I was wondering if someone could help me formalize the below construction.
Suppose I know that some set $T$ is dense on the $\mathbb R$eal closed interval $[a,b]$. The definition that I would like to work with is:
$T \subseteq [a,b]$ is dense in $[a,b]$ if for any $c \lt d \in [a,b]$, there is a $t\in T$ such that $t \in (c,d)$.
Next, suppose I have the following set: $R=\{(c_n,d_n)\ |\ \exists n \in \mathbb N: c_n=b-\frac{b-a}{n} \text{ and } d_n=b-\frac{b-a}{n+1}\}$.
I want to use the following function: $G:R \to T$, where the mapping rule is $(c_n,d_n) \mapsto x_n$ such that $x_n \in T \text{ and } x_n \in (c_n,d_n)$.
However, I am unsure whether or not the function $G$ exists. The main problem I see here is that, for a given input $(c_n,d_n)$, how do I know that $G$ only maps to a single element (a requirement for a function)? By assumption, there is at least one such element $x_n \in T \text { and } x_n\in (c_n,d_n)$, but there may be many (infinitely) more. As far as I know, the implied existential statement, "...$\exists t \in T$..." has no upper bound on the number of objects satisfying the given property...it simply asserts that there is at least one.
Is the only way around this issue to invoke the Axiom of Choice, which will allow me to posit the existence of a choice function? In particular, this choice function $K_n$ would allow me to choose precisely one of the possibly many $t \in T$ that lay within $(c_n,d_n)$. If this is true, would it be appropriate to recast my function $G$ in the following way?
$G:R \to T$, where the mapping rule is $(c_n,d_n) \mapsto K_n\left((c_n,d_n) \right)$
Are there other ways to go about this construction that do not require the Axiom of Choice?
1st Edit
For completeness, with this updated version of $G$, we can show that the set $T$ is, at a minimum, countably infinite. To see this, we will show that there is a bijection from a subset of $T$ to $\mathbb N$.
Firstly, because $\forall n \in \mathbb N: (c_n,d_n) \cap (c_{n+1},d_{n+1}) = \emptyset$, we must have that $G$ is injective. Also, $\forall n \in \mathbb N: [c_n,d_n] \subset [a,b]$, which means that $R \subseteq [a,b]$ (so $G$ is well-defined).
Next, by construction, it is straightforward to see that $R$, the domain of $G$, can be put in a bijection with $\mathbb N$ in the obvious way, which, together with the injectivity claim, shows that $G[R]$ can be put in a bijection with $\mathbb N$ (once again, in the obvious way).
But $G[R] \subseteq T$, so we see that we have a subset of $T$ which can be put in a bijection with $\mathbb N$. Therefore, $T$ is, at a minimum, countably infinite.
More generally, we show that any set that is dense in a non-empty subset of $\mathbb R$ is at least countably infinite.
2nd Edit
In the comments, Asaf Karagila stated that to prove a set is simply 'infinite' does not require the Axiom of Choice. Here is how. In this context 'infinite' (rather than 'Dedekind Infinite') simply means 'not infinite' where we define 'finite' as:
A set $A$ is finite if there exists a bijection $f: A \rightarrow \{1,...,n\},$ for some positive integer $n$
Returning to our initial problem, to show that $T$ is infinite, we will proceed by contradiction. Suppose, instead, that $T$ is finite. Therefore, there exists some $N$ such that $T$ can be put in a bijection with the set $\{1,2,\cdots,N\}$.
Now, consider the following length $\ell = \frac{b-a}{N+1}$. Consider the following list of subintervals: \begin{align}s_1=\left(a,a+1\cdot \frac{b-a}{N+1}\right),s_2=\left(a+1\cdot \frac{b-a}{N+1},a+2\cdot \frac{b-a}{N+1}\right),\cdots, \\s_N=\left(a+(N-1)\cdot \frac{b-a}{N+1},a+N\cdot \frac{b-a}{N+1}\right), s_{N+1}=\left(a+N\cdot \frac{b-a}{N+1},b\right)\end{align}.
In each of these intervals, there is a $t_j \in s_j$. All of these subintervals are disjoint. Therefore, there are a total of $N+1$ elements in $T$, which is a contradiction. $T$ must be infinite...no Axiom of Choice required.
No. The axiom of choice is necessary here.
Indeed, it is consistent that there is a dense set of the reals which does not have any countably infinite subsets. This was shown to be the case in Cohen's first model of the failure of the axiom of choice.
Since $R$, as you defined it, is very clearly countable, this would be impossible, as the function you are trying to define is injective.