Proof of Intermediate value theorem in non-standard analysis.

Question

A real function is continuous on $[a,b]$ such that either $f(a) > 0 >f(b)$ or $f(a) < 0 < f(b)$, then prove that $f(c) = 0$ for some $c \in (a,b)$.

---

Proof :-

We divide $[a,b]^*$ in $H$ equal parts, where $H$ is a positive infinite hyperinteger.

We get,

$$a, a +\delta, a+2\delta,\ ... \ , a+ H\delta = b$$

Let $a + K\delta$ be the last partition point at which $f(a+ K\delta) <0$,

Thus

$$f(a+K\delta) < 0 \le f(a+(K+1)\delta)$$

Since $f$ is continous $f(a+K\delta) \approx f(a+(K+1)\delta)$, thus $f(a+K\delta) \approx 0$,

Let $c = st(a+ K\delta)$

Therefore $f(c)= st(f(a+ K\delta)) =0 $

---

I have two questions,

1. Shouldn't it be $f^*(x)$ everywhere instead of $f(x)$ because a real function won't have a hyperreal number in its domain. Like it should be $f^*(a+K\delta) <0$ in the third line because $a+K\delta$ is not a real number it is a hyperreal number.
2. Why does $f(a+K\delta) \approx f(a+(K+1)\delta)$ ? Can't it be $f(a+K\delta) = 1000$ and $f(a+(K+1)\delta) = 2000$ ? Then those two are not infinitely close, right?

Answer 1

Concerning 1, the convention is to delete the asterisk on functions but keep them on sets. So whenever $f$ is applied to a nonzero infinitesimal or other hyperreal number not in $\mathbb R$, what is always meant is the natural extension ${}^\ast\! f$.

As for 2, the values of $f$ at have opposite sign at infinitely close points. Since the values are infinitely close, they must be infinitesimal by the triangle inequality. More precisely, $0\leq f(a+(K+1)\delta)\leq f(a+(K+1)\delta)-f(a+K\delta)$ which is infinitesimal by continuity. Therefore $f(a+(K+1)\delta)$ itself is infinitesimal.

Answer 2

For variety...

Because $f$ is continuous, the outer two equalities hold. Because $\delta$ is infinitesimal, the inner equality holds.

$$\operatorname{st}(f(a + K \delta))=f(\operatorname{st}(a + K \delta)) = f(\operatorname{st}(a + (K+1) \delta)) =\operatorname{st}(f(a + (K+1) \delta)) $$