The magic of existential transfer

Question

Yesterday I finished grading the final exam in a course on infinitesimal calculus taught to 130 freshmen. One of the problems on the exam was to show that if a function $f$ is differentiable at $c\in\mathbb{R}$ and satisfies $f'(c)>0$ then there is a point $x>c$ such that $f(x)>f(c)$. One of the students provided an original solution. If $\alpha>0$ is infinitesimal then $st(\frac{f(c+\alpha)-f(c)}{\alpha})>0$ and it easily follows that $f(c+\alpha)-f(c)>0$. Hence the hyperreal $x=c+\alpha$ satisfies $f(x)>f(c)$ and therefore the following formula is satisfied over the hyperreals: $$(\exists x)\;f(x)>f(c).$$ Now by existential transfer the same formula holds over the reals. Therefore there exists $x\in\mathbb{R}$ such that $f(x)>f(c)$, QED.

Are there additional examples of existential transfer at this level?

Answer 1

Let $f: \mathbb{N} \to \mathbb{R}$ be a (first-order-definable) function of finite support $\{1, 2, \dots, m \}$ (i.e. a finite sequence), and transfer it to $^* f$ from $^* \mathbb{N}$ to $^* \mathbb{R}$. Then since there exists $n$ such that $f(n)$ is minimised, there must also exist $m$ in $^* \mathbb{N}$ such that $^*f(m)$ is minimised. This can be used to make a cute proof of the intermediate value theorem.

We'll first need the definition of a continuous function. Namely, $f: [a,b] \to \mathbb{R}$ is continuous at $x \in \mathbb{R}$ iff $^*f(y)$ is infinitesimally close to $^*f(x)$ for all $y$ infinitesimally close to $x$.

Let $f: [a,b] \to \mathbb{R}$ be continuous. Let $x: \{0, 1, 2, \dots, k\} \to \mathbb{R}$ be defined by $x_i = \frac{b-a}{i} + a$ - that is, making a partition of the interval $[a, b]$.

Extend $f$ to $^* f$ from $^*[a,b]$ to $^*\mathbb{R}$, and extend $x$ likewise.

Consider $\{ z: \text{st}(^*f(x_z)) > 0 \}$, a set of hyperfinite integers. This has a minimal element $\bar{z}$, say.

Now $\text{st}(^*f(x_{\bar{z}})) \geq 0$ by construction.

On the other hand, if $\text{st}(^*f(x_{\bar{z}})) > 0$ then we would have $x_{\bar{z}-1}$ being infinitesimally less than $x_{\bar{z}}$, so $^*f(x_{\bar{z}-1})$ would have the same standard part as $\text{st}(^*f(x_{\bar{z}}))$ - that is, greater than zero - contradicting the minimality of $\bar{z}$.

Hence $\text{st}(^*f(x_{\bar{z}})) = 0$, so $f(\text{st}(x_{\bar{z}})) = 0$ (again by continuity).

Answer 2

You can color an infinite planar. map with 4 colors such that no adjacent things have the same color.

For each pair of countries, $c_1$ and $c_2$, you have a statement of the form $c_1$ is adjacent to $c_2$, or $c_1$ is not adjacent to $c_2$. Take each of these statements, along with the statement "the graph can be colored by four colors". Sense each finite set of statements is satisfiable by a finite portion of the graph, there is a nonstandard graph that satisfies all of those statements by the saturation principle. Sense this hyperfinite graph will contain every node of the original infinite graph, the original graph will also be four colorable.