Intersection of Empty Set/Paradoxical Implications

Theorem
Consider the Intersection of Empty Set:
 * $\displaystyle \mathbb S = \O \implies \bigcap \mathbb S = \mathbb U$

Although it appears counter-intuitive, the reasoning is sound.

This result is therefore classed as a veridical paradox.

declares, in Section $5$ of his of $1960$ that:


 * There is no profound problem here; it is merely a nuisance to be forced always to be making qualifications and exceptions just because some set somewhere along some construction might turn out to be empty. There is nothing to be done about this; it is just a fact of life.

However, later in that same work (in Section $9$, in the context of indexed families of sets) he says that:


 * ... an empty intersection does not make sense.

a sentiment which is repeated in the $2008$ collaboration with, , p. $457$.

In, the author recognizes the result, but does not adopt it:


 * If one has a given large set $X$ that is specified at the outset of the discussion to be one's "universe of discourse," and one considers only subsets of $X$ throughout, it is reasonable to let $\displaystyle \bigcap_{A \mathop \in \AA} A = X$ when $\AA$ is empty. Not all mathematicians follow this convention, however. To avoid difficulty, we shall not define the intersection when $\AA$ is empty.

accepts this result, but cautiously:


 * A natural mnemonic for these extreme cases is that $\bigcap \SS$ "grows larger" as $\SS$ "grows smaller", and $\bigcup \SS$ grows smaller as $\SS$ grows smaller. No other convention is possible, but the case $\SS = \O$ will often be treated redundantly by itself in definitions and proofs, as a reminder of the null case.

dismiss it casually:
 * As usual, we adopt the convention that in case $A = \O$ the expression $\bigcap A$ is defined only in case we work with the subsets of an underlying set $X$. In this case we put $\bigcap A = X$.

and do not even acknowledge that there may be a problem in the first place:
 * ''In the case where $I = \O$, we have
 * $\displaystyle \bigcap_{i \mathop \in I} A_i = X$;''
 * ''this moreover is the only case where $X$ plays a role; in fact, if $I$ is not empty, clearly we have:
 * $\displaystyle \bigcap_{i \mathop \in I} A_i = \set {x \mid x \in A_i \text { for every } i \in I}$.