2
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ You could try to show $(0,1)~\text{connected}\implies[0,1]~\text{connected}$ $\endgroup$ – Henry Apr 20 '16 at 18:57
  • $\begingroup$ $U$ 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. $\endgroup$ – Bungo Apr 20 '16 at 19:02
5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Well, I want to deduce from the implication that $[0,1]$ is connected. So, I cannot use it. $\endgroup$ – johnny09 Apr 20 '16 at 19:01
  • $\begingroup$ This question might be a step in a proof that $[0,1]$ is connected, which would make your proof circular... $\endgroup$ – Wojowu Apr 20 '16 at 19:02
  • $\begingroup$ @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/… $\endgroup$ – user223391 Apr 20 '16 at 19:04
1
$\begingroup$

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).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.