I am trying to understand the proof below, taken from Elementary Analysis, second edition, by Kenneth Ross (pages 182-183). The assertion being proven is that a function on an interval is continuous if and only if its graph is path-connected.
I searched for "continuous function graph path connected" and found a couple posts that seemed relevant. They will feature in my second question.
First question: The part of the proof that I still cannot fathom at all is the two sentences at the end of the first image and the beginning of the second, beginning "Now we apply the Intermediate Value Theorem again" and ending with "so we have verified IVP." My understanding is as follows: we are trying to show that $f$ satisfies the IVP (described in lower part of first image). We know that $x(t)$ is continuous, but we cannot assume that $f$ is continuous.
I am trying to talk myself into understanding the logic involved in the sentence "Since $f(x(\alpha)) = f(a)$ and $f(x(\beta)) = f(b)$, we have $f(x(t)) = c$ for some $t$ in $(\alpha, \beta)$." The story I am trying to tell myself goes something like "we know $\alpha$ leads to the number $f(a)$, and $\beta$ leads to the number $f(b)$, so consider a number $c$ between $f(a)$ and $f(b)$. Then there must be... what?" We know the function $x$ is continuous, but we cannot assume that $f$ is continuous. So why must there be a $t \in (\alpha, \beta)$ that produces the number $c$? I wonder if the function composition aspect of this is messing with me somehow.
Second question: I am a little embarrassed to admit that I've probably spent close to ten hours trying to understand this proof. When I searched for similar questions on this site, I found a couple posts (here, and Nate Eldredge's answer here) where the answers given are so simple that I frankly have trouble believing that those proofs are equivalent to the one I've included below. Are they? I noticed that both are considering a function $f$ on a closed interval, whereas the proof I've been toiling over is valid for any interval.
(I also found an answer here that seemed relevant, but the answer is about as long as the proof I've been trying to understand, so I was reluctant to spend who knows how many more hours trying to unravel this proof.)
Thanks.
If $f$ is continuous, then for each $(a, f(a))$ and $(b, f(b))$ in $G, \gamma(t)= (t, f(t))$ is a path connecting these points. Note that $\gamma$ is continuous by Proposition 21.2
For the converse, assume $G$ is path-connected. Consider $a, b$ in $I$ where $a<b$. Since $G$ is path-connected, there is a continuous function $\gamma:[0,1] \to G$ satisfying $\gamma(0)=(a, f(a))$ and $\gamma(1)=(b, f(b))$. By Proposition $21.2$, we can write $\gamma(t)=(x(t), y(t))$ where $x$ and $y$ are continuous real-valued functions on $[0,1]$. Of course, $y(t)=f(x(t))$ for all $t$. First, we claim $$\{(z, f(z)): z \in[a, b]\} \subseteq \gamma([0,1])\tag{1}$$ Since $x(t)$ is continuous on $[0,1], x(0)=a$ and $x(1)=b$, the Intermediate Value Theorem 18.2 for continuous functions shows that if $z$ is in $(a, b)$, then $x(t)=z$ for some $t \in(0,1)$. Thus $(z, f(z))=(x(t), f(x(t)))=\gamma(t)$. Since the cases $z=a$ and $z=b$ are trivial, this implies (1).
Next we show $f$ satisfies IVP, the intermediate value property:
$(\mathrm{IVP})$ If $a, b \in I, a<b$ and $c$ is between $f(a)$ and $f(b),$ then $f(x)=c$ for some $x \in[a, b]$.
Let $\alpha=\sup \{t \in[0,1] : x(t) \le a\}$ and $\beta=\inf \{t \in[0,1] : x(t) \ge b\}$. Thus $0 \le \alpha < \beta \leq 1$. Since $x(t)$ is continuous, we see that $x([\alpha, \beta])=[a, b]$. Now we apply the Intermediate Value Theorem again, this time to $f(x(t))$ on $[\alpha,\beta]$. Since $f(x(\alpha))=f(a)$ and $f(x(\beta))=f(b)$. we have $f(x(t))=c$ for some $t$ in $(\alpha, \beta)$. For this $t$ we have $x(t) \in[a, b]$, so we have verified IVP.
Now assume $f$ is not continuous on $I$. Then there exists $x_0\in I$, $\epsilon>0$, and a sequence $\left(x_n\right)$ in $I$ that converges to $x_0$ satisfying $\left|f\left(x_n\right)-f\left(x_0\right)\right|>\epsilon$ for all $n$. Then $f\left(x_n\right)>f\left(x_0\right)+\epsilon>f\left(x_0\right)$ for infinitely many $n$, or else $f\left(x_0\right)<f\left(x_0\right)-\epsilon<f\left(x_0\right)$ for infinitely many $n$. Passing to a subsequence, we may assume $f\left(x_n\right)>f\left(x_0\right) + \epsilon>f\left(x_0\right)$ for all $n$ say. By the IVP, for each $n$ there is $y_n$ between $x_n$ and $x_0$ so that $f\left(y_n\right)=f\left(x_0\right)+\epsilon$. Then $\left(y_n, f\left(y_n\right)\right)$ is a sequence in $G$ satisfying $$\lim_n y_n=x_0\text{ and } \lim \left(y_n, f\left(y_n\right)\right)=\left(x_0, f\left(x_0\right)+\epsilon\right).\tag{2}$$ Moreover, there exist $a<b$ in $I$ such that $x_0$ and all $y_n$ belong to $[a, b]$. We apply (1) to this $a$ and $b$ and $\gamma$ described there. So, for each $n \in \Bbb N$ there exists $t_n \in[0,1]$ so that $\gamma\left(t_n\right)=\left(y_n, f\left(y_n\right)\right)$. Passing to a subsequence $\left(t_{n_k}\right)_{k \in \Bbb N}$ of $\left(t_n\right)$, we may assume that $\lim_k t_{n_k}=t_0$ for some $t_0$ in $[0,1]$. Since $\gamma$ is continuous, we conclude from (2) that $$\gamma\left(t_0\right)=\lim _k \gamma\left(t_{n_k}\right)=\lim_k\left(y_{n_k}, f\left(y_{n_k}\right)\right)=\left(x_0, f\left(x_0\right)+\epsilon\right).$$ Since this limit is not in $G$, and $\gamma([0,1]) \subset G$, we have a contradiction and $f$ is continuous on $I$.
First Answer You are correct that we don't know that $f$ is continuous. But note that $f(x(t)) = y(t)$, the $y$ coordinate of the curve $\gamma$, which we do know is continuous. So while we cannot apply the Intermediate Value Theorem to $f$ directly, we can apply it to $f(x(t))$, and that is what the author means.
Second Answer "Equivalent" isn't really a concept here. They are all valid proofs of the result (at least for compact intervals - but it isn't hard to extend the result from compact intervals to general ones).
HaHa's post is just the basic idea. The post does not give details. Nate Eldredge's proof gives more details, but chooses a simpler route than the proof here. It does depend on already knowing that if $f$ is not continuous at $x$, then there must be a sequence $(x_k)$ converging to $x$ with $f(x_k)$ not converging to $f(x)$. Nate demonstrates that this is impossible, so $f$ must be continuous.
There are always multiple ways to prove things. Nate has found a nicer one than Kenneth Ross, but that does not make Kenneth Ross's proof invalid.