KindiakMath

When Integers Converge

Here’s the key question: when sequences of integers converge, are their limits integers? The answer is yes, but we need to put in a bit of effort to make that happen.

Problem 1. Let K \subseteq \mathbb Z be nonempty and bounded above. Prove that K has a maximum element.

(Click for Solution)

Solution. Since K \subseteq \mathbb Z \subseteq \mathbb R is bounded above, there exists a real number r such that for any x \in K, x \leq r \leq |r|. Use the Archimedean property to find M \in \mathbb N such that x \leq |r| \leq M \Rightarrow -x \geq -M.

Define L := M + (-K) = \{M + (-x) : x \in K\} \subseteq \mathbb Z. Since K is nonempty, so is L. Furthermore, for any x \in K,

M + (-x) \geq M + (-M) = 0.

Thus, L \subseteq \mathbb N. By the well-ordering principle, L contains a minimum element M + (-m) so that for any x \in K,

M + (-x) \geq M + (-m) \quad \iff \quad x \leq m.

By construction, m \in K, so that m = \max K, as required.

Problem 2. Let x be a real number. Prove that there exists a unique integer, denoted \lfloor x \rfloor, such that \lfloor x \rfloor \leq x <\lfloor x \rfloor + 1.

The map \lfloor \cdot \rfloor : \mathbb R \to \mathbb Z is called the floor function.

(Click for Solution)

Solution. Fix x \in \mathbb R. For existence, consider the set

K := \{n \in \mathbb Z : n \leq x\}.

Let’s first check that K is nonempty. If x \geq 0, then 0 \in K and we are done. If x < 0, then -x > 0. Use the Archimedean property to find n_0 \in \mathbb Z such that -x \leq n_0 \Rightarrow -n_0 \leq x. Thus, -n_0 \in K.

Since K is bounded above and nonempty, by Problem 1, K has a unique maximum element \lfloor x \rfloor. This implies that \lfloor x \rfloor + 1 \notin K. Thus, \lfloor x \rfloor \leq x < \lfloor x \rfloor + 1, as required.

Problem 3. Let x_n be a \mathbb Z-sequence that converges to x. Prove that x \in \mathbb Z.

(Click for Solution)

Solution. Suppose otherwise that x \notin \mathbb Z. Then x - \lfloor x \rfloor \in (0, 1). Define \epsilon := \min\{x - \lfloor x \rfloor, 1 - (x - \lfloor x \rfloor ) \} > 0. Observe that

\epsilon \leq -x + \lfloor x \rfloor + 1\quad \Rightarrow \quad x + \epsilon \leq \lfloor x \rfloor + 1

and

-\epsilon \geq -(x - \lfloor x \rfloor ) \quad \Rightarrow \quad x -\epsilon \geq \lfloor x \rfloor.

Since x_n \to x, find N \in \mathbb N such that for n > N,

|x_n - x| < \epsilon \quad \Rightarrow \quad \lfloor x \rfloor \leq x - \epsilon < x_n < x + \epsilon \leq \lfloor x \rfloor + 1.

Subtracting \lfloor x \rfloor on both sides, 0 < x_n - \lfloor x \rfloor < 1. Since x_n and \lfloor x \rfloor are integers, x_n - \lfloor x \rfloor \in \mathbb Z, a contradiction.

—Joel Kindiak, 5 Jan 25, 1825H

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