# Proof of $[0,1]~\text{disconnected}\implies(0,1)~\text{disconnected}$ [duplicate]

I want to prove the following implication

$$[0,1]~\text{disconnected}\implies(0,1)~\text{disconnected}.$$

My try: Suppose $[0,1]=U\cup V$ with $U,V$ open, disjoint and nonempty.

Using the subspace topology of $\mathbb{R}$ we also have $U=U'\cap[0,1]$ and $V=V'\cap[0,1]$ where $U',V'$ are open in $\mathbb{R}$.

We have $(0,1)=(0,1)\cap[0,1]=(U'\cap(0,1))\cup(V'\cap(0,1))$. This is a union of open sets since $(0,1)$ is an open interval.

How can I prove that this is also a union of disjoint sets?

Will this $U\cap V=(U'\cap[0,1])\cap(V\cap[0,1])=U'\cap V'\cap[0,1]=\emptyset$ be sufficient for showing the disjoint-requirement?

Also, I do not know how to start for showing that $(0,1)$ is union of nonempty sets.

I would appreciate any help.

## marked as duplicate by user223391, Leucippus, zz20s, Daniel W. Farlow, Claude LeiboviciApr 21 '16 at 5:14

• You could try to show $(0,1)~\text{connected}\implies[0,1]~\text{connected}$ – Henry Apr 20 '16 at 18:57
• $U$ and $V$ are disjoint. Since $U' \cap (0,1) \subset U$ and $V' \cap (0,1) \subset V$, it follows that $U' \cap (0,1)$ and $V' \cap (0,1)$ are disjoint. You need to show that they are both nonempty. – Bungo Apr 20 '16 at 19:02

Well, $[0,1]$ is connected so the implication follows trivially. Also, $[0,1]$ is disconnected implies that the Earth is flat, $2+2=5$ and anything else you'd like.
• Well, I want to deduce from the implication that $[0,1]$ is connected. So, I cannot use it. – johnny09 Apr 20 '16 at 19:01
• This question might be a step in a proof that $[0,1]$ is connected, which would make your proof circular... – Wojowu Apr 20 '16 at 19:02
• @johnny09 If you want to prove that $[0,1]$ is connected, it's not that hard to do directly. See here: math.stackexchange.com/questions/339401/… – user223391 Apr 20 '16 at 19:04
You just take off the points $0,1$.
$(0,1)=(U\setminus\{0,1\})\cup(V\setminus\{0,1\})$. Prove $U\setminus\{0,1\}$ and $V\setminus\{0,1\}$ are open in $(0,1)$ (follows almost trivially), non-empty (trivial) and disjoint (more trivial).