Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.