Here is the problem as it is written in the text:

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Let $\displaystyle f(t,x)$ be defined and continuous for $\displaystyle a\leq t\leq b$ and $\displaystyle x\in\mathbb{R}^n$. The purpose of this exercise is to show that the problem $\displaystyle dx/dt=f(t,x)$, $\displaystyle x(a)=x_0$, has a solution on the interval $\displaystyle t\in[a,c]$ for some $\displaystyle c>a$. Perform the operations as follows: Divide $\displaystyle [a,b]$ into $\displaystyle n$ equal parts: $\displaystyle a=t_0,t_1,...,t_n=b$, and define a continuous function $\displaystyle x_n$ inductively by

$\displaystyle \bigg\{\begin{array}{l}x_n'(t)=f(t_i,x_n(t_i)),~~t _i<t<t_{i+1}\\x_n(a)=x_0\end{array}$

Put $\displaystyle \Delta_n(t)=x_n'(t)-f(t,x_n(t))$, so that

$\displaystyle x_n(t)=x_0+\int_a^t f(s,x_n(s))+\Delta_n(s)\,ds$

Use the Arzela-Ascoli Theorem to find a convergent subsequence of the $\displaystyle x_n$. Show that the limit satisfies $\displaystyle dx/dt=f(t,x)$ and $\displaystyle x(a)=x_0$.

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The version of the A-A Thm that seems the most applicable here is the following:

Let $\displaystyle A$ be a compact set. If $\displaystyle \mathcal{B}\subset\mathcal{C}(A,\mathbb{R}^n)$ is (uniformly) equicontinuous and pointwise bounded, then every sequence in $\displaystyle \mathcal{B}$ has a uniformly convergent subsequence.

Letting $\displaystyle x_n=\mathcal{B}$, I don't know how to show equicontinuity or pointwise boundedness. I know that each individual $\displaystyle x_n$ is bounded and uniformly continuous, but that doesn't mean that they're pointwise bounded or equicontinuous (because there are infinitely many $\displaystyle x_n$).

EDIT: I've figured out what to do after applying the A-A Theorem. However, I still don't know how to prove the criteria for using the A-A Theorem, so I still need help with that.

Also, I'm not sure where that constant $\displaystyle c$ came from. It pops up and then disappears almost immediately, never to be mentioned again. Maybe it's a typo.

Please help!