5.7 *sets of Points of Discontinuity (Project) 291Proof. The following proof is a little sketchy. You are asked to fill in the
missing details. Let A = { a 1 , a2, · · · , an,···} be a countable set. We'll assume
A is infinite; the finite case is left to you. Define f : IR -t IR as follows, using
decimal notation:\:/x E IR, f(x) = O.x1x2X3 · · · Xn ···,where Xi= {
0
1~ff x < ai } ·
1 X ?_ ai
That is, f(x) is the infinite decimal whose digits Xi are 0 or 1 according to
whether x < ai or x ?. ai.
Then (justify the following):
(a) f is bounded.
(b) \:/x, y E IR, x:::; y::::} Vi EN, Xi :::; Yi·
( c) f is monotone increasing.
(d) Suppose x < y. Then f(x) = f(y) <==> (x, y] contains no points of A.
(e) Lemma: Let x E IR. Choose any no EN and let 6 = min{lx - ail : ai f:. x,
i = 1, 2,-··no}. Then \:/y E N 0 (x), the first n 0 digits of f(x) and f(y) agree.
(f) CLAIM: f is continuous on Ac. To prove this, suppose x E Ac.
Case 1: x is not a cluster point of A. Then :H > 0 3 N 0 (x) contains no
point of A. Then by ( d) above, f is constant on some neighborhood of x, so f
is continuous at x.
Case 2: x is a cluster point of A. Let € > 0. Choose n 0 E N 3 1/10n° < €.
Let 6 = min{lx - ail: i = 1 ,2,-· ·no}. By (e) above, Ix -yl < 6::::} the first
no digits of f(x) and f(y) agree so Ix - YI < 6::::} lf(x) - f(y)I < 1/10n° < €.
Therefore, f is continuous at x.
(g) CLAIM: f is discontinuous at every point of A. To prove this, consider a
fixed ai EA.
Case 1: ai is not a cluster point of An (-oo, ai)· Then :H > 0 3 (ai -6, ai)
contains no point of A. Let x E (ai -6, ai)-Show that the decimals representing
f(x) and f(ai) differ only in the ith place; the ith digit of f(x) is 0 while that of
f(ai) is 1. Thus, f(x) = f(ai) - l~i, and so lim_ f(x) = f(ai) - l~i f:. f(ai)·
x-+-an
Therefore, f is not continuous at ai.
Case 2: ai is a cluster point of An (-oo, ai)· Let c > 0. Choose no EN 3
1/10n° < €. Let 6 = min{laj - ail : j = 1, 2, ···no, j f:. j}.
Let ai - 6 < x < ai. Since ai is a cluster point of An (-oo, ai), :Jj f:. i 3
aj E (x, ai)· But j > no by definition of 6. Thus, f(x), f(ai), and f(aj) are
decimals that agree in the first n 0 places, but the ith digit off (ai) is 1 while
1
that of f(aj) is 0. Therefore, f(ai) - f(x) ?. f(ai) - f(aj) ?. lQi, so \:/x E
1
(ai - 6, ai), f(x) :::; f(ai) - lQi. Thus, x~1!.1; f(x) < f(ai)· Therefore, f is not
continuous at ai. •