Intervals are connected and the only connected sets in $mathbb{R}$












3












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















3












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














3












3








3


3



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















8












$begingroup$

$newcommand{cl}{operatorname{cl}}$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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

$newcommand{cl}{operatorname{cl}}$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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















8












$begingroup$

$newcommand{cl}{operatorname{cl}}$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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














8












8








8





$begingroup$

$newcommand{cl}{operatorname{cl}}$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.






share|cite|improve this answer









$endgroup$



$newcommand{cl}{operatorname{cl}}$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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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