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The "standard topology" (put in quotations because I haven't verified that it is in fact a topology yet) is defined by $$\mathcal U \in \mathcal O_\text{standard} \iff \forall p\in \mathcal U \exists r\in \Bbb R^+ : B_r(p)\subseteq \mathcal U$$

Now I'm checking whether this is a topology and I'm getting hung up on the very first property.

If $\emptyset\in \mathcal O_\text{standard}$, then for every $p\in \emptyset$, there exists $\dots$ But there is no $p\in \emptyset$ for us to check the rest of the condition. So how are we allowed to conclude that $\emptyset \in \mathcal O_\text{standard}$?

Michael Hardy
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user355875
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    It is true that every $p$ in the empty set has a certain property, for there are no $p$ in the empty set (jargon, the assertion is vacuously true). – André Nicolas Jul 24 '16 at 00:50
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    This is an example of a "vacuous quantification". A universally quantified statement over the empty set is usually taken as true by convention. – Carl Mummert Jul 24 '16 at 00:51
  • So we just define this as a true statement? – user355875 Jul 24 '16 at 00:52
  • Yes, basically. Similarly, if we said "there exists a $p$ in the empty set such that for all $r > 0$, $B_r(p) \subseteq U$", we would treat that as a false statement. This convention has the benefit that it preserves the property that the negation of a universally quantified statement is existentially quantified, and vice versa. – Carl Mummert Jul 24 '16 at 01:10
  • Wikipedia: Vacuous truth – Martin Sleziak Jul 24 '16 at 03:46

3 Answers3

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To put a different slant what others have said, the statement

"Every $x \in U$ has the property $p(x)$" is equivalent to

"There is no $x\in U$ which lacks the property $p(x)$."

In the case of the empty set $\varnothing$, we are therefore allowed to say that "Every $p\in \varnothing$ has the property that there exists ... etc." because you cannot find an $x\in \varnothing$ which doesn't have the property. (In fact, you cannot find any $x\in \varnothing$ at all!)

A second informal way to look at it : if every $x\in A$ has property $p(x)$, and every $x\in B$ has property $p(x)$, then intuitively you would want to say that every $x\in A \cap B$ has property $p(x)$. Extending this intuition to the case where $A$ and $B$ have no elements in common $(A\cap B = \varnothing)$ leads us to the idea of vacuous universal quantification.

user326210
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    Actually, if every $x\in A$ has property $p(x)$, or every $x\in B$ has property $p(x)$, then either way intuitively you would want to say that every $x\in A \cap B$ has property $p(x)$. – Colin McLarty Jul 24 '16 at 09:10
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This has in fact nothing to do with topology: it is just basic logic. A statement of the form "for all $x$ in $X$, then...", where $X$ is empty, is vacuously true.

Alex Provost
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You can unbind the quantifiers:

$$ \forall x\in S: P(x) \quad\equiv \quad\forall x: (x \in S) \implies P(x) $$ $$ \exists x\in S: P(x) \quad\equiv \quad\exists x: (x \in S) \wedge P(x) $$

so the formula you wrote means the same thing as

$$ \forall p : p \in \mathcal U \implies \left( \exists r : r \in \Bbb R^+ \wedge B_r(p)\subseteq \mathcal U \right) $$

so the case of $\mathcal{U} = \varnothing$ no longer has you quantifying over an empty domain. And since the antecedent is always false, the conditional is always (vacuously) true.

In fact, we always have

  • $\forall x \in \varnothing : P(x)$ is a tautology
  • $\exists x \in \varnothing : P(x)$ is a contradiction

Some approaches to doing logic actually reject empty domains, and use rules of deduction that only work for nonempty domains, such as:

$$ (\forall x \in \mathcal{D} : P(x)) \implies (\exists x \in \mathcal{D} : P(x)) $$

This should look familiar, as it is quite similar to the objection you had!

However, this implication does not hold without the hypothesis that $\mathcal{D}$ is nonempty. An example of a corrected deduction rule is:

$$ (\forall x \in \mathcal{D} : P(x)) \implies (\mathcal{D} = \varnothing) \vee (\exists x \in \mathcal{D} : P(x)) $$

So keep this in mind — the rules of logic you have learned, be it formally or informally, may actually be wrong for the case where empty domains of quantification are allowed, so until you are used to the more general case, always treat such possibilities with care.