$[0, 1]setminus mathbb Q$ can't be exhausted by Jordan sets
$begingroup$
We wish to show that $[0, 1]setminus mathbb Q$ can't be exhausted by Jordan sets. What I mean by this is that there's no nested sequence of Jordan sets $J_1 subseteq J_2 subseteq dots$ such that $bigcup_{k=1}^{infty}J_k = [0,1]setminusmathbb Q$
Jordan in this sense means Jordan Measurable (boundary is negligible)
Sadly, the fact that the boundary of the union is a subset of the union of the boundaries is only true for finite union, else this would be piece of cake.
How would you tackle this? It may be related to improper integration
general-topology measure-theory improper-integrals
asked Jan 1 at 23:34
Oria GruberOria Gruber
6,53432460
$endgroup$
add a comment |
$begingroup$
We wish to show that $[0, 1]setminus mathbb Q$ can't be exhausted by Jordan sets. What I mean by this is that there's no nested sequence of Jordan sets $J_1 subseteq J_2 subseteq dots$ such that $bigcup_{k=1}^{infty}J_k = [0,1]setminusmathbb Q$
Jordan in this sense means Jordan Measurable (boundary is negligible)
Sadly, the fact that the boundary of the union is a subset of the union of the boundaries is only true for finite union, else this would be piece of cake.
How would you tackle this? It may be related to improper integration
general-topology measure-theory improper-integrals
asked Jan 1 at 23:34
Oria GruberOria Gruber
6,53432460
$endgroup$
add a comment |
$begingroup$
We wish to show that $[0, 1]setminus mathbb Q$ can't be exhausted by Jordan sets. What I mean by this is that there's no nested sequence of Jordan sets $J_1 subseteq J_2 subseteq dots$ such that $bigcup_{k=1}^{infty}J_k = [0,1]setminusmathbb Q$
Jordan in this sense means Jordan Measurable (boundary is negligible)
Sadly, the fact that the boundary of the union is a subset of the union of the boundaries is only true for finite union, else this would be piece of cake.
How would you tackle this? It may be related to improper integration
general-topology measure-theory improper-integrals
asked Jan 1 at 23:34
Oria GruberOria Gruber
6,53432460
$endgroup$
We wish to show that $[0, 1]setminus mathbb Q$ can't be exhausted by Jordan sets. What I mean by this is that there's no nested sequence of Jordan sets $J_1 subseteq J_2 subseteq dots$ such that $bigcup_{k=1}^{infty}J_k = [0,1]setminusmathbb Q$
Jordan in this sense means Jordan Measurable (boundary is negligible)
Sadly, the fact that the boundary of the union is a subset of the union of the boundaries is only true for finite union, else this would be piece of cake.
How would you tackle this? It may be related to improper integration
general-topology measure-theory improper-integrals
general-topology measure-theory improper-integrals
asked Jan 1 at 23:34
Oria GruberOria Gruber
6,53432460
asked Jan 1 at 23:34
Oria GruberOria Gruber
6,53432460
edited Jan 1 at 23:49
Oria Gruber
asked Jan 1 at 23:34
Oria GruberOria Gruber
6,53432460
asked Jan 1 at 23:34
Oria GruberOria Gruber
6,53432460
asked Jan 1 at 23:34
Oria GruberOria Gruber
6,53432460
6,53432460
add a comment |
add a comment |
2 Answers
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Better answer than the above one as I don't invoke measures.
Notice that $X = [0,1]setminus mathbb Q$ doesn't have any interior points, $text{int}(X) = emptyset$, so if $J subset X$ then also $text{int}(J) = emptyset$.
This means that $J subseteq partial J$. Now, if $J$ is Jordan measurable, then by definition $partial J$ is negligible, so $J$ is negligible.
Thus if $X = bigcup_{k=1}^{infty}J_k$ then it follows that $X$ is negligible, since it's a countable union of negligible sets, and this is a contradiction, since $X$ is not negligible.
answered Jan 2 at 17:17
Oria GruberOria Gruber
6,53432460
$endgroup$
add a comment |
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Note that the inner Jordan measure of any subset of $A:=[0,1]setminusmathbb{Q}$ is $0$, since $A$ does not contain any positive length intervals. Hence, any Jordan subset of $A$ is of Jordan measure $0$, and hence of Lebesque measure $0$. Hence, any countable union of such sets is still of Lebesque measure $0$. Since $A$ has Lebesque measure $1$, this union can not equal $A$.
answered Jan 2 at 0:02
SmileyCraftSmileyCraft
3,481516
$endgroup$
$begingroup$
I would have prefered to solve this without invoking measures as we didn't get there yet. I don't know what you mean by inner measure
$endgroup$
– Oria Gruber
Jan 2 at 9:09
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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$begingroup$
Better answer than the above one as I don't invoke measures.
Notice that $X = [0,1]setminus mathbb Q$ doesn't have any interior points, $text{int}(X) = emptyset$, so if $J subset X$ then also $text{int}(J) = emptyset$.
This means that $J subseteq partial J$. Now, if $J$ is Jordan measurable, then by definition $partial J$ is negligible, so $J$ is negligible.
Thus if $X = bigcup_{k=1}^{infty}J_k$ then it follows that $X$ is negligible, since it's a countable union of negligible sets, and this is a contradiction, since $X$ is not negligible.
answered Jan 2 at 17:17
Oria GruberOria Gruber
6,53432460
$endgroup$
add a comment |
$begingroup$
Better answer than the above one as I don't invoke measures.
Notice that $X = [0,1]setminus mathbb Q$ doesn't have any interior points, $text{int}(X) = emptyset$, so if $J subset X$ then also $text{int}(J) = emptyset$.
This means that $J subseteq partial J$. Now, if $J$ is Jordan measurable, then by definition $partial J$ is negligible, so $J$ is negligible.
Thus if $X = bigcup_{k=1}^{infty}J_k$ then it follows that $X$ is negligible, since it's a countable union of negligible sets, and this is a contradiction, since $X$ is not negligible.
answered Jan 2 at 17:17
Oria GruberOria Gruber
6,53432460
$endgroup$
add a comment |
$begingroup$
Better answer than the above one as I don't invoke measures.
Notice that $X = [0,1]setminus mathbb Q$ doesn't have any interior points, $text{int}(X) = emptyset$, so if $J subset X$ then also $text{int}(J) = emptyset$.
This means that $J subseteq partial J$. Now, if $J$ is Jordan measurable, then by definition $partial J$ is negligible, so $J$ is negligible.
Thus if $X = bigcup_{k=1}^{infty}J_k$ then it follows that $X$ is negligible, since it's a countable union of negligible sets, and this is a contradiction, since $X$ is not negligible.
answered Jan 2 at 17:17
Oria GruberOria Gruber
6,53432460
$endgroup$
Better answer than the above one as I don't invoke measures.
Notice that $X = [0,1]setminus mathbb Q$ doesn't have any interior points, $text{int}(X) = emptyset$, so if $J subset X$ then also $text{int}(J) = emptyset$.
This means that $J subseteq partial J$. Now, if $J$ is Jordan measurable, then by definition $partial J$ is negligible, so $J$ is negligible.
Thus if $X = bigcup_{k=1}^{infty}J_k$ then it follows that $X$ is negligible, since it's a countable union of negligible sets, and this is a contradiction, since $X$ is not negligible.
answered Jan 2 at 17:17
Oria GruberOria Gruber
6,53432460
answered Jan 2 at 17:17
Oria GruberOria Gruber
6,53432460
answered Jan 2 at 17:17
Oria GruberOria Gruber
6,53432460
answered Jan 2 at 17:17
Oria GruberOria Gruber
6,53432460
6,53432460
add a comment |
add a comment |
$begingroup$
Note that the inner Jordan measure of any subset of $A:=[0,1]setminusmathbb{Q}$ is $0$, since $A$ does not contain any positive length intervals. Hence, any Jordan subset of $A$ is of Jordan measure $0$, and hence of Lebesque measure $0$. Hence, any countable union of such sets is still of Lebesque measure $0$. Since $A$ has Lebesque measure $1$, this union can not equal $A$.
answered Jan 2 at 0:02
SmileyCraftSmileyCraft
3,481516
$endgroup$
$begingroup$
I would have prefered to solve this without invoking measures as we didn't get there yet. I don't know what you mean by inner measure
$endgroup$
– Oria Gruber
Jan 2 at 9:09
add a comment |
$begingroup$
Note that the inner Jordan measure of any subset of $A:=[0,1]setminusmathbb{Q}$ is $0$, since $A$ does not contain any positive length intervals. Hence, any Jordan subset of $A$ is of Jordan measure $0$, and hence of Lebesque measure $0$. Hence, any countable union of such sets is still of Lebesque measure $0$. Since $A$ has Lebesque measure $1$, this union can not equal $A$.
answered Jan 2 at 0:02
SmileyCraftSmileyCraft
3,481516
$endgroup$
$begingroup$
I would have prefered to solve this without invoking measures as we didn't get there yet. I don't know what you mean by inner measure
$endgroup$
– Oria Gruber
Jan 2 at 9:09
add a comment |
$begingroup$
Note that the inner Jordan measure of any subset of $A:=[0,1]setminusmathbb{Q}$ is $0$, since $A$ does not contain any positive length intervals. Hence, any Jordan subset of $A$ is of Jordan measure $0$, and hence of Lebesque measure $0$. Hence, any countable union of such sets is still of Lebesque measure $0$. Since $A$ has Lebesque measure $1$, this union can not equal $A$.
answered Jan 2 at 0:02
SmileyCraftSmileyCraft
3,481516
$endgroup$
Note that the inner Jordan measure of any subset of $A:=[0,1]setminusmathbb{Q}$ is $0$, since $A$ does not contain any positive length intervals. Hence, any Jordan subset of $A$ is of Jordan measure $0$, and hence of Lebesque measure $0$. Hence, any countable union of such sets is still of Lebesque measure $0$. Since $A$ has Lebesque measure $1$, this union can not equal $A$.
answered Jan 2 at 0:02
SmileyCraftSmileyCraft
3,481516
answered Jan 2 at 0:02
SmileyCraftSmileyCraft
3,481516
answered Jan 2 at 0:02
SmileyCraftSmileyCraft
3,481516
answered Jan 2 at 0:02
SmileyCraftSmileyCraft
3,481516
3,481516
$begingroup$
I would have prefered to solve this without invoking measures as we didn't get there yet. I don't know what you mean by inner measure
$endgroup$
– Oria Gruber
Jan 2 at 9:09
add a comment |
$begingroup$
I would have prefered to solve this without invoking measures as we didn't get there yet. I don't know what you mean by inner measure
$endgroup$
– Oria Gruber
Jan 2 at 9:09
$begingroup$
I would have prefered to solve this without invoking measures as we didn't get there yet. I don't know what you mean by inner measure
$endgroup$
– Oria Gruber
Jan 2 at 9:09
$begingroup$
I would have prefered to solve this without invoking measures as we didn't get there yet. I don't know what you mean by inner measure
$endgroup$
– Oria Gruber
Jan 2 at 9:09
add a comment |
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