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Let $\langle X,\rho \rangle$ be a metric space and $f:\emptyset\to X$ a function. Since $\emptyset$ is compact, I know that $f$ is uniformly continuous. But can it be proven by vacuous truth? It's the same argument as for continuity, or something change because the alteration of the quantification?

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The alteration of quantification doesn't affect the vacuous truth here. To see this, write down the definition of uniform continuity:

$$ \forall \epsilon, \exists \delta, \forall x, y \in \emptyset : |x - y| < \delta \implies |f(x) - f(y)| < \epsilon $$

For a fixed $\epsilon > 0$, any $\delta > 0$ would work. The implication $|x - y| < \delta \implies |f(x) - f(y)| < \epsilon$ is vacuously true because there are no $x, y$ to satisfy $|x - y| < \delta$ in $\emptyset$ no matter what $\delta$ is chosen.

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Don't you really want: $\forall\varepsilon>0:\exists\delta>0:\forall x,y\in\emptyset: \dots$ – Thomas Andrews Feb 11 '13 at 23:42
@ThomasAndrews Added. Better than leaving it implicit. Thanks. – Ayman Hourieh Feb 11 '13 at 23:45

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