Thread: Pathwise connected proof

Ok, so I would like to know how to prove the following proof. If anyone has any insight I would greatly appreciate it.

Suppose is pathwise connected and \rightarrow R" alt="f\rightarrow R" /> is continuous. Prove that is an interval.

Is it enough to create a function that is continuous and then show that the function lives on an interval?

For example, where and . Then show that the function lives on the interval [0,1]?

You have to show that this works for any arbitrary function. Picking a particular function (as you did) will not suffice.

You are really being asked to show two things:

1) If is pathwise connected, is a space, and \to X" alt="f\to X" /> is continuous, then is pathwise connected.

2) A subset of is pathwise connected if and only if it is an interval.

Give it a try.

I have the proof of part 2 in my notes that I obtained from previous guidance. So I guess part one is what I struggle with. Its not hard for me to believe that it is true but I don't know how to get started. Maybe a proof by contradiction?

Recall the definition of a pathwise connected set : we say that is pathwise connected if for points , there exists a continuous function such that and . Our goal will be to produce such a function for our .

You have two important facts: is pathwise connected, so given , there exists a continuous function such that and ; and \to X" alt="f\to X" /> is continuous. How can you come up with connecting two points?

Not to butt in, but I feel as though this is simpler. We (and by we I mean you, zebra, and I) have already proven that a path connected space is connected. So, is connected and so by continuity is connected. And, surely you know every connected subspace of is an interval.