1. 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 $D$ is pathwise connected and $f\rightarrow R" alt="f\rightarrow R" /> is continuous. Prove that $f(D)$ 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, $\phi (t)=(1-t)\alpha +t\beta$ where $\phi (0)=\alpha$and $\phi (1)=\beta$. Then show that the function lives on the interval [0,1]?

2. 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 $D$ is pathwise connected, $X$ is a space, and $f\to X" alt="f\to X" /> is continuous, then $f(D)$ is pathwise connected.

2) A subset of $\mathbb{R}$ is pathwise connected if and only if it is an interval.

Give it a try.

3. 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?

4. Recall the definition of a pathwise connected set $X$: we say that $X$ is pathwise connected if for points $a,b\in X$, there exists a continuous function $\gamma:[0,1]\to X$ such that $\gamma(0)=a$ and $\gamma(1)=b$. Our goal will be to produce such a function for our $X$.

You have two important facts: $D$ is pathwise connected, so given $a,b\in D$, there exists a continuous function $\gamma_D:[0,1]\to D$ such that $\gamma(0)=a$ and $\gamma(1)=b$; and $f\to X" alt="f\to X" /> is continuous. How can you come up with $\gamma:[0,1]\to f(X)$ connecting two points?

5. 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, $D$ is connected and so by continuity $f(D)$ is connected. And, surely you know every connected subspace of $\mathbb{R}$ is an interval.