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Show that if $F = \emptyset$, then the statement $x\in\bigcap F$ will be true no matter what $x$ is

I know that $x\in\bigcap F = \forall A \in F, x\in A$

But how can $x$ be in any set, much less 'all' sets in $F$ when $F$ it has no sets in it?

I know that $x\in\bigcup F$ is false because it means that $\exists A \in F$ so that $x \in A$ and that can't be true because no set 'exists' in $F$.

So how can $x\in\bigcap F$ be always true if there is no set in $F$ in which $x$ can be?

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  • $\begingroup$ @graydad F is a family of sets, $x\in\bigcap F$ means that $x$ is in every set in $F$, right? So for each set 'A' in $F$, x is in A. $\endgroup$ – Juanma Eloy Jan 29 '15 at 4:31
  • $\begingroup$ @JMoravitz I disagree: $x \in F$ is false, as you say, but $x \in \cap F$ is true, because $\cap F$ is the intersection of all the sets in $F$. Adding sets to $F$ makes $\cap F$ smaller, not bigger. $\endgroup$ – Gregory J. Puleo Jan 29 '15 at 4:32
  • $\begingroup$ @Gregory J. Puleo But F is empty. There are no sets to intersect. So how can x belong in the intersection of all sets in F if there isn't anything in F where x can belong? $\endgroup$ – Juanma Eloy Jan 29 '15 at 4:34
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    $\begingroup$ So, as I'm understanding then, $x\in \bigcap \mathcal{F} \Leftrightarrow (A\in\mathcal F \Rightarrow x\in A)$. In that definition then, the right side is vacuously true since $\forall A, A\notin \mathcal{F}$. $\endgroup$ – JMoravitz Jan 29 '15 at 4:37
  • $\begingroup$ @ziggystarman There are no sets in the intersection of which $x$ does not belong. Without the existence of a counter example, the universal statement is true, though vacuously. $\endgroup$ – Graham Kemp Jan 29 '15 at 4:48
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The statement $$x\in\bigcap F$$ means $$\hbox{for all $S\in F$ we have $x\in S$}.\tag{$*$}$$ Now, what is the truth value of a statement $$\hbox{for all $S\in F$,}\ldots\langle\hbox{whatever}\rangle$$ when $F$ is the empty set? The accepted answer is that such a statement is always true, regardless of the "whatever".

To some extent this is a convention, and serves as part of our (mathematical) understanding of the word "all". It has not always been so: if you look at mathematical and philosophical sources from the 19th century, you can find a good deal of discussion on this. But this is the way it is taken nowadays, and probably the main reason is the following.

Assume for the time being that $F$ is not empty. Then hopefully it is clear that the two statements $$\hbox{for all $S\in F$, statement $X$ is true}\tag1$$ and $$\hbox{there is no $S\in F$ for which statement $X$ is false}\tag2$$ are logically equivalent: that is, they are really just two ways of saying the same thing. Now $(2)$ is clearly true when $F$ is the empty set (there is no $S$ in $F$ at all, so there is certainly no $S$ in $F$ for which statement $X$ is false). So if we want the equivalence of $(1)$ and $(2)$ to also hold for the empty set - that is, we want to treat the empty set as "nothing special" - then we have to agree that $(1)$ is also true when $F$ is empty.

In terms of your question, the conclusion of all this is that if $F$ is empty, then $(*)$ is true, regardless of the value of $x$.

Hope this helps!

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What you have encountered is the vacuous truthiness of an empty universe.

A universal statement is falsified if there exists a counterexample within the domain of discourse.   However, there are no counterexamples within an empty set, so whatever is asserted about all its members is true.

$$\forall y\in \emptyset ( P(y) ) \iff \neg \exists x\in \emptyset (P(y))$$

Or in your case $\forall A \in \emptyset (x\in A) \iff \neg\exists A\in \emptyset (x\notin A)$

There are no sets in the empty set of which $x$ is not a member, so $x$ is a member of all sets in the emptyset.

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