Download PROBLEM SET 9 (1) There is always a bi

PROBLEM SET 9
(1) There is always a bi-infinite path, and in fact we can write down an explicit vertex in one.
Label the path of length n with numbers {0, . . . , n} such that i ∼ j ⇔ |i − j| = 1, and consider
the vertex in the ultraproduct corresponding to the sequence bn/2c of vertices in the factors. For
any fixed k, there are only finitely coordinates in which this vertex is within distance k of a vertex of
degree zero. Thus, by the fundamental theorem of ultraproducts, it does not have distance at most
k from such a vertex in the ultraproduct, so it lies on a bi-infinite path.
(2) We always have S ⊆ S ∗ if we identify R with a subset of R∗ . Thus, if S ∗ is the set of infinitesimals,
then the only possibilities are S = ∅ and S = {0}. However, ∅∗ = ∅ and {0}∗ = {0}. (For the latter,
note that that [(xi )i∈I ] ∈ {0}∗ iff {i ∈ I : xi = 0} ∈ F , and that condition means [(xi )i∈I ] = 0.)
Because there exist nonzero infinitesimals, the set of infinitesimals cannot be of the form S ∗ .
(3) First, suppose f is continuous according to the ε-δ definition. In other words, for all ε > 0, there
exists δ > 0 such that for all y ∈ R, |x − y| < δ implies |f (x) − f (y)| < ε. Given any fixed ε and
the corresponding δ, this is a first-order statement about R, and hence the corresponding statement
holds in R∗ . In particular, if x ≈ y, then |x − y| is smaller than every positive real number and hence
so is |f (x) − f (y)|, so f (x) ≈ f (y).
Conversely, suppose f is not continuous according to the ε-δ definition. Then there is some
sequence x1 , x2 , . . . ∈ R and some ε ∈ R with ε > 0 such that the xi converge to x but |f (x) − f (xi )| >
ε for each i. Then consider the element y := [(x1 , x2 , . . . )] ∈ R∗ . It satisfies y ≈ x but f (y) 6≈ f (x).
(4) Suppose y ≈ x. We have y 2 = (y − x + x)2 = (y − x)2 + 2(y − x)x + x2 , and (y − x)2 ≈ 0, so
y 2 ≈ 2(y − x)x + x2 . Thus, y 2 ≈ x2 iff 2(y − x)x ≈ 0. If x is not infinite, then |x| is bounded by some
real number, and thus 2(y − x)x is infinitesimal because y − x is, so f is continuous at all non-infinite
x.
However, it is discontinuous at x when x is infinite. Specifically, we can take y = x + 1/x, since
1/x is infinitesimal, and then y 2 ≈ 2 + x2 , rather than x2 .
(5) We’ll use these branching quantifiers to create a formula saying there’s an injective function that
is not surjective. This is equivalent to having an infinite domain and is therefore not a first-order
property. Let c be a constant symbol, and consider the formula
∀w ∃x
(((w = y) ↔ (x = z)) ∧ ¬(x = c)).
∀y ∃z
In terms of functions f and g instantiating the quantifiers via f (w) = x and g(y) = z, this formula
says w = y iff f (w) = g(y), so f and g must be the same function and it must be injective. However,
it also says c is not in the function’s image.
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