Limit with Rational Epsilon and Delta

Theorem
Let $\openint a b$ be an open real interval.

Let $c \in \openint a b$.

Let $f: \openint a b \setminus \set c \to \R$ be a real function.

Let $L \in \R$.

Suppose that:


 * $\forall \epsilon > 0 \in \Q_{>0}: \exists \delta \in \Q_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$

Then the limit of $f$ exists as $x$ tends to $c$, and is equal to $L$.

Proof
Recall the definition of a limit of a real function:


 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$

Denote by $\map P {\epsilon,\delta}$ the proposition that the above statement holds.

Suppose that $\map P {\epsilon,\delta}$ holds for all rational $\epsilon$ in the interval of interest:


 * $\forall \epsilon \in \Q_{>0}: \exists \delta \in \Q_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$

Let $\epsilon_0 > 0$ be a real number.

From Between two Real Numbers exists Rational Number, there is some rational $\epsilon_0'$ such that $0 < \epsilon_0' < \epsilon_0$.

Then $\size {\map f x - L} < \epsilon_0'$ implies $\size {\map f x - L} < \epsilon_0$.

Thus $\map P {\epsilon,\delta}$ holding for all rational $\epsilon$ implies $\map P {\epsilon,\delta}$ for all real $\epsilon$.

Now suppose $\map P {\epsilon,\delta}$ has been established for some rational $\epsilon$ and some real $\delta_0$.

From Between two Real Numbers exists Rational Number, there is some rational $\delta_0'$ such that $\size {x-c} < \delta_0' < \delta_0$.

Then $\size {x-c} < \delta_0$ is implied by $\size {x-c} < \delta_0'$

This means $\map P {\epsilon,\delta_0'}$ implies $\map P {\epsilon,\delta_0}$ holds.

We conclude:


 * $\map P {\epsilon,\delta}$ holds for all rational $\epsilon$ and $\delta$ in the interval of interest

is a stronger statement than:


 * $\map P {\epsilon,\delta}$ holds for all real $\epsilon$ and $\delta$ in the interval of interest.