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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.

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

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