88
5. CONTINUOUS FUNCTIONS
(6) The Ruler (Thomae’s) function
8
<0, if x 62 Q
f (x) = 1
m
m
: , if x 2 Q with x = , n 2 N, in lowest terms
n
n
n
is continuous at each irrational and discontinuous at each rational (graph is on
Page 127).
Proof.
(a) Suppose c 2 Q. Let (xn) be a sequence of irrationals with lim(xn) = c.
Then lim f (xn) = 0 6= f (c). thus f (x) is discontinuous at x = c.
(b) Suppose c 2 R\Q and ✏ > 0. By the Archimedean Property,
1
9 n0 2 N 3
< ✏. Now 9 only a finite number of rationals in (c
n0
with denominator less than n0.
Choose > 0 3 (c
than n0. then
x 2 R and |x
c| <
, c + ) contains no rationals with denominator less
=) |f (x)
f (c)| = |f (x)| = f (x) 
Thus f is continuous at x = c.
⇣1⌘
(7) f (x) = x sin
. Is the discontinuity at x = 0 removable?
x
⇣1⌘
Yes, since lim x sin
= 0.
x!0
x
Proof.
⇣1⌘
|f (x) 0| = x sin
 |x| ! 0 as x ! 0.
x
Then
8
⇣1⌘
<
x sin
, if x 6= 0
f (x) =
x
:0,
if x = 0
is continuous on R.
1, c + 1)
1
< ✏.
n0
⇤
⇤