Injectivity and Monotonicity of Continuous Function
Have you ever wonder if a function is injective, then it is (strictly) monotonic. Obviously, if we do not assume that the function is continuous, then a simple piecewise function can be constructed to give a counterexample. If continuity is assumed and the domain is an interval, then the answer is affirmative. That means injectivity and (strict) monotonicity is equivalent.
Let us recall some definition. Let I be an interval.
Definition 1. A function f : I ® R is injective if for any x, y in I, f (x) = f (y) implies that x = y.
Definition 2. A function f : I ® R is (strictly) monotonic if it is either (strictly) increasing or (strictly) decreasing. That means for all x > y in I, either f (x) > f (y) or f (x) < f (y).
We state our assertion as follows.
Theorem 3. If I is an interval and f : I ® R is continuous and injective, then f is (strictly) monotonic.
First we shall prove the following observation. For the ease of exposition, we shall use 'monotonic' interchangebly with 'strictly monotonic' and 'increasing' with 'strictly increasing'.
Proposition 4. Suppose I is an interval and f : I ® R is continuous and injective.
Then for any x, y and z in I with x < y < z either f (x) < f (y) < f (z) or f (x) > f (y) > f (z).
Hence we have
(i) If f (x) < f (y) or f (x) < f (z) or f (y) < f (z), then f (x) < f (y) < f (z).
(ii) If f (x) > f (y) or f (x) > f (z) or f (y) > f (z), then f (x) > f (y) > f (z).
Proof. Suppose x < y < z. Then (x, y)Ç(y, z) = Æ. Since f is injective, this implies that
f ((x, y))Ç f ((y, z)) = Æ. We have then the following possibilities regarding f (x), f (y) and f (z):
Case (1) f (x) < f (y) and f (y) < f (z).
Case (2) f (x) < f (y) and f (y) > f (z).
Case (3) f (x) > f (y) and f (y) < f (z).
Case (4) f (x) > f (y) and f (y) > f (z).
Then by the IntermediateValue Theorem, since I is an interval, we have the following conclusions according to each case above:
(1) ( f (x), f (y) ) Í f ((x, y)) and ( f (y), f (z)) Í f ((y, z));
(2) ( f (x), f (y) ) Í f ((x, y)) and ( f (z), f (y)) Í f ((y, z));
(3) ( f (y), f (x) ) Í f ((x, y)) and ( f (y), f (z)) Í f ((y, z));
(4) ( f (y), f (x) ) Í f ((x, y)) and ( f (z), f (y)) Í f ((y, z));
Case (2) implies that ( f (x), f (y) ) Ç ( f (z), f (y)) = (max(f (x), f (z)), f (y))¹ Æ. But
( f (x), f (y) ) Ç ( f (z), f (y))Í f ((x, y))Ç f ((y, z)) = Æ and so ( f (x), f (y) ) Ç ( f (z), f (y)) = Æ contradicting ( f (x), f (y) ) Ç ( f (z), f (y)) ¹ Æ. Thus Case (2) is not possible.
Similarly case (3) implies that ( f (y), f (x) ) Ç ( f (y), f (z)) = ( f (y), min(f (x), f (z)))¹ Æ. But ( f (y), f (x) ) Ç ( f (y), f (z))Í f ((x, y))Ç f ((y, z)) = Æ and so ( f (y), f (x) ) Ç ( f (y), f (z)) = Æ contradicting ( f (y), f (x) ) Ç ( f (y), f (z)) ¹ Æ. Thus Case (3) is not possible. Therefore, we are left with cases (1) and (4). That is to say , f (x) < f (y) < f (z) or f (x) > f (y) > f (z). This completes the proof of the proposition.
Proof of Theorem 3.
Suppose for some x1, x2 in I with x1 < x2 , we have that f (x1) < f (x2). We shall show that then f is (strictly) increasing, i.e., for any y, z in I with y < z , f (y) < f (z).
If x1 = y and x2 = z, then we have nothing to show since f (x1) < f (x2). If only one of y or z is equal to either x1 or x2 , then by Proposition 4 part (i) f (y) < f (z). It remains to see the same conclusion can be reached when y and z are distinct from both x1 or x2. By the total ordering on R, we have the following six possibilities:
Case (1) y < z < x1 < x2;
Case (2) y < x1 < z < x2;
Case (3) y < x1 < x2 < z;
Case (4) x1 < y < z < x2 ;
Case (5) x1 < y < x2 < z;
Case (6) x1 < x2 < y < z.
For cases (1), (2) and (3), applying Proposition 4 Part (i), we obtained f (y) < f (x1) using the inequality y < x1 < x2 and the supposition f (x1) < f (x2). Applying Proposition 4 Part (i) again, we have then f (y) < f (z) since f (y) < f (x1) and either y < x1 < z or y < z < x1.
For cases (4) and (5) since x1 < y < x2 and f (x1) < f (x2), applying Proposition 4 Part (i), we get f (y) < f (x2). Then applying Proposition 4 Part (i) again we get f (y) < f (z) since we now have f (y) < f (x2) and either y < z < x2 or y < x2 < z.
For case (6) Applying Proposition 4 part (i) gives us f (x2) < f (y). Therefore, applying again Proposition 4 Part (i) we get f (y) < f (z) since x2 < y < z. Hence f is (strictly) increasing.
Similarly if we have f (x1) > f (x2), we can show that for any y, z in I with y < z, we have that f (y) > f (z). We only have to reverse the inequality in the images in the above proceeding and use Proposition 4 Part (ii) instead of Part (i). This means that f is (strictly) decreasing.
Therefore, f is (strictly) monotonic. This completes the proof of Theorem 3.
Example of a discontinuous function which is injective but not monotonic.
Define g : R ® R by g(x) = x if x is rational and g(x) = -x if x is irrational. Then g is not continuous and g is injective but not monotonic.
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