196 METHODS OF MATHEMATICAL PROOF, PART II Chapter 6
We deal next with union. To prove S u T is open, let x E S u T.
To prove that x is an interior point of S u T, we must produce 6 > 0
such that N(x; 6) c S u T. Since x is either in S or T (but we don't
know which one), and since our hypotheses deal with properties of S
and T individually, this proof is clearly a candidate for division into
cases. Our strategy is to divide the proof into the two cases x E S and
x E T, and see whether, in each case, we can arrive at the desired con-
clusion. Now if x E S, then since S is open, there exists a positive 6,
such that N(x; 6,) c S. But since S c S u T, we may let 6 = 6, for this
case, noting that N(x; 6) = N(x; 6,) E S E S u T, so that x is, indeed,
an interior point of S u T. Similarly, if x E T, then 36, > 0 such that
N(x; 6,) E T. Letting 6 = we conclude again in this case that
N(x; 6) = N(x; 6,) c T c S u T, as desired. 0The theorem that union and intersection of two open sets is open is gen-
erally taught in a course in advanced calculus and/or elementary topology.
If you have not already seen these proofs, then it is worth noting that their
complexity is fairly representative of many proofs in courses at that level.
Two final remarks: (1) You should realize now that techniques that were
stressed individually in Chapters 4 and 5 (e.g., cases, specialization, choose)
are beginning to be used, several at a time, in individual proofs. (2) You
should know that examples such as 7, 8, and 9 illustrate that an important
aspect of proof writing is the careful interpretation of definitions.
EPSILON-DELTA PROOFS
Let us now turn to the problem of writing epsilon-delta proofs that L =
lirn,,, f(x). Such proofs may arise in connection with specific functions,
as in 25 = lirn,,, x2 or Ma + B = lirn,,, (Mx + B), or may be required
for the purpose of establishing a more general theorem, such as "if
lirn,,, f(x) = L, and lirn,,, g(x) = L,, then lirn,,, (f(x) + g(x)) = L, +
L, ."
A typical epsilon-delta proof is structured as follows. Begin by letting
E > 0 be given. The crux of the proof is to define, in terms of this E, a 6 > 0
having the property that, whenever x is a real number satisfying the inequal-
ities 0 < Ix - a1 < 6 (i.e., x is within 6 of a, but x # a), then its correspon-
ding f(x) satisfies (f(x) - LI < E [i.e., f(x) is within E of L]. The problem,
inevitably, is how to choose 6. The key for many specific functions is to
look for a relationship between the quantities lx - a1 and I f(x) - LI. Note,
in particular, that if a positive constant k can be found such that 1 f (x) - LI 5
klx - a1 for all x within some neighborhood N(a; 6,) of a (with the pos-
sible exception of x = a itself), then 6 = min (6 ,, ~/k) will do the job,
ForifO < Ix - a1 < 6, then0 < Ix - a1 < 6, andso 1 f(x) - LI < klx - a( <
(&/6)(1x - al) c (~/6)(6) = E, as desired. Let us look now at some examples.