Why can we define sequences in this fashion?

[Oats]

I am familiar with the following formulation of the principle of recursive definition.

Let $E$ be a set, $e \in E$ and $h: E \longrightarrow E$. Then there is a unique function $f: \mathbb{N} \longrightarrow E$ for which $f(1) = e$ and for all $n \in \mathbb{N}$, $f(n + 1) = h(f(n))$.

Now, in certain proofs in analysis, there are times where a recursive definition for a function is used. Here are two examples.

If $p$ is any limit point of a set $E$, then every neighborhood of $p$ contains infinitely many points of $E$.

$\textbf{Proof:}$ Let $p$ be a limit point of $E$, let $\epsilon > 0$. We construct a sequence in $E$ as follows. Since $p$ is a limit point of $E$, there exists a point of $E$, call it $x_1$, such that $x_1 \in B_{\epsilon}(p)$. Now, assuming $x_1, \dotsc, x_n$ have been defined, we consider the open ball $B_{\frac{1}{2}d(p,x_n)}(p)$. Then there is a point $x_{n + 1} \in E$ such that $x_{n+1} \in B_{\frac{1}{2}d(p,x_n)}(p) \subseteq B_{\epsilon}(p)$. Note that for $1 \leq i \leq n$ we have that $x_{n + 1} \neq x_{i}$ since $d(x_{n+1}, p) \neq d(x_i, p)$. Then the range of this sequence, $\{x_n \mid n\in \mathbb{N}\}$ is then an infinite subset of $B_{\epsilon}(p) \cap E$ $\textbf{QED}$.

If $\{a_n\}$ is a sequence and $p \in X$ is a point such that for every open set $U$ containing $p$, the set $\{n \in \mathbb{N} \mid a_n \in U\}$ is infinite, then the sequence $\{a_n\}$ has a subsequence that converges to $p$.

$\textbf{Proof}$ $B_1(p)$ is an open set that contains $a_n$ for infinitely many $n$. Choose $n_1$ so that $a_{n_1} \in B_1(p)$. Then $B_{\frac{1}{2}}(p)$ contains $a_n$ for infinitely many $n$, so choose $n_2$ so that $a_{n_2} \in B_{\frac{1}{2}}(p)$ and $n_2 > n_1$. Having defined $n_1, \dotsc, n_k$ in this fashion, choose $n_{k+1}$ such that $a_{n_{k+1}} \in B_{\frac{1}{k+1}}(p)$ and $n_{k+1} > n_{k}$. Clearly this will converge to $p$. $\textbf{QED}$.

My issue here is understanding the explicit connection between the statement of definition by recursion, and its actual application in the two aforementioned proofs? What would the set be? The function on the set? As an additional question, I was wondering how exactly we could even choose the points specified above. How is it one could actually choose which of the points to associate to each $n$ to form a sequence, especially when, for each $n$ there could be infinitely many points. Would this require the axiom of choice just to define functions/sequences in this manner? Thanks in advance for any response.

 

[fresh_42]

$f : \mathbb{N} \longrightarrow E$ is given by $n \mapsto x_n$ and $h : E \longrightarrow E$ by $h(x_n)= x_{n+1}$. The function $f$ in the definition of the recursion is simply the numbering, which is usually noted by an index without explicitly relate to $f$ and the function $h$ is the successor function, i.e. defined by the process in the proof, which finds the next element of the sequence. Since there is nowhere any uniqueness necessary, it doesn't matter, which of all possible functions $h$ are described, i.e. the choice can be made as soon as the existence of an element $x_{n+1}$ is guaranteed - any choice. The existence is given by the definition of a limit point $p$.

The axiom of choice is not used here, because the definition limit point already gives such elements $x_n$ and any will do. AC would mean that we had arbitrary many sets and we had to select an element out of each of these sets, but here we have only one set. We only need it to be non empty.

 

[FactChecker]

Something seems strangely missing in the two example proofs. The first example talks about "every neighborhood of p", but the proof never proposes an arbitrary neighborhood of p. The second example talks about "every open set U containing p", but the proof never proposes an arbitrary open set U. Are you sure that you copied the proofs correctly?

That being said, recursive proofs are often done without specifying functions h and f as mentioned in your statement of the recursive principle. Unless that principle is stated as a Lemma or theorem which is invoked in a proof, I wouldn't expect the proof to exactly match the principle as stated.