Why is the complement of any perfect totally disconnected subset of $mathbb{R}$ a countable union of disjoint...
$begingroup$
The cantor set $C$ is obtained by repeatedly removing the middle $1/3$, starting from the interval $[0,1]$. Since the number of intervals removed in each step of construction is finite, $[0,1] backslash C$ is the union of only countable many disjoint intervals.
In a topology book, there is a proof of the fact that a perfect totally disconnected subset on $Bbb{R}$ is homeomorphic to $C$. In the proof, it is claimed that if a set $A$ is perfect and totally disconnected, then $Rbackslash A$ is consisted of countably many disjoint intervals. This is true if $A=C$, but I don't know why it is true for such $A$ in general. Is there a simple explanation for this claim? Can't there be uncountably many intervals in $Rbackslash A$?
real-analysis general-topology cantor-set
asked Jan 7 at 12:50
Ma JoadMa Joad
829316
$endgroup$
add a comment |
$begingroup$
The cantor set $C$ is obtained by repeatedly removing the middle $1/3$, starting from the interval $[0,1]$. Since the number of intervals removed in each step of construction is finite, $[0,1] backslash C$ is the union of only countable many disjoint intervals.
In a topology book, there is a proof of the fact that a perfect totally disconnected subset on $Bbb{R}$ is homeomorphic to $C$. In the proof, it is claimed that if a set $A$ is perfect and totally disconnected, then $Rbackslash A$ is consisted of countably many disjoint intervals. This is true if $A=C$, but I don't know why it is true for such $A$ in general. Is there a simple explanation for this claim? Can't there be uncountably many intervals in $Rbackslash A$?
real-analysis general-topology cantor-set
asked Jan 7 at 12:50
Ma JoadMa Joad
829316
$endgroup$
1
$begingroup$
That's confusing. Every open subset of reals is a union of countably many disjoint intervals. This follows because connected components are open intervals and every connected component is a connected component of some rational.
$endgroup$
– freakish
Jan 7 at 13:08
1
$begingroup$
@freakish Presumably "countable" is used in the sense of "countably infinite".
$endgroup$
– Robert Israel
Jan 7 at 13:11
add a comment |
$begingroup$
The cantor set $C$ is obtained by repeatedly removing the middle $1/3$, starting from the interval $[0,1]$. Since the number of intervals removed in each step of construction is finite, $[0,1] backslash C$ is the union of only countable many disjoint intervals.
In a topology book, there is a proof of the fact that a perfect totally disconnected subset on $Bbb{R}$ is homeomorphic to $C$. In the proof, it is claimed that if a set $A$ is perfect and totally disconnected, then $Rbackslash A$ is consisted of countably many disjoint intervals. This is true if $A=C$, but I don't know why it is true for such $A$ in general. Is there a simple explanation for this claim? Can't there be uncountably many intervals in $Rbackslash A$?
real-analysis general-topology cantor-set
asked Jan 7 at 12:50
Ma JoadMa Joad
829316
$endgroup$
The cantor set $C$ is obtained by repeatedly removing the middle $1/3$, starting from the interval $[0,1]$. Since the number of intervals removed in each step of construction is finite, $[0,1] backslash C$ is the union of only countable many disjoint intervals.
In a topology book, there is a proof of the fact that a perfect totally disconnected subset on $Bbb{R}$ is homeomorphic to $C$. In the proof, it is claimed that if a set $A$ is perfect and totally disconnected, then $Rbackslash A$ is consisted of countably many disjoint intervals. This is true if $A=C$, but I don't know why it is true for such $A$ in general. Is there a simple explanation for this claim? Can't there be uncountably many intervals in $Rbackslash A$?
real-analysis general-topology cantor-set
real-analysis general-topology cantor-set
asked Jan 7 at 12:50
Ma JoadMa Joad
829316
asked Jan 7 at 12:50
Ma JoadMa Joad
829316
edited Jan 7 at 22:34
Ma Joad
asked Jan 7 at 12:50
Ma JoadMa Joad
829316
asked Jan 7 at 12:50
Ma JoadMa Joad
829316
asked Jan 7 at 12:50
Ma JoadMa Joad
829316
829316
1
$begingroup$
That's confusing. Every open subset of reals is a union of countably many disjoint intervals. This follows because connected components are open intervals and every connected component is a connected component of some rational.
$endgroup$
– freakish
Jan 7 at 13:08
1
$begingroup$
@freakish Presumably "countable" is used in the sense of "countably infinite".
$endgroup$
– Robert Israel
Jan 7 at 13:11
add a comment |
1
$begingroup$
That's confusing. Every open subset of reals is a union of countably many disjoint intervals. This follows because connected components are open intervals and every connected component is a connected component of some rational.
$endgroup$
– freakish
Jan 7 at 13:08
1
$begingroup$
@freakish Presumably "countable" is used in the sense of "countably infinite".
$endgroup$
– Robert Israel
Jan 7 at 13:11
1
1
$begingroup$
That's confusing. Every open subset of reals is a union of countably many disjoint intervals. This follows because connected components are open intervals and every connected component is a connected component of some rational.
$endgroup$
– freakish
Jan 7 at 13:08
$begingroup$
That's confusing. Every open subset of reals is a union of countably many disjoint intervals. This follows because connected components are open intervals and every connected component is a connected component of some rational.
$endgroup$
– freakish
Jan 7 at 13:08
1
1
$begingroup$
@freakish Presumably "countable" is used in the sense of "countably infinite".
$endgroup$
– Robert Israel
Jan 7 at 13:11
$begingroup$
@freakish Presumably "countable" is used in the sense of "countably infinite".
$endgroup$
– Robert Israel
Jan 7 at 13:11
add a comment |
1 Answer
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So first of all every open subset of $mathbb{R}$ is a countable union of disjoint intervals. I leave this as an exercise.
If you ask why $mathbb{R}backslash A$ is not a finite union of intervals then the argument goes as follows: assume that $U=mathbb{R}backslash A$ is a finite union of open intervals. Then $A$ is a finite union of closed intervals. Here I treat singletons ${x}$ as closed intervals as well. So write down $A=bigcup_{i=1}^n C_i$ where each $C_i$ is a closed interval and they are pairwise disjoint.
If any of $C_j$ is a singleton then $A$ cannot be perfect. Indeed, let $C_j={x}$. Since there are only finitely many $C_i$'s then some small neighbourhood $V$ of $x$ cannot intersect any of $C_i$ except for $C_j$. Otherewise either $x$ belongs to some bigger interval (contradiction with $C_j$ being a singleton) or there are infinitely many $C_i$ "converging" to $x$ (contradiction with finite number of $C_i$). Therefore $x$ is isolated and so $A$ is not perfect. Contradiction.
It follows that all $C_i$ are proper intervals. But each $C_i$ is a connected component of $A$. That contradicts $A$ being totally disconnected.
answered Jan 7 at 13:07
freakishfreakish
11.8k1629
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add a comment |
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$begingroup$
So first of all every open subset of $mathbb{R}$ is a countable union of disjoint intervals. I leave this as an exercise.
If you ask why $mathbb{R}backslash A$ is not a finite union of intervals then the argument goes as follows: assume that $U=mathbb{R}backslash A$ is a finite union of open intervals. Then $A$ is a finite union of closed intervals. Here I treat singletons ${x}$ as closed intervals as well. So write down $A=bigcup_{i=1}^n C_i$ where each $C_i$ is a closed interval and they are pairwise disjoint.
If any of $C_j$ is a singleton then $A$ cannot be perfect. Indeed, let $C_j={x}$. Since there are only finitely many $C_i$'s then some small neighbourhood $V$ of $x$ cannot intersect any of $C_i$ except for $C_j$. Otherewise either $x$ belongs to some bigger interval (contradiction with $C_j$ being a singleton) or there are infinitely many $C_i$ "converging" to $x$ (contradiction with finite number of $C_i$). Therefore $x$ is isolated and so $A$ is not perfect. Contradiction.
It follows that all $C_i$ are proper intervals. But each $C_i$ is a connected component of $A$. That contradicts $A$ being totally disconnected.
answered Jan 7 at 13:07
freakishfreakish
11.8k1629
$endgroup$
add a comment |
$begingroup$
So first of all every open subset of $mathbb{R}$ is a countable union of disjoint intervals. I leave this as an exercise.
If you ask why $mathbb{R}backslash A$ is not a finite union of intervals then the argument goes as follows: assume that $U=mathbb{R}backslash A$ is a finite union of open intervals. Then $A$ is a finite union of closed intervals. Here I treat singletons ${x}$ as closed intervals as well. So write down $A=bigcup_{i=1}^n C_i$ where each $C_i$ is a closed interval and they are pairwise disjoint.
If any of $C_j$ is a singleton then $A$ cannot be perfect. Indeed, let $C_j={x}$. Since there are only finitely many $C_i$'s then some small neighbourhood $V$ of $x$ cannot intersect any of $C_i$ except for $C_j$. Otherewise either $x$ belongs to some bigger interval (contradiction with $C_j$ being a singleton) or there are infinitely many $C_i$ "converging" to $x$ (contradiction with finite number of $C_i$). Therefore $x$ is isolated and so $A$ is not perfect. Contradiction.
It follows that all $C_i$ are proper intervals. But each $C_i$ is a connected component of $A$. That contradicts $A$ being totally disconnected.
answered Jan 7 at 13:07
freakishfreakish
11.8k1629
$endgroup$
add a comment |
$begingroup$
So first of all every open subset of $mathbb{R}$ is a countable union of disjoint intervals. I leave this as an exercise.
If you ask why $mathbb{R}backslash A$ is not a finite union of intervals then the argument goes as follows: assume that $U=mathbb{R}backslash A$ is a finite union of open intervals. Then $A$ is a finite union of closed intervals. Here I treat singletons ${x}$ as closed intervals as well. So write down $A=bigcup_{i=1}^n C_i$ where each $C_i$ is a closed interval and they are pairwise disjoint.
If any of $C_j$ is a singleton then $A$ cannot be perfect. Indeed, let $C_j={x}$. Since there are only finitely many $C_i$'s then some small neighbourhood $V$ of $x$ cannot intersect any of $C_i$ except for $C_j$. Otherewise either $x$ belongs to some bigger interval (contradiction with $C_j$ being a singleton) or there are infinitely many $C_i$ "converging" to $x$ (contradiction with finite number of $C_i$). Therefore $x$ is isolated and so $A$ is not perfect. Contradiction.
It follows that all $C_i$ are proper intervals. But each $C_i$ is a connected component of $A$. That contradicts $A$ being totally disconnected.
answered Jan 7 at 13:07
freakishfreakish
11.8k1629
$endgroup$
So first of all every open subset of $mathbb{R}$ is a countable union of disjoint intervals. I leave this as an exercise.
If you ask why $mathbb{R}backslash A$ is not a finite union of intervals then the argument goes as follows: assume that $U=mathbb{R}backslash A$ is a finite union of open intervals. Then $A$ is a finite union of closed intervals. Here I treat singletons ${x}$ as closed intervals as well. So write down $A=bigcup_{i=1}^n C_i$ where each $C_i$ is a closed interval and they are pairwise disjoint.
If any of $C_j$ is a singleton then $A$ cannot be perfect. Indeed, let $C_j={x}$. Since there are only finitely many $C_i$'s then some small neighbourhood $V$ of $x$ cannot intersect any of $C_i$ except for $C_j$. Otherewise either $x$ belongs to some bigger interval (contradiction with $C_j$ being a singleton) or there are infinitely many $C_i$ "converging" to $x$ (contradiction with finite number of $C_i$). Therefore $x$ is isolated and so $A$ is not perfect. Contradiction.
It follows that all $C_i$ are proper intervals. But each $C_i$ is a connected component of $A$. That contradicts $A$ being totally disconnected.
answered Jan 7 at 13:07
freakishfreakish
11.8k1629
edited Jan 7 at 17:32
answered Jan 7 at 13:07
freakishfreakish
11.8k1629
answered Jan 7 at 13:07
freakishfreakish
11.8k1629
answered Jan 7 at 13:07
freakishfreakish
11.8k1629
11.8k1629
add a comment |
add a comment |
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That's confusing. Every open subset of reals is a union of countably many disjoint intervals. This follows because connected components are open intervals and every connected component is a connected component of some rational.
$endgroup$
– freakish
Jan 7 at 13:08
1
$begingroup$
@freakish Presumably "countable" is used in the sense of "countably infinite".
$endgroup$
– Robert Israel
Jan 7 at 13:11