Runnin’ on Empty

Suppose A is any set (it can be a set of familiar objects, like real numbers, it can be the empty set, or it can even be a set of something silly and/or pointless).

Is it true that A \cup \emptyset = A?  If so, why? If not, why not?  What does a proof for your answer look like?

Is it true that A \cap \emptyset = A?  If so, why?  If not, why not?  What does a proof for your answer look like?

Lastly, a question I raised this past week: what is the set A \times \emptyset?  Using our definition, we know that A \times \emptyset = \left\{(x,y) : x \in A \text{ and } y \in \emptyset\right\}.  Of course, this means that the element y doesn’t exist, and so we might instead write A \times \emptyset = \left\{(x, \,\,): x \in A\right\}.

This seems perfectly reasonable, and I, too, was tempted to draw this very same conclusion.  However, it is wrong.   Technically speaking, it violates the definition of Cartesian product.  By specifying the second “entry” in the ordered pair (x, \,\,) as “nothing” we are no longer talking about an ordered pair!   An ordered pair is, by definition, a pair of things that exist.  Hence, there are no ordered pairs in the set A \times \emptyset, which means that A \times \emptyset = \dots  Ready for it?  Have you figured it out yet??  That’s right!  A \times \emptyset = \emptyset!

Also, as many of you guessed, when you have two finite sets A and B (that is, two sets that each have finite cardinality), then the cardinality of their cross-product should be \left|A \times B \right| = \left|A\right|\cdot\left|B\right|.  Assuming this is true, and letting B = \emptyset what do we learn about the cardinality of A \times \emptyset?  In turn, what does this suggest must be true about the set A \times \emptyset?

So now I leave you with something to prove, and I’d like you to prove it a specific way.  Given that proofs by contradiction are discussed in sections 2.3 and 2.4 of our textbook, this arrives at an ideal time.

Write a proof by contradiction that if A is any set then A \times \emptyset = \emptyset.

This is not an officially assigned proof problem, but handing in a correct solution with your actual proof problems just might add some points to your grade.  How many points?  At the very least I promise to add \left|A \times \emptyset\right| bonus points, if not more.


2 thoughts on “Runnin’ on Empty

  1. Your conclusion or instinct is correct, but the order of your thoughts is (perhaps) a bit off. Assuming the claim A \cup \emptyset = A = \emptyset \cup A is, in fact, true for all sets A, THEN it would be natural to conclude that “the empty set acts like the additive identity!”

    However, starting off with the assertion “the empty set should act like an additive identity” is unjustified; you would be claiming that something very specific is true before having checked it.

    Said more succinctly: its not that the empty set “acts like zero” causes A \cup \emptyset = A, rather, its that A \cup \emptysety = A causes the empty set to “act like zero.”

    You should consult and ponder Exploration 1 on page 50; it addresses this idea (and others).

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s