Let $(q_n)_{n\in\mathbb{N}}, q_n\in\mathbb{Q}$ be a Cauchy-series, and $\displaystyle\lim_{n\to\infty} q_n = q\in \mathbb{R}$.
Given a formula $A[x]$ where $x$ is a variable, let $\forall n\in \mathbb{N}: A[q_n]$ be true.
Question: When does $A[q]$ hold?
I'm specially interested in this particular strategy: Let $\forall x\in\mathbb{R}:B[x]$ be to prove.
If it's true, $\forall x\in\mathbb{Q}:B[x]$ follows. Since $\mathbb{Q}$ is countable, we can inductively prove this statement. Now we can define for every real number a sequence $(q_n)_{n\in\mathbb{N}}$ as above, and if we show $\forall n\in \mathbb{N}: A[q_n] \implies A[q]$, we'd have our proof.
Is there some manageable criteria for when this strategy works?
Just as example:
One such criteria is that if $A[a]$ holds and if there's a neighborhood $U$ of $a$ for which $\forall x\in U: A[x]$ holds. I.e.:
If $\varphi$ is a formula and $U$ is a neighborhood, the following equivalence holds:
$\forall x\in \mathbb{Q}:(\varphi(x) \land \exists U: \forall y\in U: \varphi(y))
\quad\Leftrightarrow\quad
\forall x\in \mathbb{R}: \varphi(x)$
I also found another one:
Let $A[x_1,..,x_n]$ be a formula that's true for all $x_1,...x_n\in\mathbb{Q}$. If all functions and relations (excluding $=$ ) are continuous, then $A$ holds for all $x_1,..,x_n\in\mathbb{R}$. The only drawback is that the continuousity has to be shown in $\mathbb{R}$ or has to be an axiom of the function/relation.
The criterion is that $\{x:A[x]\text{ holds}\}$ must be a closed set. This can be stated directly as
Another option: let $\overline A[x]$ mean that $A[x]$ does not hold.