Lemma 8.4.5: Let $E$ be a non-empty subset of the real line with $\sup (E)<\infty$ (i.e., $E$ is bounded from above). Then there exists a sequence $\left(a_n\right)_{n=1}^{\infty}$ whose elements $a_n$ all lie in $E$, such that $\lim _{n \rightarrow \infty} a_n=\sup (E)$.
Proof: For each positive natural number $n$, let $X_n$ denote the set \begin{align*} X_n:=\{x \in E: \sup (E)-1 / n \leq x \leq \sup (E)\} \end{align*} Since $\sup (E)$ is the least upper bound for $E$, then $\sup (E)-1 / n$ cannot be an upper bound for $E$, and so $X_n$ is non-empty for each $n$. Using the axiom of choice (or the axiom of countable choice), we can then find a sequence $\left(a_n\right)_{n=1}^{\infty}$ such that $a_n \in$ $X_n$ for all $n \geq 1$. In particular $a_n \in E$ for all $n$, and $\sup (E)-1 / n \leq a_n \leq \sup (E)$ for all $n$. But then we have $\lim _{n \rightarrow \infty} a_n=\sup (E)$ by the squeeze test.
My thought: I understand that the Axiom of Choice is invoked because we intend to pick countably many elements $a_1,a_2,\ldots$ from countably many sets $X_1,X_2,\ldots$ such that $X_n=\{x \in E: \sup (E)-1 / n \leq x \leq \sup (E)\}$ for each $n\in \mathbb{N}$ and there is no explicit rule how to pick these elements. However, I could also consider an alternative proof without invoking the notation of countably many sets:
We know that $\sup (E)$ is the least upper bound of $E$, and hence for every $\epsilon>0$, there exists $x \in E$ such that \begin{align*} \sup (E)-\epsilon \leq x \leq \sup (E) \end{align*} We can construct a sequence $\left(a_n\right)_{n=1}^{\infty}$ as follows:
- Step 1: Let $\epsilon=1$. By the definition of $\sup (E)$, there exists $x_1 \in E$ such that $\sup (E)-1 \leq x_1 \leq \sup (E)$. Set $a_1=x_1$.
- Step 2: Let $\epsilon=\frac{1}{2}$. There exists $x_2 \in E$ such that $\sup (E)-\frac{1}{2} \leq x_2 \leq \sup (E)$. Set $a_2=x_2$.
- Continuing in this manner, for each $n \geq 1$, let $\epsilon=\frac{1}{n}$. There exists $x_n \in E$ such that $\sup (E)-\frac{1}{n} \leq x_n \leq \sup (E)$. Set $a_n=x_n$.
By this construction, we have $\sup (E)-\frac{1}{n} \leq a_n \leq \sup (E)$ for all $n \geq 1$. Therefore, $\lim _{n \rightarrow \infty} a_n=\sup (E)$ by the squeeze theorem.
My question: Does the alternative proof still rely on the Axiom of Choice? That is, as we explicitly state the existence of each element $x_n$ and define $a_n = x_n$ for $n\in \mathbb{N}$, it seems to me that no Axiom of Choice is needed. If I am wrong and the Axiom of Choice is still invoked, does it mean that the Axiom of Choice is always needed when constructing a sequence with infinitely many terms?
A sequence needs to be defined by a finite formula. So to define the sequence $(a_n)$, you need to write down a statement which specifies the values of the $a_n$'s simultaneously, all at once.
For example, you have given an explicit formula that simultaneously defines each of the nonempty sets $X_n$, one for each $n$. That works just fine. Therefore, you have a sequence of sets $X_n$. The definition of that sequence did not require the axiom of choice.
Your "alternate definition" for the sequence $(a_n)$ uses infinitely many formulas: for each natural number $n$, you have a formula which applies existential instantiation to choose an element from the nonempty set $X_n$. So you have infinitely many statements, each with a separate application of existential instantiation. But that doesn't count: you are not allowed to define a sequence by infinitely many applications of existential instantiation, nor by any conjunction of infinitely many separate defining statements.
To define the sequence $(a_n)$, you need choose one $a_n$ from each of the nonempty sets $X_n$ simultaneously. That's exactly what the axiom of countable choice lets you do. Think of it as simultaneous existential instantiation.