I was reading Pugh's Real Analysis:

I've found this in the beginning of the book:

A class is a collection of sets. The sets are members of the class. For example we could consider the class $\mathcal{E}$ of sets of even natural numbers. Is the set $\{2,15\}$ a member of $\mathcal{E}$? No. How about the singleton $\{6\}$? Yes. How about the empty set? $\color{red}{\text{Yes, each element of the empty set is even.}}$

How's it possible that each element of the empty set is even when the empty set doesn't have any elements?

Edit: I've read the comments and the answer and I was thinking something quite quite different: If I have a set that has no elements, then the absence of elements would make the task of assigning a property to one of it's elements impossible. Is that feasible?

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    $\begingroup$ If you can't find an odd element in the empty set (and, you can't), it follows that every element of the empty set is even. Of course, it's also true that every element of the empty set is odd. $\endgroup$ – Gerry Myerson Jun 21 '13 at 9:39
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    $\begingroup$ It's even worse: All elements of the empty set have blue eyes. $\endgroup$ – Christian Blatter Jun 21 '13 at 10:51
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    $\begingroup$ @ChristianBlatter Now Limp Bizkit's song makes sense: $$\text{No one knows what it's like}\\ \text{To be the bad man}\\ \text{To be the sad man}\\ \text{Behind}\color{blue}{\text{ blue eyes}}$$ $\endgroup$ – Red Banana Jun 21 '13 at 10:57
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    $\begingroup$ @ChristianBlatter Not to mention that the elements of the empty set are very odd. $\endgroup$ – egreg Jun 21 '13 at 11:44
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    $\begingroup$ x is a member of the set with no members. $\endgroup$ – user 726941 Jun 21 '13 at 12:07

Each element of the empty set is even can be paraphrased as if $x$ is an element of the empty set, then $x$ is even:

$$\forall x\Big(x\in\varnothing\to x\text{ is even}\Big)\;.\tag{1}$$

How could you show that this was false? You’d have to show that there was some $x\in\varnothing$ that was not even. And you can’t do this: you can’t find any $x$ in the empty set, let alone one that is even. Since you can’t show that $(1)$ is false, it must be true.

To restate the argument in slightly different terms, the statement

if $x$ is an element of the empty set, then $x$ is even

imposes a condition on elements of the empty set, but the empty set has no elements, so it doesn’t actually impose a condition on anything. Thus, nothing can violate it: no object is an element of the empty set, so no object is even a candidate to violate the requirement of being even.

The usual terminology is that the statement $(1)$ is vacuously true: it’s true because it doesn’t actually impose a requirement on anything. Note that you could replace $x\text{ is even}$ in $(1)$ with pretty much any statement about $x$, and the resulting sentence would be vacuously true by essentially the same argument.

  • $\begingroup$ Hello, Brian. Thanks for your answer. Did you see the edit? Why does that heuristic does not work? $\endgroup$ – Red Banana Jun 21 '13 at 9:48
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    $\begingroup$ @BandeiraGustavo: Basically because it doesn’t fit with the way we want our logic to work. We want the negation of $\forall x(\varphi(x))$, i.e., $\neg\forall x(\varphi(x))$, to be $\exists x(\neg\varphi(x))$. Thus, the negation of every element of the empty set is even ought to be there is an element of the empty set that is not even. The latter statement is certainly false, so we want the former statement to be true. $\endgroup$ – Brian M. Scott Jun 21 '13 at 9:56

Because there are no elements to witness otherwise. This is called vacuous truth in mathematics.

Statements of the form "For every $x$ ..." are false if and only if there is a counterexample. The statement "For every $x$, if $x\in\varnothing$ then $x$ is even" has no counterexamples.

  • $\begingroup$ I think, you can extend your answer with saying that for any formula $F$, the sentence "for every $x\in \emptyset$ $F(x)$ holds" is true $\endgroup$ – Ilya Jun 21 '13 at 9:45
  • $\begingroup$ @egreg: doesn't the same argument apply? There are not $x\in \emptyset$ that violate existence of $y$. $\endgroup$ – Ilya Jun 21 '13 at 11:05
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    $\begingroup$ @Ilya Beware of the Jabberwocky, my son! Er, beware of the empty set. $\endgroup$ – egreg Jun 21 '13 at 11:41
  • $\begingroup$ @Ilya: You are correct. $\endgroup$ – Cameron Buie Jun 22 '13 at 0:30
  • $\begingroup$ @egreg: I'm not sure why you think that that statement is false. Can you explain? $\endgroup$ – Cameron Buie Jun 22 '13 at 0:34

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