# 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

You're probably not looking for the answer still after a year, but for those who come across this in the future, it may be helpful. You are pretty much right about your proof, but here is a more formal way of writing it.

That is that "If $$a, then the subspace $$[a,b]$$ of $$E^1$$ is connected".

Let $$a and let $$[a,b]$$ be a subspace of $$R$$ with the $$E^{1}$$ topology. For the sake of contradiction, assume that $$[a,b]$$ is not connected. Then $$[a,b]= U \cup V$$, where $$U$$ and $$V$$ are nonempty disjoint open sets in $$[a,b]$$. Without loss of generality we may assume that $$b \in V$$. Because $$U$$ is nonempty and bounded above (by $$b$$), the axiom of completeness states that $$U$$ has a least upper bound $$s$$. We will prove that $$s$$ is not an element of either $$U$$ or $$V$$, and this will yield a contradiction. Suppose $$s \in U$$. Since $$U$$ is open, there exists an $$\epsilon> 0$$ such that $$B_{\epsilon}(s) \subseteq U$$. Thus, since $$s+ \frac{\epsilon}{2} \in U$$ and $$s, there exists an element in $$U$$ that is greater than $$s$$. Therefore, $$s$$ is not an upper bound for $$U$$. This is a contradiction. Thus, $$s \not \in U$$. Now suppose $$s \in V$$. Then, because $$V$$ is open, there exists an $$\epsilon>0$$ such that $$B_{\epsilon}(s) \subseteq V$$. Then $$s- \frac{\epsilon}{2}$$ is an upper bound for $$U$$. But $$s- \frac{\epsilon}{2}, which is a contradiction because $$s$$ is the \emph{least} upper bound for $$U$$. So $$s \not \in V$$. Thus, $$s \not \in U \cup V$$, which is a contradiction. Therefore, $$[a,b]$$ is connected.

• I think it would be clearer for the second part, if you added the argument that if $s-\frac{\epsilon}{2}$ is not an upper bound for $U$, then there must be a $p\in U$ such that $s-\frac{\epsilon}{2} < p$ and since $s=\sup{U}$, $p\in(s-\frac{\epsilon}{2},s)\subseteq B_{\epsilon}(s) \subseteq V$ which cannot be the case since $U$ and $V$ are disjoint. – üzeyir Jun 10 at 16:43

TheoremA topological space $$X$$ is connected iff every continuous function $$f: X \to\{±1\}$$ is a constant function.

ProofLet $$X$$ be a connected space and $$f: X \to \{±1\}$$ a continuous function. We want to show that $$f$$ is a constant function. If $$f$$ is non-constant, then it is onto. Let $$A = f^-{1}(1)$$ and $$B = f^{-1}(-1)$$. Then $$A$$ and $$B$$ are disjoint non-empty subsets of $$X$$ such that $$A$$ and $$B$$ are both open and closed subsets of $$X$$ and $$X = A U B$$.(Why?). This is a contradiction. Therefore $$f$$ is constant.

Conversely, let us assume that $$X$$ is not connected. Therefore there exist two disjoint proper non-empty subsets $$A$$ and $$B$$ in $$X$$ such that $$A$$ and $$B$$ are both open and closed in $$X$$ and $$X = AU B$$. Now we define a map $$f: X \to \{±1\}$$ as

$$f (x) = \begin{cases} 1,&x\in A\\-1,& x \in B\end{cases}$$.

Then $$f: X \to\{±1\}$$ is a continuous non-constant function. (Why?) This completes the proof.

Now I am giving you a shorter proof of your problem.

Suppose $$j$$ is any interval if $$f: J \to\{±1\}$$ is an onto Continuous function, then there exist $$x,y\in J$$ such that $$f(x) = -1$$ and $$f (y) = 1$$. By the intermediate value theorem, there exists $$z$$ between $$x$$ and $$y$$ such that $$f(z) = 0$$, a contradiction to our assumption that $$f$$ takes only the values $$\pm 1$$. Hence no such $$f$$ exists and hence $$J$$ is connected. It follows that any interval is connected.

• I'm sorry but I couldn't really follow the proof. What is $f$ here? And why showing that no onto continuous function from an interval J to $\{\pm 1\}$ exists shows that $J$ is connected? – P-addict May 13 at 11:53
• I have edited. Now you can see. – Ryszard Ebgelking May 13 at 12:30
• Here $j$ is any interval. – Ryszard Ebgelking May 13 at 12:32