The question says:
Let $I = [0,1]$. Suppose $I = U \bigcup V$ is a separation of $I,$ and that $0 \in U.$ Show that $V = \emptyset.$
My trial:
I got a hint if $V \neq \emptyset,$ then I can use the following problem:
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 $W \bigcap [0,a) = \emptyset.$
But I still confused about how to use this problem, could anyone help me please?
Let $A=\{x\in[0,1]\mid[0,x]\subset U\}$. Since $0\in U$ and $U$ is an open subset of $[0,1]$, $A\neq\emptyset$. And $A$ is bounded, since it is a subset of $[0,1]$. Therefore, it makes sense to talk about $s=\sup A$.
Sunce $s\in[0,1]=U\cup V$, $s\in U$ or $s\in V$. But, since $V$ is an open subset of $[0,1]$, if $s\in V$, then $(s-\varepsilon,s]\subset V$ for some $\varepsilon>0$. Therefore, $(s-\varepsilon,s]$ would have no element of $U$. But this goes agains the definition os $s$ as $\sup\{x\in[0,1]\mid[0,x]\subset U\}$. So, $s\in U$. But then $[0,s]\subset U$.
And we cannot have $s<1$ because then, since $s\in U$ and $U$ is open, $[0,s+\varepsilon)\subset U$ for some $\varepsilon>0$ and so $s+\varepsilon\in A$, which is impossible, since $s+\varepsilon>s=\sup A$.
So, $s=1$, and therefore $V=\emptyset$, which is impossible.