# Show that there is a number $a \in (0,1]$ such that $U = [0,a) \bigcup W.$

Let $$U \subseteq I$$ be an open set (but not all of $$I$$), and assume that $$0 \in U.$$ Show that there is a number $$a \in (0,1]$$ such that $$U = [0,a) \bigcup W,$$ where $$W$$ is also open in $$I$$ and $$W \bigcap [0,a) = \emptyset.$$

My trial:

I know that every open set in $$\mathbb{R}$$ is a union of disjoint open intervals. I also know that we have 3 types of intervals an interval that is totally outside the interval $$[0,1]$$ from left, an interval that extends before and after $$0$$ by a small number say $$\frac{-1}{n}$$ and $$\frac{1}{n}$$ where $$n \geq 1$$ and a third interval that $$(\frac{1}{n}, 1]$$ but I do not know how to choose $$a$$ and $$W$$ in all these cases. I also know that a space $$X$$ is connected if the only separations of $$X$$ are the trivial ones(I am not sure how this may help here). Could anyone help me formulate a rigorous proof for this please?

• What exactly is $I$? – blat Feb 14 at 9:34
• @blat the unit interval – Secretly Feb 14 at 10:29

Since $$U$$ is open and $$0 \in U$$ there exists $$t>0$$ such that $$[0,t) \subseteq U$$. Let $$a=\sup \{t: [0,t) \subseteq U\}$$. Let $$W=(a,1]\cap U$$. Can you verify that this satisfies the requirements?
[You will have to show that $$a \notin U$$. Prove this by contradiction].
• If $a \in U$ then $[0,a+\epsilon) \subseteq U$ for some $\epsilon >0$ because $U$ is open. So this would lead to a contradiction to the definition of $a$ as a supremum. @Secretly – Kavi Rama Murthy Feb 14 at 10:30
• Yes. $W$ is open in $[0,1]$ by the definition of subspace topology. – Kavi Rama Murthy Feb 14 at 10:36