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?

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    $\begingroup$ What exactly is $I$? $\endgroup$ – blat Feb 14 at 9:34
  • $\begingroup$ @blat the unit interval $\endgroup$ – 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].

  • $\begingroup$ No I can not .... I am not understanding your solution .... could you provide details please? $\endgroup$ – Secretly Feb 14 at 10:27
  • $\begingroup$ 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 $\endgroup$ – Kavi Rama Murthy Feb 14 at 10:30
  • $\begingroup$ Can I use the subspace topology definition in the solution? $\endgroup$ – Secretly Feb 14 at 10:33
  • $\begingroup$ Yes. $W$ is open in $[0,1]$ by the definition of subspace topology. $\endgroup$ – Kavi Rama Murthy Feb 14 at 10:36

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