6.3 EXISTENCE AND UNIQUENESS (OPTIONAL) 215Y, = Y, n U = Y, n (X u Y,)
= (Y, n X) u (Y, n Y,)
= % u (Yl n Y2)
= Y, n Y,
so that Y, E Y,, by Example 9, Article 4.1. A completely analogous
argument, with the roles of Y, and Y, reversed, shows Y, E Y,, so that
we may conclude Y, = Y,, as desired.EXAMPLE 4 A real number u is called a least upper bound of a set S 5 R
if and only if (i) x I u for all x E S and (ii) To every P > 0, there cor-
responds y E S such that y > u - 8. Prove that a subset S of R has
at most one least upper bound.
Solution Suppose u, and u, are both least upper bounds for S. We claim
u, = u, and will proceed to use an argument by contrapositive to prove
it. If u, # u,, we may assume with no loss of generality that u, < u,.
We will apply the technique of specialization to part (ii) of the definition,
letting #I = Hu, - 24,). Then, by (ii), there exists y E S such that y >
u2 - Q = u2 - 3(u2 - u,) = (ul + u,) > (ul + ul) = ul. But y > U,
and y E S contradicts property (i) for u,. Thus u, = u,, as desired. With
uniqueness thus established, we often denote the least upper bound of a
set S, when it exists, by lub S. 0
An important property of limit of a function at a point a is that such
limits are unique when they exist, provided that f is defined in an open in-
terval containing a. Our general approach to proving this uniqueness is
the same abstract one taken in Examples 3 and 4. But due to the underlying
complexity, involving the epsilon-delta definition of limit, of the assumption
that L, and L, both satisfy the definition of L = lim,,, f (x), we must resort
to a different approach to prove that L, = L,. Recall from Example 5,
Article 6.2, that if a is a real number having the property (Vp > O)(lal c p),
then a = 0. We use this result in the next example.
EXAMPLE 5 Prove that if L, and L, are real numbers, both satisfying the
definition of L = lim,,, f(x), where f is defined in an open interval con-
taining a, then L, = L,.Solution Using the result of Example 5, Article 6.2, we attempt to show L, =
L, by the following approach. Let p > 0 be given; we claim IL, - L,I < p.
If we can prove this, we may conclude L, = L,, by the cited example.
Now since L, = lim,,, f(x), then corresponding to any positive real
number, and in particular to our given p divided by 2, there exists 6, > 0
such that (f(x) - L,I < 42 whenever 0 < (x - a( < 6,. Similarly for L,,
there exists 6, > 0 such that 1 f (x) - L,( < p/2 whenever 0 -c Ix - a1 < 6,.
Now since f is defined in some open interval containing a, there exists