Mixing
Proving that any closed subset $U subset mathbb{R}$ is a countable intersection of open sets
$begingroup$
I got stuck in this proof.
First, we know that if $U$ is closed, then the complement $U^c$ is open.
Therefore, what I tried to do is use one of my previous results in another question. This is, that every open set in $mathbb{R}$ is a countable union of disjoint open intervals.
Using the previous statement, simple set theory and De Morgan's Laws, we get
$$U = (U^c)^c = bigg( bigcup_{xin U^c} I_xbigg)^c = bigg( bigcap_{xin U^c} I^c_xbigg)$$
where $I_x$ are the intervals of the open set $U^c$. My problem is that we know that the intervals $I_x$ of $U^c$ are open, and, therefore the intervals $I^c_x$ are closed, which is the opposite of what I was trying to obtain.
I know there's another answer. Nevertheless, it is completely different to what I am trying to do. I want to know why this method does not work.
real-analysis general-topology analysis proof-verification
asked Jan 25 at 9:38
The BoscoThe Bosco
613212
$endgroup$
|
show 6 more comments
$begingroup$
I got stuck in this proof.
First, we know that if $U$ is closed, then the complement $U^c$ is open.
Therefore, what I tried to do is use one of my previous results in another question. This is, that every open set in $mathbb{R}$ is a countable union of disjoint open intervals.
Using the previous statement, simple set theory and De Morgan's Laws, we get
$$U = (U^c)^c = bigg( bigcup_{xin U^c} I_xbigg)^c = bigg( bigcap_{xin U^c} I^c_xbigg)$$
where $I_x$ are the intervals of the open set $U^c$. My problem is that we know that the intervals $I_x$ of $U^c$ are open, and, therefore the intervals $I^c_x$ are closed, which is the opposite of what I was trying to obtain.
I know there's another answer. Nevertheless, it is completely different to what I am trying to do. I want to know why this method does not work.
real-analysis general-topology analysis proof-verification
asked Jan 25 at 9:38
The BoscoThe Bosco
613212
$endgroup$
1
$begingroup$
You have another problem there, $U^c$ may not be countable.
$endgroup$
– Klaus
Jan 25 at 9:44
$begingroup$
That is fine. The theorem works anyways. It is for any set of the reals.
$endgroup$
– The Bosco
Jan 25 at 9:47
$begingroup$
But you asked for a countable intersection. $bigcap_{xin U^c} I^c_x$ is not a countable intersection unless it is empty.
$endgroup$
– Klaus
Jan 25 at 9:51
$begingroup$
Why is that? I fail to see it
$endgroup$
– The Bosco
Jan 25 at 9:53
1
$begingroup$
No, check their indices. They always have an intersection over countably many sets ($k = 1$ to $infty$). You intersect over uncountably many sets, but claim you want a countable intersection.
$endgroup$
– Klaus
Jan 25 at 10:06
|
show 6 more comments
$begingroup$
I got stuck in this proof.
First, we know that if $U$ is closed, then the complement $U^c$ is open.
Therefore, what I tried to do is use one of my previous results in another question. This is, that every open set in $mathbb{R}$ is a countable union of disjoint open intervals.
Using the previous statement, simple set theory and De Morgan's Laws, we get
$$U = (U^c)^c = bigg( bigcup_{xin U^c} I_xbigg)^c = bigg( bigcap_{xin U^c} I^c_xbigg)$$
where $I_x$ are the intervals of the open set $U^c$. My problem is that we know that the intervals $I_x$ of $U^c$ are open, and, therefore the intervals $I^c_x$ are closed, which is the opposite of what I was trying to obtain.
I know there's another answer. Nevertheless, it is completely different to what I am trying to do. I want to know why this method does not work.
real-analysis general-topology analysis proof-verification
asked Jan 25 at 9:38
The BoscoThe Bosco
613212
$endgroup$
I got stuck in this proof.
First, we know that if $U$ is closed, then the complement $U^c$ is open.
Therefore, what I tried to do is use one of my previous results in another question. This is, that every open set in $mathbb{R}$ is a countable union of disjoint open intervals.
Using the previous statement, simple set theory and De Morgan's Laws, we get
$$U = (U^c)^c = bigg( bigcup_{xin U^c} I_xbigg)^c = bigg( bigcap_{xin U^c} I^c_xbigg)$$
where $I_x$ are the intervals of the open set $U^c$. My problem is that we know that the intervals $I_x$ of $U^c$ are open, and, therefore the intervals $I^c_x$ are closed, which is the opposite of what I was trying to obtain.
I know there's another answer. Nevertheless, it is completely different to what I am trying to do. I want to know why this method does not work.
real-analysis general-topology analysis proof-verification
real-analysis general-topology analysis proof-verification
asked Jan 25 at 9:38
The BoscoThe Bosco
613212
asked Jan 25 at 9:38
The BoscoThe Bosco
613212
edited Jan 25 at 9:44
The Bosco
asked Jan 25 at 9:38
The BoscoThe Bosco
613212
asked Jan 25 at 9:38
The BoscoThe Bosco
613212
asked Jan 25 at 9:38
The BoscoThe Bosco
613212
613212
1
$begingroup$
You have another problem there, $U^c$ may not be countable.
$endgroup$
– Klaus
Jan 25 at 9:44
$begingroup$
That is fine. The theorem works anyways. It is for any set of the reals.
$endgroup$
– The Bosco
Jan 25 at 9:47
$begingroup$
But you asked for a countable intersection. $bigcap_{xin U^c} I^c_x$ is not a countable intersection unless it is empty.
$endgroup$
– Klaus
Jan 25 at 9:51
$begingroup$
Why is that? I fail to see it
$endgroup$
– The Bosco
Jan 25 at 9:53
1
$begingroup$
No, check their indices. They always have an intersection over countably many sets ($k = 1$ to $infty$). You intersect over uncountably many sets, but claim you want a countable intersection.
$endgroup$
– Klaus
Jan 25 at 10:06
|
show 6 more comments
1
$begingroup$
You have another problem there, $U^c$ may not be countable.
$endgroup$
– Klaus
Jan 25 at 9:44
$begingroup$
That is fine. The theorem works anyways. It is for any set of the reals.
$endgroup$
– The Bosco
Jan 25 at 9:47
$begingroup$
But you asked for a countable intersection. $bigcap_{xin U^c} I^c_x$ is not a countable intersection unless it is empty.
$endgroup$
– Klaus
Jan 25 at 9:51
$begingroup$
Why is that? I fail to see it
$endgroup$
– The Bosco
Jan 25 at 9:53
1
$begingroup$
No, check their indices. They always have an intersection over countably many sets ($k = 1$ to $infty$). You intersect over uncountably many sets, but claim you want a countable intersection.
$endgroup$
– Klaus
Jan 25 at 10:06
1
1
$begingroup$
You have another problem there, $U^c$ may not be countable.
$endgroup$
– Klaus
Jan 25 at 9:44
$begingroup$
You have another problem there, $U^c$ may not be countable.
$endgroup$
– Klaus
Jan 25 at 9:44
$begingroup$
That is fine. The theorem works anyways. It is for any set of the reals.
$endgroup$
– The Bosco
Jan 25 at 9:47
$begingroup$
That is fine. The theorem works anyways. It is for any set of the reals.
$endgroup$
– The Bosco
Jan 25 at 9:47
$begingroup$
But you asked for a countable intersection. $bigcap_{xin U^c} I^c_x$ is not a countable intersection unless it is empty.
$endgroup$
– Klaus
Jan 25 at 9:51
$begingroup$
But you asked for a countable intersection. $bigcap_{xin U^c} I^c_x$ is not a countable intersection unless it is empty.
$endgroup$
– Klaus
Jan 25 at 9:51
$begingroup$
Why is that? I fail to see it
$endgroup$
– The Bosco
Jan 25 at 9:53
$begingroup$
Why is that? I fail to see it
$endgroup$
– The Bosco
Jan 25 at 9:53
1
1
$begingroup$
No, check their indices. They always have an intersection over countably many sets ($k = 1$ to $infty$). You intersect over uncountably many sets, but claim you want a countable intersection.
$endgroup$
– Klaus
Jan 25 at 10:06
$begingroup$
No, check their indices. They always have an intersection over countably many sets ($k = 1$ to $infty$). You intersect over uncountably many sets, but claim you want a countable intersection.
$endgroup$
– Klaus
Jan 25 at 10:06
|
show 6 more comments
1 Answer
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$begingroup$
Every open interval $(a,b)$ is a countable union of closed (but not disjoint) intervals, for instance $bigcup_{nge 3}[a+(b-a)/n,b-(b-a)/n]$. And a countable union of countable unions is a countable union. So every open set in $Bbb R$ is a countable union of closed intervals. Take it from there.
answered Jan 25 at 10:18
TonyKTonyK
43.5k357136
$endgroup$
add a comment |
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$begingroup$
Every open interval $(a,b)$ is a countable union of closed (but not disjoint) intervals, for instance $bigcup_{nge 3}[a+(b-a)/n,b-(b-a)/n]$. And a countable union of countable unions is a countable union. So every open set in $Bbb R$ is a countable union of closed intervals. Take it from there.
answered Jan 25 at 10:18
TonyKTonyK
43.5k357136
$endgroup$
add a comment |
$begingroup$
Every open interval $(a,b)$ is a countable union of closed (but not disjoint) intervals, for instance $bigcup_{nge 3}[a+(b-a)/n,b-(b-a)/n]$. And a countable union of countable unions is a countable union. So every open set in $Bbb R$ is a countable union of closed intervals. Take it from there.
answered Jan 25 at 10:18
TonyKTonyK
43.5k357136
$endgroup$
add a comment |
$begingroup$
Every open interval $(a,b)$ is a countable union of closed (but not disjoint) intervals, for instance $bigcup_{nge 3}[a+(b-a)/n,b-(b-a)/n]$. And a countable union of countable unions is a countable union. So every open set in $Bbb R$ is a countable union of closed intervals. Take it from there.
answered Jan 25 at 10:18
TonyKTonyK
43.5k357136
$endgroup$
Every open interval $(a,b)$ is a countable union of closed (but not disjoint) intervals, for instance $bigcup_{nge 3}[a+(b-a)/n,b-(b-a)/n]$. And a countable union of countable unions is a countable union. So every open set in $Bbb R$ is a countable union of closed intervals. Take it from there.
answered Jan 25 at 10:18
TonyKTonyK
43.5k357136
answered Jan 25 at 10:18
TonyKTonyK
43.5k357136
answered Jan 25 at 10:18
TonyKTonyK
43.5k357136
answered Jan 25 at 10:18
TonyKTonyK
43.5k357136
43.5k357136
add a comment |
add a comment |
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1
$begingroup$
You have another problem there, $U^c$ may not be countable.
$endgroup$
– Klaus
Jan 25 at 9:44
$begingroup$
That is fine. The theorem works anyways. It is for any set of the reals.
$endgroup$
– The Bosco
Jan 25 at 9:47
$begingroup$
But you asked for a countable intersection. $bigcap_{xin U^c} I^c_x$ is not a countable intersection unless it is empty.
$endgroup$
– Klaus
Jan 25 at 9:51
$begingroup$
Why is that? I fail to see it
$endgroup$
– The Bosco
Jan 25 at 9:53
1
$begingroup$
No, check their indices. They always have an intersection over countably many sets ($k = 1$ to $infty$). You intersect over uncountably many sets, but claim you want a countable intersection.
$endgroup$
– Klaus
Jan 25 at 10:06