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$A$ is an interval $\implies$ $A$ is pathwise connected.

This kind of goes off one of my previous general questions about path connectedness. I've tried to formalize my attempt at proving this a bit:

My definition of path connectedness says that $A\subset X$ is pathwise connected if $\forall x,y\in A$ there exists a continuous path $\gamma:[a,b]\rightarrow A$ with $\gamma(a)=x$ and $\gamma(b)=y$.

Attempt at a proof:

Let $A\subset\mathbb{R}$ be an interval. Without loss of generality, let $A=[a,b]$ for some $a,b\in\mathbb{R}$: $a<b$. Then, $\forall c,d\in A$, $a\leq c<d\leq b$. Consider $\gamma:[0,1]\rightarrow [a,b]: \gamma(t)=c+t(d-c).$ Then, $\gamma(0)=c$ and $\gamma(1)=d$. Since $c,d\in A$ were arbitrary, $\implies$ A is path connected.

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That appears to be correct. In particular, you are showing that an interval is convex (from which it follows immediately that it is path-connected) –  you Mar 10 '12 at 2:27
    
Intervals according to Rudin are open. So, by intervals do you mean, closed intervals, firstly? –  user21436 Mar 10 '12 at 2:28
    
Good point, although the proof doesn't really rely on any assumption about wether the endpoints are contained in the interval or not –  you Mar 10 '12 at 2:32
    
Should I write "let $A=(a,b)$ WLOG"? –  Emir Mar 10 '12 at 2:37
    
@Emir, what is your question??? Your proof is correct... –  zapkm Mar 10 '12 at 2:42

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The proof is correct, but you have to justify the "without loss of generality", as you just dealt with closed bounded intervals. Maybe you could add a line saying that the argument if we treat other types of intervals.

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A possible modification of the argument that takes care of all cases is as follows: without deciding what kind of interval $A$ is, pick $c,d \in A$ that you want to show are connected by a path. Then without loss of generality, assume $A = [c,d]$ –  ronno Dec 23 '12 at 18:38

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