disjoint, but their union is $[0,2]$, which is connected. It is important that $Acap U$ and $Asetminus U$ are both relatively open subsets of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:23
$begingroup$
$Acup Uneemptysetne Asetminus Uimplies Acup U ne Asetminus U$?I am a little bit confused with the notation
$endgroup$
– Mathematics
Nov 17 '12 at 6:37
$begingroup$
@Mathematics: No, it doesn’t. But it’s true that $Acup Une Asetminus U$, simply because $Unevarnothing$. I think that you’re making this more complicated than it really is; all I’m saying is that if $A$ is not connected, there must be an open $U$ in $Bbb R$ such that $Acap U$ and $Asetminuscl U$ are non-empty relatively open subsets of $A$ whose union is $A$ $-$ in other words, a disconnection of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:40
add a comment |
$begingroup$
Actually is it true that if two sets are disjoint, no matter it is open or not it is disconnected. As by the definition, it seems to requires that 2 sets are open
$endgroup$
– Mathematics
Nov 17 '12 at 6:21
$begingroup$
@Mathematics: $[0,1)$ and $[1,2]$ are disjoint, but their union is $[0,2]$, which is connected. It is important that $Acap U$ and $Asetminus U$ are both relatively open subsets of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:23
$begingroup$
$Acup Uneemptysetne Asetminus Uimplies Acup U ne Asetminus U$?I am a little bit confused with the notation
$endgroup$
– Mathematics
Nov 17 '12 at 6:37
$begingroup$
@Mathematics: No, it doesn’t. But it’s true that $Acup Une Asetminus U$, simply because $Unevarnothing$. I think that you’re making this more complicated than it really is; all I’m saying is that if $A$ is not connected, there must be an open $U$ in $Bbb R$ such that $Acap U$ and $Asetminuscl U$ are non-empty relatively open subsets of $A$ whose union is $A$ $-$ in other words, a disconnection of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:40
$begingroup$
Actually is it true that if two sets are disjoint, no matter it is open or not it is disconnected. As by the definition, it seems to requires that 2 sets are open
$endgroup$
– Mathematics
Nov 17 '12 at 6:21
$begingroup$
Actually is it true that if two sets are disjoint, no matter it is open or not it is disconnected. As by the definition, it seems to requires that 2 sets are open
$endgroup$
– Mathematics
Nov 17 '12 at 6:21
$begingroup$
@Mathematics: $[0,1)$ and $[1,2]$ are disjoint, but their union is $[0,2]$, which is connected. It is important that $Acap U$ and $Asetminus U$ are both relatively open subsets of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:23
$begingroup$
@Mathematics: $[0,1)$ and $[1,2]$ are disjoint, but their union is $[0,2]$, which is connected. It is important that $Acap U$ and $Asetminus U$ are both relatively open subsets of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:23
$begingroup$
$Acup Uneemptysetne Asetminus Uimplies Acup U ne Asetminus U$?I am a little bit confused with the notation
$endgroup$
– Mathematics
Nov 17 '12 at 6:37
$begingroup$
$Acup Uneemptysetne Asetminus Uimplies Acup U ne Asetminus U$?I am a little bit confused with the notation
$endgroup$
– Mathematics
Nov 17 '12 at 6:37
$begingroup$
@Mathematics: No, it doesn’t. But it’s true that $Acup Une Asetminus U$, simply because $Unevarnothing$. I think that you’re making this more complicated than it really is; all I’m saying is that if $A$ is not connected, there must be an open $U$ in $Bbb R$ such that $Acap U$ and $Asetminuscl U$ are non-empty relatively open subsets of $A$ whose union is $A$ $-$ in other words, a disconnection of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:40
$begingroup$
@Mathematics: No, it doesn’t. But it’s true that $Acup Une Asetminus U$, simply because $Unevarnothing$. I think that you’re making this more complicated than it really is; all I’m saying is that if $A$ is not connected, there must be an open $U$ in $Bbb R$ such that $Acap U$ and $Asetminuscl U$ are non-empty relatively open subsets of $A$ whose union is $A$ $-$ in other words, a disconnection of $A$.
$endgroup$
– Brian M. Scott
Nov 17 '12 at 6:40
add a comment |
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1
$begingroup$
Maybe some of the discussion here is helpful? Set $A$ interval in $mathbb{R}implies$ connected
$endgroup$
– yunone
Nov 17 '12 at 5:56
1
$begingroup$
In Principles of Mathematical Analysis by Rudin, a proof for the same is given.
$endgroup$
– Gautam Shenoy
Nov 17 '12 at 5:58