Periodic points of hyperbolic toral automorphism are dense on the torus

Question

I am having a difficult time following the proof of Proposition 4.2 (see image below) from Devaney's An Introduction to Chaotic Dynamical Systems (2e) on p. 192.

Now, from topology I know that a subset $A \subseteq X$ of a topological space $(X, \tau)$ is dense in $X$ iff. the closure $\bar{A}$ of $A$ is equal the space $X$, i.e. $\bar{A} = X$. And the closure is the set of all contact\adherent points and the closure can also be expressed as the union of all points in $A$ and the set of all limit points of $A$ (the derived set).

But even knowing this I cannot follow the proof below at all.

What I particularly do no understand:

(i) We have to prove $\overline{Per(L_{A})} = T$. In other words, $ \overline{Per(L_{A})} \subseteq T$ and $ T \subseteq \overline{Per(L_{a})}$, right? So it seems that the proof does $ T \subseteq \overline{Per(L_{a})}$ but not $ \overline{Per(L_{a})} \subseteq T$. Why?

(ii) So in showing $ T \subseteq \overline{Per(L_{a})}$, the proof assumes $p \in T$ and then it shows $p \in \overline{Per(L_{a})}$ i.e. that $p$ is an adherent point of $Per(L_{a})$. How is this then equivalent to showing that $p$ is a periodic point of $L_{a}$? Also why do we choose $p \in T$ to have rational coordinates? And why is the phrase in the proof "Such points are clearly dense in T, for we may take $k$ arbitrarily large"? Is it because $\mathbb{Q}$ is dense in $\mathbb{R}$ and then $\mathbb{Q} \times \mathbb{Q}$ is dense in $\mathbb{R} \times \mathbb{R}$?

(iii) And then I am completely lost on the last paragraph of the proof where it is shown that $p$ is actually periodic with period less than or equal to $k^{2}$.

For clarification of the notation used in the proposition. $L_{A}$ is the hyperbolic toral automorphism defined by:

Let $L(x) = A \cdot x$ where $A$ is a $2 \times 2$ matrix satisfying (i) All entries are integers; (ii) $\det(A) = \pm 1$; $A$ is hyperbolic, meaning that none of its eigenvalues have absolute value one. The map induced on $T$ by $A$ is called a hyperbolic toral automorphism and is denoted by $L_{A}$.

The $2$-torus $T$ is defined setting $T$ as the set of all equivalence classes of all points in the plane whose coordinates differ by integers. Formally, let $T$ be the set of all equivalence classes under the equivalence relation $\sim \subseteq \mathbb{R}^{2} \times \mathbb{R}^{2}$ defined by $(x, y) \sim (x', y')$ if and only if $x - x'$ and $y - y'$ are integers.

Answer 1

Regarding point (i), keep in mind that the closure is not an absolute concept, it is relative to a particular topological space. To say this another way, one does not define "the closure of a set $A$", one defines instead "the closure of a subset $A$ of a topological space $X$". By definition, this is a subset of $X$. Namely, given a topological space $X$ and a subset $A \subset X$, the closure of $A$ in $X$ is a subset $\overline A \subset X$, defined in one of many equivalent ways (the union of $A$ with its limit points in $X$; the intersection of all closed subsets of $X$ that contain $A$; ...); different textbooks will pick different ones of these equivalent ways as "the" definition. So, the closure of $A$ in $X$ is a subset of $X$ by definition.

In the context of your question, on the left hand side of the equation $\overline{\text{Per}(L_A)}=T$, that bar on top means "closure in $T$". So, by definition of closure in $T$, $\overline{\text{Per}(L_A)}$ is a subset of $T$.

Regarding points (ii) and (iii), what's in these two paragraphs is a proof that every point in $T$ with rational coordinates is a periodic point (plus one quick sentence about density). To put this another way, letting $T(\mathbb Q)$ denote the set of all $p \in T$ with rational coordinates, what's proved in these two paragraphs is that $$T(\mathbb Q) \subset \text{Per}(L_A) $$ Let me assume that you can understand why that is true by reading those two paragraphs.

How now does one finish the proof of the proposition?

There are two further steps.

First, $T(\mathbb Q)$ is dense in $T$, i.e. $\overline{T(\mathbb Q)}=T$ (again, that means closure in $T$). As you say, this is a consequence of the fact that $\mathbb Q$ is dense in $\mathbb R$ and therefore $\mathbb Q^2$ is dense in $\mathbb R^2$. (This is encapsulated in one quick sentence: "Such points are clearly dense in $T$...")

Second, one applies an exercise in topology: Given a topological space $X$ and subsets $A \subset B \subset X$, if $A$ is dense in $X$ then $B$ is dense in $X$. Now apply this to $T(\mathbb Q) \subset \text{Per}(L_A) \subset T$.

Returning now to point (iii), the proof that each $p \in T(\mathbb Q)$ is periodic uses an exercise in set theory: for any finite set $B$ and any bijection $f : B \to B$ (i.e. for any permutation of $B$), every point of $B$ is a periodic point of $f$. In point (iii), fixing a positive integer $k$, the finite set $B$ is the set of all points in $T(\mathbb Q)$ whose two coordinates are rational numbers expressed with denominator equal to $k$ and with numerator in the set $\{0,...,k\}$. This set is finite, in fact its cardinality is exactly equal to $k^2$. What (iii) is showing is that $L_A$ restricts to a bijection of this set ("This means that $L_A$ permutes these points").