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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

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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)2012-03-10
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    Intervals according to Rudin are open. So, by intervals do you mean, closed intervals, firstly?2012-03-10
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    Good point, although the proof doesn't really rely on any assumption about wether the endpoints are contained in the interval or not2012-03-10
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    Should I write "let $A=(a,b)$ WLOG"?2012-03-10
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    @Emir, what is your question??? Your proof is correct...2012-03-10

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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]$2012-12-23