Every open subset $Usubseteqmathbb{R}$ is countable union of disjoint open intervals.
$begingroup$
Theorem. Every open subset $Usubseteqmathbb{R}$ is countable union of disjoint open intervals.
I was looking for proofs for this result and I came to this interesting post: Any open subset of $Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs].
Among all the proofs I started from the simplest one: the one written by G.T. https://math.stackexchange.com/a/1949873/554978.
However, I am not convinced of the proof that $I_x$ is an interval. I do not know if I did not understand correctly or the proof in question is not valid, so I propose one myself and I would like you to tell me which one is valid.
Definition. An interval is a subset $Isubseteqmathbb{R}$ such that, for all $a<c<b$ in $mathbb{R}$, if $a,bin I$ then $cin I$.
Let $xin U$ and we suppose that $xinmathbb{Q}$, then define begin{align} I_x = bigcuplimits_{substack{Itext{ an open interval} \ x~in~I~subseteq~U}} I,end{align}
we must prove that $I_x$ is an interval. About that let $a,bin I_x$ such that $a<c<b$, then we must prove that $cin I_x$.
Since $a,bin I_x$, then $a,bin I$ for same open interval $I$ which contains $x$. If $a$ and $b$ belong to the same $I$, since $a<c<b$ and $I$ is an interval, $cin I$, therefore $cin I_x$.
Now, we denote with $I_a$ the open interval of $I_x$ which contains $a$, but not $b$ and we denote with $I_b$ the open interval of $I_x$ which contains $b$, but not $a$.
First case: $[c=x]$.
If $c=x$, then $cin I_x$ by definition of $I_x$;
Second case: $[c<x]$.
If $c<x$, then either $a<c<x<b$ or $a<c<b<x$.
$(i)$ If $a<c<x<b$, since $ain I_a$ and $xin I_a$ and $I_a$ is an interval, then $cin I_a$, therefore $cin I_x$.
$(ii)$ If $a<c<b<x$, since $xin I_a$ and $ain I_a$, and $I_a$ is an interval we have that $bin I_a$, absurd.
Third case $[c>x]$.
If $c>x$, then either $a<c<x<b$ or $x<a<c<b$.
$(i)$ If $a<c<x<b$, since $ain I_a$, $xin I_a$ and $I_a$ is an interval we have that $cin I_a$ therefore $cin I_x$.
$(ii)$ If $x<a<c<b$, since $xin I_b$ and $bin I_b$, we have that $ain I_b$ absurd.
Then in general $cin I_x$, this prove that $I_x$ is an interval.
Thanks!
real-analysis analysis proof-verification
edited Dec 15 '18 at 12:08
Brahadeesh
6,47942363
asked Dec 15 '18 at 9:37
Jack J.Jack J.
3442419
$endgroup$
add a comment |
$begingroup$
Theorem. Every open subset $Usubseteqmathbb{R}$ is countable union of disjoint open intervals.
I was looking for proofs for this result and I came to this interesting post: Any open subset of $Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs].
Among all the proofs I started from the simplest one: the one written by G.T. https://math.stackexchange.com/a/1949873/554978.
However, I am not convinced of the proof that $I_x$ is an interval. I do not know if I did not understand correctly or the proof in question is not valid, so I propose one myself and I would like you to tell me which one is valid.
Definition. An interval is a subset $Isubseteqmathbb{R}$ such that, for all $a<c<b$ in $mathbb{R}$, if $a,bin I$ then $cin I$.
Let $xin U$ and we suppose that $xinmathbb{Q}$, then define begin{align} I_x = bigcuplimits_{substack{Itext{ an open interval} \ x~in~I~subseteq~U}} I,end{align}
we must prove that $I_x$ is an interval. About that let $a,bin I_x$ such that $a<c<b$, then we must prove that $cin I_x$.
Since $a,bin I_x$, then $a,bin I$ for same open interval $I$ which contains $x$. If $a$ and $b$ belong to the same $I$, since $a<c<b$ and $I$ is an interval, $cin I$, therefore $cin I_x$.
Now, we denote with $I_a$ the open interval of $I_x$ which contains $a$, but not $b$ and we denote with $I_b$ the open interval of $I_x$ which contains $b$, but not $a$.
First case: $[c=x]$.
If $c=x$, then $cin I_x$ by definition of $I_x$;
Second case: $[c<x]$.
If $c<x$, then either $a<c<x<b$ or $a<c<b<x$.
$(i)$ If $a<c<x<b$, since $ain I_a$ and $xin I_a$ and $I_a$ is an interval, then $cin I_a$, therefore $cin I_x$.
$(ii)$ If $a<c<b<x$, since $xin I_a$ and $ain I_a$, and $I_a$ is an interval we have that $bin I_a$, absurd.
Third case $[c>x]$.
If $c>x$, then either $a<c<x<b$ or $x<a<c<b$.
$(i)$ If $a<c<x<b$, since $ain I_a$, $xin I_a$ and $I_a$ is an interval we have that $cin I_a$ therefore $cin I_x$.
$(ii)$ If $x<a<c<b$, since $xin I_b$ and $bin I_b$, we have that $ain I_b$ absurd.
Then in general $cin I_x$, this prove that $I_x$ is an interval.
Thanks!
real-analysis analysis proof-verification
edited Dec 15 '18 at 12:08
Brahadeesh
6,47942363
asked Dec 15 '18 at 9:37
Jack J.Jack J.
3442419
$endgroup$
add a comment |
$begingroup$
Theorem. Every open subset $Usubseteqmathbb{R}$ is countable union of disjoint open intervals.
I was looking for proofs for this result and I came to this interesting post: Any open subset of $Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs].
Among all the proofs I started from the simplest one: the one written by G.T. https://math.stackexchange.com/a/1949873/554978.
However, I am not convinced of the proof that $I_x$ is an interval. I do not know if I did not understand correctly or the proof in question is not valid, so I propose one myself and I would like you to tell me which one is valid.
Definition. An interval is a subset $Isubseteqmathbb{R}$ such that, for all $a<c<b$ in $mathbb{R}$, if $a,bin I$ then $cin I$.
Let $xin U$ and we suppose that $xinmathbb{Q}$, then define begin{align} I_x = bigcuplimits_{substack{Itext{ an open interval} \ x~in~I~subseteq~U}} I,end{align}
we must prove that $I_x$ is an interval. About that let $a,bin I_x$ such that $a<c<b$, then we must prove that $cin I_x$.
Since $a,bin I_x$, then $a,bin I$ for same open interval $I$ which contains $x$. If $a$ and $b$ belong to the same $I$, since $a<c<b$ and $I$ is an interval, $cin I$, therefore $cin I_x$.
Now, we denote with $I_a$ the open interval of $I_x$ which contains $a$, but not $b$ and we denote with $I_b$ the open interval of $I_x$ which contains $b$, but not $a$.
First case: $[c=x]$.
If $c=x$, then $cin I_x$ by definition of $I_x$;
Second case: $[c<x]$.
If $c<x$, then either $a<c<x<b$ or $a<c<b<x$.
$(i)$ If $a<c<x<b$, since $ain I_a$ and $xin I_a$ and $I_a$ is an interval, then $cin I_a$, therefore $cin I_x$.
$(ii)$ If $a<c<b<x$, since $xin I_a$ and $ain I_a$, and $I_a$ is an interval we have that $bin I_a$, absurd.
Third case $[c>x]$.
If $c>x$, then either $a<c<x<b$ or $x<a<c<b$.
$(i)$ If $a<c<x<b$, since $ain I_a$, $xin I_a$ and $I_a$ is an interval we have that $cin I_a$ therefore $cin I_x$.
$(ii)$ If $x<a<c<b$, since $xin I_b$ and $bin I_b$, we have that $ain I_b$ absurd.
Then in general $cin I_x$, this prove that $I_x$ is an interval.
Thanks!
real-analysis analysis proof-verification
edited Dec 15 '18 at 12:08
Brahadeesh
6,47942363
asked Dec 15 '18 at 9:37
Jack J.Jack J.
3442419
$endgroup$
Theorem. Every open subset $Usubseteqmathbb{R}$ is countable union of disjoint open intervals.
I was looking for proofs for this result and I came to this interesting post: Any open subset of $Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs].
Among all the proofs I started from the simplest one: the one written by G.T. https://math.stackexchange.com/a/1949873/554978.
However, I am not convinced of the proof that $I_x$ is an interval. I do not know if I did not understand correctly or the proof in question is not valid, so I propose one myself and I would like you to tell me which one is valid.
Definition. An interval is a subset $Isubseteqmathbb{R}$ such that, for all $a<c<b$ in $mathbb{R}$, if $a,bin I$ then $cin I$.
Let $xin U$ and we suppose that $xinmathbb{Q}$, then define begin{align} I_x = bigcuplimits_{substack{Itext{ an open interval} \ x~in~I~subseteq~U}} I,end{align}
we must prove that $I_x$ is an interval. About that let $a,bin I_x$ such that $a<c<b$, then we must prove that $cin I_x$.
Since $a,bin I_x$, then $a,bin I$ for same open interval $I$ which contains $x$. If $a$ and $b$ belong to the same $I$, since $a<c<b$ and $I$ is an interval, $cin I$, therefore $cin I_x$.
Now, we denote with $I_a$ the open interval of $I_x$ which contains $a$, but not $b$ and we denote with $I_b$ the open interval of $I_x$ which contains $b$, but not $a$.
First case: $[c=x]$.
If $c=x$, then $cin I_x$ by definition of $I_x$;
Second case: $[c<x]$.
If $c<x$, then either $a<c<x<b$ or $a<c<b<x$.
$(i)$ If $a<c<x<b$, since $ain I_a$ and $xin I_a$ and $I_a$ is an interval, then $cin I_a$, therefore $cin I_x$.
$(ii)$ If $a<c<b<x$, since $xin I_a$ and $ain I_a$, and $I_a$ is an interval we have that $bin I_a$, absurd.
Third case $[c>x]$.
If $c>x$, then either $a<c<x<b$ or $x<a<c<b$.
$(i)$ If $a<c<x<b$, since $ain I_a$, $xin I_a$ and $I_a$ is an interval we have that $cin I_a$ therefore $cin I_x$.
$(ii)$ If $x<a<c<b$, since $xin I_b$ and $bin I_b$, we have that $ain I_b$ absurd.
Then in general $cin I_x$, this prove that $I_x$ is an interval.
Thanks!
real-analysis analysis proof-verification
real-analysis analysis proof-verification
edited Dec 15 '18 at 12:08
Brahadeesh
6,47942363
asked Dec 15 '18 at 9:37
Jack J.Jack J.
3442419
edited Dec 15 '18 at 12:08
Brahadeesh
6,47942363
asked Dec 15 '18 at 9:37
Jack J.Jack J.
3442419
edited Dec 15 '18 at 12:08
Brahadeesh
6,47942363
edited Dec 15 '18 at 12:08
Brahadeesh
6,47942363
edited Dec 15 '18 at 12:08
Brahadeesh
6,47942363
6,47942363
asked Dec 15 '18 at 9:37
Jack J.Jack J.
3442419
asked Dec 15 '18 at 9:37
Jack J.Jack J.
3442419
asked Dec 15 '18 at 9:37
Jack J.Jack J.
3442419
3442419
add a comment |
add a comment |
1 Answer
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$begingroup$
The third case is solved incorrectly.
Third case $[c>x]$
If $c>x$, then either $a<c<x<b$ or $x<a<c<b$.
It should be $a < x < c < b$, not $a < c < x < b$.
So, the proof of the first part should now be:
(i) If $a < x < c < b$, since $b in I_b$, $x in I_b$ and $I_b$ is an interval we have that $c in I_b$ and therefore $c in I_x$.
With this correction, your proof is alright. Well done! :)
answered Dec 15 '18 at 12:07
BrahadeeshBrahadeesh
6,47942363
$endgroup$
add a comment |
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$begingroup$
The third case is solved incorrectly.
Third case $[c>x]$
If $c>x$, then either $a<c<x<b$ or $x<a<c<b$.
It should be $a < x < c < b$, not $a < c < x < b$.
So, the proof of the first part should now be:
(i) If $a < x < c < b$, since $b in I_b$, $x in I_b$ and $I_b$ is an interval we have that $c in I_b$ and therefore $c in I_x$.
With this correction, your proof is alright. Well done! :)
answered Dec 15 '18 at 12:07
BrahadeeshBrahadeesh
6,47942363
$endgroup$
add a comment |
$begingroup$
The third case is solved incorrectly.
Third case $[c>x]$
If $c>x$, then either $a<c<x<b$ or $x<a<c<b$.
It should be $a < x < c < b$, not $a < c < x < b$.
So, the proof of the first part should now be:
(i) If $a < x < c < b$, since $b in I_b$, $x in I_b$ and $I_b$ is an interval we have that $c in I_b$ and therefore $c in I_x$.
With this correction, your proof is alright. Well done! :)
answered Dec 15 '18 at 12:07
BrahadeeshBrahadeesh
6,47942363
$endgroup$
add a comment |
$begingroup$
The third case is solved incorrectly.
Third case $[c>x]$
If $c>x$, then either $a<c<x<b$ or $x<a<c<b$.
It should be $a < x < c < b$, not $a < c < x < b$.
So, the proof of the first part should now be:
(i) If $a < x < c < b$, since $b in I_b$, $x in I_b$ and $I_b$ is an interval we have that $c in I_b$ and therefore $c in I_x$.
With this correction, your proof is alright. Well done! :)
answered Dec 15 '18 at 12:07
BrahadeeshBrahadeesh
6,47942363
$endgroup$
The third case is solved incorrectly.
Third case $[c>x]$
If $c>x$, then either $a<c<x<b$ or $x<a<c<b$.
It should be $a < x < c < b$, not $a < c < x < b$.
So, the proof of the first part should now be:
(i) If $a < x < c < b$, since $b in I_b$, $x in I_b$ and $I_b$ is an interval we have that $c in I_b$ and therefore $c in I_x$.
With this correction, your proof is alright. Well done! :)
answered Dec 15 '18 at 12:07
BrahadeeshBrahadeesh
6,47942363
answered Dec 15 '18 at 12:07
BrahadeeshBrahadeesh
6,47942363
answered Dec 15 '18 at 12:07
BrahadeeshBrahadeesh
6,47942363
answered Dec 15 '18 at 12:07
BrahadeeshBrahadeesh
6,47942363
6,47942363
add a comment |
add a comment |
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