3
$\begingroup$

$A$ 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. Then, $\forall c,d\in A$, $a\leq c. 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.

  • 1
    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
  • 0
    Intervals according to Rudin are open. So, by intervals do you mean, closed intervals, firstly?2012-03-10
  • 0
    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
  • 0
    Should I write "let $A=(a,b)$ WLOG"?2012-03-10
  • 0
    @Emir, what is your question??? Your proof is correct...2012-03-10

1 Answers 1