5
$\begingroup$

Theorem. Every open subset $U\subseteq\mathbb{R}$ 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 $I\subseteq\mathbb{R}$ such that, for all $a<c<b$ in $\mathbb{R}$, if $a,b\in I$ then $c\in I$.

Let $x\in U$ and we suppose that $x\in\mathbb{Q}$, then define \begin{align} I_x = \bigcup\limits_{\substack{I\text{ an open interval} \\ x~\in~I~\subseteq~U}} I,\end{align} we must prove that $I_x$ is an interval. About that let $a,b\in I_x$ such that $a<c<b$, then we must prove that $c\in I_x$.

Since $a,b\in I_x$, then $a,b\in I$ 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, $c\in I$, therefore $c\in I_x$.

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 $c\in I_x$ 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 $a\in I_a$ and $x\in I_a$ and $I_a$ is an interval, then $c\in I_a$, therefore $c\in I_x$.

$(ii)$ If $a<c<b<x$, since $x\in I_a$ and $a\in I_a$, and $I_a$ is an interval we have that $b\in I_a$, 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 $a\in I_a$, $x\in I_a$ and $I_a$ is an interval we have that $c\in I_a$ therefore $c\in I_x$.

$(ii)$ If $x<a<c<b$, since $x\in I_b$ and $b\in I_b$, we have that $a\in I_b$ absurd.

Then in general $c\in I_x$, this prove that $I_x$ is an interval.

Thanks!

$\endgroup$

1 Answer 1

3
$\begingroup$

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! :)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .