Chapter 4 Problem 19
Suppose f is a real function with domain R1 which has the intermediate value property:
If f (a) < c < f (b), then f (x) = c for some x between a and b. Suppose also for every
rational r, that the set of all x with f (x) = r is closed. Prove that f is continuous. Hint: If
xn → x0 but f (xn ) > r > f (x0 ) for some r and all n, then f (tn ) = r for some tn between x0
and xn ; thus tn → x0 . Find a contradiction.
Before we attack this problem head on, we need to prove a lemma.
Lemma. Let f be a real function defined on R1 . Suppose f has a simple discontinuity. (See
Rudin, page 94.) Then f does not have the intermediate value property (IVP).
Proof. Let x0 be a point at which f has a simple discontinuity. Let f (x0 ) = D. Then
limx→x−0 f (x0 ) = A exists and limx→x+0 f (x0 ) = B, where either A 6= D or B =
6 D. (If
limx→x−0 f (x0 ) = D = limx→x+0 f (x0 ) then the function is continuous at x0 .)
Case 1 A 6= D .
Part (a). A < D. Choose 1 = 1/3|D − A|. Then δ1 exists such that
x ∈ (x0 − δ1 , x0 ) =⇒ |f (x) − A| < 1 .
Next, choose c such that A < D − 1 < c < D. If x ∈ (x0 − δ1 , x0 ), then f (x) < c < D.
Assuming the IVP, there must be some xc such that x0 − δ1 < xc < x0 and f (xc ) = c. But
this is a contradiction, since
xc ∈ (x0 − δ1 , x0 ) =⇒ |f (xc ) − A| < 1 .
Hence the assumption that the IVP applies is false.
Part (b). A > D. Choose 2 = 1/3|D − A|. Then δ2 exists such that
x ∈ (x0 − δ2 , x0 ) =⇒ |f (x) − A| < 2 .
Next, choose c such that D < c < D + 2 < A. If x ∈ (x0 − δ2 , x0 ), then D < c < f (x).
Assuming the IVP, there must be some xc such that x0 − δ2 < xc < x0 and f (xc ) = c. But
this is a contradiction, since
xc ∈ (x0 − δ2 , x0 ) =⇒ |f (xc ) − A| < 2 .
Hence the assumption that the IVP applies is false again.
Case 2 is for B 6= D; the proof is almost identical except for some minor tweeks. Thus
for a real function f defined on R1 , a simple discontinuity is inconsistent with f
having the intermediate value property.
Now we proceed with the proof of the original problem. But we rephrase the theorem
as follows. Suppose f is a real function with domain R1 and f has the intermediate value
property. If f is not continuous, then there exists a rational r such that the set x with
f (x) = r is not closed.
Proof. Since, by assumption, f is not continuous, but it has the intermediate value property,
f cannot have a discontinuity of the first, or simple, kind. Suppose f is not continuous at
x0 . This means that for some > 0 and for all δ > 0, there exists points x such that
0 < |x − x0 | < δ
and
|f (x) − f (x0 )| ≥ .
Case 1. Suppose that for some δ > 0, 0 < 0 < |x − x0 | < δ, implies that f (x) > f (x0 ) − .
That is, for all x sufficiently close to x0 , we have that f (x) ≥ f (x0 ) + . We then first
choose δ1 < δ. Then for x0 < x1 < x0 + δ1 , f (x1 ) ≥ f (x0 ). Then, by the IVP, for any
rational number r such that f (x0 ) < r < f (x1 ), there exists x1r such that x0 < x1r < x1
and f (x1r ) = r. Next, choose x2 < x1r such that f (x2 ) ≥ f (x0 ) + . Then there exists x2r
such that x0 < x2r < x2 and f (x2r ) = r. Having chosen: x0 < x(n−1)r < · · · < x2r < x1r
where f (x(n−1)r ) = · · · f (x2r ) = f (x1r ) = r, we can now choose xn < x(n−1)r such that
f (xn ) ≥ f (x0 ) + . There there exists, xnr such that xnr < xn and f (xnr ) = r. Thus, we
have established a sequence {xnr } such that the image under f of each is r and the sequence
converges to x0 . Thus x0 is a limit point of this sequence but is not a element of the sequence.
Therefore the set xnr is not closed.
Case 2. Suppose that for some δ > 0, 0 < |x − x0 | < δ, implies that f (x) − f (x0 ) ≤ .
That is, for all x sufficiently close to x0 , we have that f (x) ≤ f (x0 ) + . The proof proceeds
analogously to that of Case 1, the major difference being that the rational number r is less
than f (x0 ), i.e. f (x0 ) − < r < f (x0 ).
Case 3. Suppose that for all δ > 0, for some x such that 0 < |x − x0 | < δ, f (x) ≥ f (x0 ) + and for some x such that 0 < |x − x0 | < δ, f (x) ≤ f (x0 ) − . In other words, there is no δ
small enough to insure that Case 1 or Case 2 applies.