# Prove that [0,1] is connected.

I wrote a proof, and I just wanted to verify if it is correct.

Proof : Suppose not. Then, $$[0,1] = U\cup V$$ for $$U$$ and $$V$$ open in $$[0,1]$$ such that $$U\cap V = \emptyset$$. Let $$0\in U$$. Consider $$s= sup\{t \mid [0,t] \subset U\}$$. Clearly, $$s\in[0,1]$$.

If $$s\in U$$, then since $$U$$ is open, $$\exists \epsilon >0$$ such that $$B(s;\epsilon) \subset U$$ . But then $$s+\epsilon \in U$$ , and $$s+\epsilon > s$$, contradicting the definition of $$s$$. So, $$s \notin U$$.

Then, $$s\in V$$. But again, since $$V$$ is open, $$\exists \epsilon >0$$ such that $$B(s;\epsilon)\subset V$$. But, since $$s$$ is the least upper bound of $$U$$, for this $$\epsilon$$, $$\exists s_{1} \in U$$ such that $$s_{1} \in B(s;\epsilon)$$. Then, $$s_{1} \in U\cap V$$, which contradicts the fact that $$U$$ and $$V$$ are disjoint.

So, our assumption is wrong and $$[0,1]$$ is connected.

• Looks good to me. – Arthur Nov 6 '18 at 14:19