This post is intended as a footnote to one that I wrote a couple of years ago about the meaning of “implies” in mathematics, which was part of a series of posts designed as an introduction to certain aspects of university mathematics.
If you are reasonably comfortable with the kind of basic logic needed in an undergraduate course, then you may enjoy trying to find the flaw in the following argument, which must have a flaw, since I’m going to prove a general statement and then give a counterexample to it. If you find the exercise extremely easy, then you may prefer to hold back so that others who find it harder will have a chance to think about it. Or perhaps I should just say that if you don’t find it easy, then I think it would be a good exercise to think about it for a while before looking at other people’s suggested solutions.
First up is the general statement. In fact, it’s a very general statement. Suppose you are trying to prove a statement and you have a hypothesis to work with. In other words, you are trying to prove the statement
Now if and are two statements, then is true if and only if either is false or is true. Hence what we are trying to prove can be rewritten as follows.
Now we can bring the inside the as long as we convert the into , so let’s do that. What we want to prove becomes this.
I’ll assume here that we haven’t done something foolish and given the name to one of the variables involved in the statement . So now I’m going to use the general rule that is equivalent to to rewrite what we want to prove as the following.
Finally, let’s rewrite what’s inside the brackets using the sign.
Every single step I took there was a logical equivalence, so the conclusion is that if you want to show that implies , your task is the same as that of finding a single such that .
Now let me give a counterexample to that useful logical principle. Let be a set of real numbers. Define the diameter of to be . I’ll write it .
Consider the following implication.
That is clearly correct: if every element of has modulus at most 1, then is contained in the interval , so clearly can’t have diameter greater than 2.
But then, by the logical principle just derived, there must be a single element of such that if that element has modulus at most 1, then the diameter of is at most 2. In other words,
But that is clearly nonsense. If all we know is that one particular element of has modulus at most 1, it can’t possibly imply that has diameter at most 2.
What has gone wrong here? If you can give a satisfactory answer, then you will have a good grasp of what mathematicians mean by “implies”.