Page 271 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Images and Inverse Images: A Research Project 257
equation between two sets, we proceed by taking an arbitrary element of each
set and trying to prove that it is an element of the other.
Suppose first that y is an arbitrary element of f (W ∩ X). By the definition of
f (W ∩ X), this means that y = f (x) for some x ∈ W ∩ X. Since x ∈ W ∩ X,
it follows that x ∈ W and x ∈ X. But now we have y = f (x) and x ∈ W,sowe
can conclude that y ∈ f (W). Similarly, since y = f (x) and x ∈ X, it follows
that y ∈ f (X). Thus, y ∈ f (W) ∩ f (X). This completes the first half of the
proof.
Now suppose that y ∈ f (W) ∩ f (X). Then y ∈ f (W), so there is some
w ∈ W such that f (w) = y, and also y ∈ f (X), so there is some x ∈ X such
that y = f (x). If only we knew that w and x were equal, we could conclude
that w = x ∈ W ∩ X,so y = f (x) ∈ f (W ∩ X). But the best we can do is to
say that f (w) = y = f (x). This should remind you of the definition of one-
to-one. If we knew that f was one-to-one, we could conclude from the fact
that f (w) = f (x) that w = x, and the proof would be done. But without this
information we seem to be stuck.
Let’s summarize what we’ve discovered. First of all, the first half of the
proof worked fine, so we can certainly say that in general f (W ∩ X) ⊆
f (W) ∩ f (X). The second half worked if we knew that f was one-to-one,
so we can also say that if f is one-to-one, then f (W ∩ X) = f (W) ∩ f (X).
But what if f isn’t one-to-one? There might be some way of fixing up the
proof to show that the equation f (W ∩ X) = f (W) ∩ f (X) is still true even if
f isn’t one-to-one. But by now you have probably come to suspect that perhaps
f (W ∩ X) and f (W) ∩ f (X) are not always equal, so maybe we should devote
some time to trying to show that the proposed theorem is incorrect. In other
words, let’s see if we can find a counterexample – an example of a function f
and sets W and X for which f (W ∩ X) = f (W) ∩ f (X).
Fortunately, we can do better than just trying examples at random. Of course,
we know we’d better use a function that isn’t one-to-one, but by examining
our attempt at a proof, we can tell more than that. The attempted proof that
f (W ∩ X) = f (W) ∩ f (X) ran into trouble only when W and X contained
elements w and x such that w = x but f (w) = f (x), so we should choose an
example in which this happens. In other words, not only should we make sure
f isn’t one-to-one, we should also make sure W and X contain elements that
show that f isn’t one-to-one.
The graph in Figure 1 shows a simple function that isn’t one-to-one. Writ-
ing it as a set of ordered pairs, we could say f ={(1, 4), (2, 5), (3, 5)}, and
f : A → B, where A ={1, 2, 3} and B ={4, 5, 6}. The two elements of A that
showthatf isnotone-to-oneare2and3,sotheseshouldbeelementsofW andX,
respectively. Why not just try letting W ={2} and X ={3}? With these choices