TITLE: $(\frac{1}{3},1)\cup(\frac{1}{5},\frac{1}{3})\cup(\frac{1}{7},\frac{1}{5})\cup\ldots$ is not a finite union of open intervals
QUESTION [6 upvotes]: Let $X=(\frac{1}{3},1)\cup(\frac{1}{5},\frac{1}{3})\cup(\frac{1}{7},\frac{1}{5})\cup\ldots \subseteq \mathbb{R}$
I have tried to show that $X$ is not a finite union of open intervals.
Does the following argument make sense?

Suppose, for contradiction, that 
$$X=(a_1,b_1)\cup\ldots\cup(a_n,b_n)$$ 
where we assume that $a_1\leq a_2 \leq \ldots \leq a_n.$
If $a_1<0,$ then there is a negative number in $X$ - which is a contradiction.
If $a_1=0,$ then $X$ contains $\frac{1}{2m+1}$ for sufficiently large $m \in \mathbb{N}$ - which is a contradiction.
If $a_1>0,$ then $X$ does not contain $\frac{1}{2m}$ for sufficiently large $m \in \mathbb{N}$ - which is a contradiction.

REPLY [2 votes]: To see a stronger statement: your $X$ has infinitely many connected components, a finite union of open intervals $F$ would have finitely many. So they are not even homeomorphic to each other.