# How to demonstrate that [0,1] is a connected set?

to a point, the demonstration is clear for me. I'll sum it up like this:

So firstly I'll try to demonstrate that I=[0,1] is not connected to get to a contradiction, which will conduct to the fact that [0,1] is actually connected.

In case [0,1] is disconnected, that means there are two open sets A and B, so that $A\cap I$, respectively $B\cap I$ that are non-empty, disjoint, and their union is I. (from the definition of disconnected sets)

Without loss of generality, let's say 1 E B we define c=sup($A\cap I$) which means c>=0 and c<=1 which conducts to the fact hat c E [0,1], so c E I which means c E ($A\cap I$)U($B\cap I$) because I is the union of $A\cap I$ and $B\cap I$ from the definition of disconnected sets this means that c E $(AUB)\cap I$ which implies that c E (AUB)

then, there are two cases case 1) c E A case II) c E B

for case I) it says that in this case cE (O,1) * Because A is open and it contains c, then there are points in $A\cap I$ which are higher than c, which contradicts the definition of supremum**

case II) is similar and conducts to a contradiciton as well

*which is where my confusion starts, how do you know for sure that it is (0,1) and not just [0,1) ? **you can know for sure that there is such a point in A(because it's an open set) but how do I know that there is such a point in $A\cap I$ ?

in my book i have this definition of connectedness

I is connected if I is not disconnected

I is disconnected if there are two open sets A & B, so that :

i) $A\cap I$ is non empty set

$B\cap I$ is non empty set

ii) ($A\cap I$) U ($A\cap I$) = I

iii) $$A\cap I)\cap (B\cap I$$ is empty set

• Your notation is confusing me. Rather than use "A^I", I think you want $A\cap I$, which displays (thanks to the magic of MathJax) as $A\cap I$. Commented Feb 18, 2016 at 1:01
• I'd say that the hardest part is to find a definition for $[0,1]$ and for " a set is connected" which doesn't imply trivially that $[0,1]$ is connected Commented Feb 18, 2016 at 1:51

Your question is a little convoluted, so I'll just answer the question in the title. I hope this helps anyway.

Suppose $[0,1]$ can be separated into $A,B$. We can suppose $1 \in B$ without loss of generality.

Let $c:=\sup A$.

Now, $c$ can't belong to $A$, because if it did it would not be $1$ and, since $A$ is open, it would imply that a greater element than $c$ would belong to $A$. But $c$ can't belong to $B$ either, because this would imply either that $c=0$ (from where $A$ would not be open) or that there would exist someone smaller than $c$ in $B$ which would still be an upper bound for $A$, since $B$ is open.

Therefore, $c$ is not in $A$ nor $B$, a contradiction.

• how can [0,1] which is closed, be separated into A and B if A and B are open? Commented Feb 18, 2016 at 1:49
• $A,B$ are open in the subspace topology. Commented Feb 18, 2016 at 2:18
• could you please explain to me what is the difference between open and open in the subspace topology? that would be greatly appreciated. thank you Commented Feb 18, 2016 at 2:46
• @meiznub If $A$ is open in the real numbers, then $A\cap[0,1]$ is called "open in the subspace topology" (where the subspace is $[0,1]$). Conversely, a set $B$ is open in the subspace topology iff there is a set $A$ that's open in the real numbers such that $B=A\cap[0,1]$. Commented Feb 18, 2016 at 2:58
• @meiznub Basically, ignoring the whole subspace topology thing: Suppose $[0,1]$ is separated into $A\cap[0,1]$ and $B\cap[0,1]$ where $A$ and $B$ are open. Assume $1\in B\cap[0,1]$. Define $c$ to be $\sup(A\cap[0,1])$ and find a contradiction. Commented Feb 18, 2016 at 3:02

An alternative answer is to show that a path-connected metric space is connected, and then showing $I$ is path-connected is fairly simple. (Then you need to think why the converse, connected implies path-connected, is not true!)

• Showing that a path-connected metric space is connected will go through the fact that $[0,1]$ is connected. Commented Feb 18, 2016 at 1:16
• It seems so. Good to remember these things Commented Feb 18, 2016 at 1:21
• @AloizioMacedo This makes me wonder: is using that result actually necessary? Or just the most common way to do it? Commented Feb 18, 2016 at 2:10