Functions with Dense Graphs
Functions with Dense Graphs Eli Dupree and Ben Mathes September 12, 2013 Most of us love an extreme example. In this note we would like to adorn the “bouquet of discontinuous functions” [1] with our favorite flower: a function with a dense graph. A classical example can be found in [3] and in chapter 9 of [5]. In this note we describe a new family of examples. Corresponding to each surjective function r : N → Q+ we construct a function fr : Q+ → Q+ with a graph dense in the first quadrant. We are letting N denote the set of natural numbers {1, 2, 3, . . .}, and Q+ denote the set of positive rational numbers. In topological conversations we say that a set D is dense in a topological space X in case every non-empty open subset of X contains an element of D. When the topological space is Q+ this property can be phrased in terms of the open intervals O(a,b) ≡ { x : a < x < b }, the basic open subsets of Q+ . A subset D of Q+ is dense if and only if O(a,b) ∩ D 6= ∅ for all a < b in Q+ . When D is a subset of Q+ × Q+ ≡ { (x, y) : x, y ∈ Q+ }, then D is dense in the first quadrant exactly when every non-empty open rectangle O(a,b) × O(c,d) ≡ (x, y) : x ∈ O(a,b) and y ∈ O(c,d) contains elements of D. Of course, since the rational quadrant is dense in the real quadrant, such a set D is also a dense subset of the real first quadrant. When you have convinced yourself that a graph { (x, f (x)) : x ∈ Q+ } is dense in the first quadrant if and only if, for each a < b in Q+ , f (O(a,b) ) is dense in Q+ , you begin to appreciate that functions with dense graphs are the extreme antithesis of continuous functions. 1
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Figure 1: Graph of fr with the classic enumeration. Assume we are given a surjective function r : N → Q+ , which we will call an enumeration of Q+ . Define the function fr : Q+ → Q+ that maps the rational m/n (written in reduced form) to rm . We will prove that fr has a graph dense in the first quadrant. With the classic enumeration 1 1 2 1 2 3 { , , , , , , . . .}, 1 2 1 3 2 1 the graph is approximated in Figure 1, where we have plotted (x, fr (x)) for x = i/104 (i = 1, . . . , 104 ). We encourage the reader to play with fr for other enumerations r, like the fascinating ones in [2]. We found a completely elementary proof (though quite intricate) that the graph of fr is dense in the case of the classic enumeration. For the general proof presented below, we rely on the prime number theorem, a calculus problem, and a short lemma. For each x ∈ Q+ , let π(x) denote the number of prime positive integers p less than or equal to x. The prime number theorem gives a handle on the rate of growth of π. A proof can be found in [4]. Prime number theorem. The function π(x) behaves asymptotically like x/ ln x, i.e. π(x) lim x = 1 x→∞
ln x
2
By assigning the following problem, we give ourselves an excuse to state the prime number theorem to calculus students. A good calculus problem. If 0 < a < b, then lim
π(bx) − π(ax) x ln x
x→∞
= b − a.
From the prime number theorem, and with c > 0, we get 1 = lim
π(cx)
x→∞ cx ln(cx)
=
1 π(cx) ln c lim (1 + ). x c x→∞ ln x ln x
(To see the second equality, multiply the right side by ln x/ ln x, and use the fact that ln(cx) = ln c + ln x.) It follows that lim
x→∞
π(cx) x ln(x)
= c,
which solves the calculus problem. A short lemma. Assume a, b ∈ Q+ with a < b. There exists N ∈ N such that, for all integers m > N , there exist pm ∈ O(am,bm) relatively prime to m. Choose N1 so that aN1 > 1, then find q > 1 so that (aN1 )q > N1 . The calculus problem implies that π(bm) − π(am) → ∞ as m → ∞, so there exists N ≥ N1 with π(bm) − π(am) > q for all m > N . This is the N that we state the existence of in the short lemma. We now prove that it does what was claimed of it. Assume that m > N . Recalling (aN1 )q > N1 convinces us that aq N1q−1 > 1, and aq mq−1 > aq N q−1 ≥ aq N1q−1 > 1. Multiply both sides of the inequality above by m to obtain (am)q > m. Our choice of N ensures that the interval O(am,bm) contains at least q primes (pi )qi=1 . Each of these primes satisfies am < pi , so that (am)q < Πqi=1 pi . If each pi is a divisor of m (i = 1, . . . , q), then we have Πqi=1 pi ≤ m, contradicting the inequality m < (am)q < Πqi=1 pi . 3
Thus there exists a prime p ∈ {p1 , . . . , pq } such that p does not divide m. This prime number has the property that p ∈ O(am,bm) , and it is relatively prime to m, so we choose pm = p to establish our lemma. Theorem. Assume r : N → Q+ is an enumeration of the positive rational numbers, and define f : Q+ → Q+ , for reduced m/n ∈ Q+ , by f(
m ) = rm . n
Then the graph of f is dense in the first quadrant. Proof. To show the graph of f dense in the first quadrant, let an interval O(u,v) in Q+ be given (0 < u < v). We need to show that f (O(u,v) ) is a dense subset of Q+ . Apply the lemma to the numbers a = 1/v and b = 1/u. We obtain N such that, for each m > N one has pm relatively prime to m with m m < pm < . v u Consequently, we have m N . We conclude that f (O(u,v) ) contains the set T ≡
n
f ( pmm ) : m > N
o
= { rm : m > N }.
Since removing finitely many points from a dense set is still dense, we see that f (O(u,v) ) contains the dense set T , so that f (O(u,v) ) is itself dense.
References [1] G. Drago, P. D. Lamberti and P. Toni, A “bouquet” of discontinuous functions for beginners in mathematical analysis, Amer. Math. Monthly 118 (9) (2011) 799-810. [2] S. P. Glasby, Enumerating the rationals from left to right, Amer. Math. Monthly 118 (9) (2011) 830-835. [3] G. Hamel, Eine Basis aller Zalen und die unstetigen Losungen der Funktionalgleichung: f (x + y) = f (x) + f (y). Math. Ann., vol. 60, pp. 459462, 1905.
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[4] M. France and G. Tenenbaum, The Prime Numbers and Their Distribution, Student Mathematical Library, Volume 6, AMS (2000). [5] A. B. Kharazishvili, Strange Functions in Real Analysis, Pure and Applied Mathematics, Chapman & Hall, 2006.
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Figure 1: Graph of fr with the classic enumeration. Assume we are given a surjective function r : N → Q+ , which we will call an enumeration of Q+ . Define the function fr : Q+ → Q+ that maps the rational m/n (written in reduced form) to rm . We will prove that fr has a graph dense in the first quadrant. With the classic enumeration 1 1 2 1 2 3 { , , , , , , . . .}, 1 2 1 3 2 1 the graph is approximated in Figure 1, where we have plotted (x, fr (x)) for x = i/104 (i = 1, . . . , 104 ). We encourage the reader to play with fr for other enumerations r, like the fascinating ones in [2]. We found a completely elementary proof (though quite intricate) that the graph of fr is dense in the case of the classic enumeration. For the general proof presented below, we rely on the prime number theorem, a calculus problem, and a short lemma. For each x ∈ Q+ , let π(x) denote the number of prime positive integers p less than or equal to x. The prime number theorem gives a handle on the rate of growth of π. A proof can be found in [4]. Prime number theorem. The function π(x) behaves asymptotically like x/ ln x, i.e. π(x) lim x = 1 x→∞
ln x
2
By assigning the following problem, we give ourselves an excuse to state the prime number theorem to calculus students. A good calculus problem. If 0 < a < b, then lim
π(bx) − π(ax) x ln x
x→∞
= b − a.
From the prime number theorem, and with c > 0, we get 1 = lim
π(cx)
x→∞ cx ln(cx)
=
1 π(cx) ln c lim (1 + ). x c x→∞ ln x ln x
(To see the second equality, multiply the right side by ln x/ ln x, and use the fact that ln(cx) = ln c + ln x.) It follows that lim
x→∞
π(cx) x ln(x)
= c,
which solves the calculus problem. A short lemma. Assume a, b ∈ Q+ with a < b. There exists N ∈ N such that, for all integers m > N , there exist pm ∈ O(am,bm) relatively prime to m. Choose N1 so that aN1 > 1, then find q > 1 so that (aN1 )q > N1 . The calculus problem implies that π(bm) − π(am) → ∞ as m → ∞, so there exists N ≥ N1 with π(bm) − π(am) > q for all m > N . This is the N that we state the existence of in the short lemma. We now prove that it does what was claimed of it. Assume that m > N . Recalling (aN1 )q > N1 convinces us that aq N1q−1 > 1, and aq mq−1 > aq N q−1 ≥ aq N1q−1 > 1. Multiply both sides of the inequality above by m to obtain (am)q > m. Our choice of N ensures that the interval O(am,bm) contains at least q primes (pi )qi=1 . Each of these primes satisfies am < pi , so that (am)q < Πqi=1 pi . If each pi is a divisor of m (i = 1, . . . , q), then we have Πqi=1 pi ≤ m, contradicting the inequality m < (am)q < Πqi=1 pi . 3
Thus there exists a prime p ∈ {p1 , . . . , pq } such that p does not divide m. This prime number has the property that p ∈ O(am,bm) , and it is relatively prime to m, so we choose pm = p to establish our lemma. Theorem. Assume r : N → Q+ is an enumeration of the positive rational numbers, and define f : Q+ → Q+ , for reduced m/n ∈ Q+ , by f(
m ) = rm . n
Then the graph of f is dense in the first quadrant. Proof. To show the graph of f dense in the first quadrant, let an interval O(u,v) in Q+ be given (0 < u < v). We need to show that f (O(u,v) ) is a dense subset of Q+ . Apply the lemma to the numbers a = 1/v and b = 1/u. We obtain N such that, for each m > N one has pm relatively prime to m with m m < pm < . v u Consequently, we have m N . We conclude that f (O(u,v) ) contains the set T ≡
n
f ( pmm ) : m > N
o
= { rm : m > N }.
Since removing finitely many points from a dense set is still dense, we see that f (O(u,v) ) contains the dense set T , so that f (O(u,v) ) is itself dense.
References [1] G. Drago, P. D. Lamberti and P. Toni, A “bouquet” of discontinuous functions for beginners in mathematical analysis, Amer. Math. Monthly 118 (9) (2011) 799-810. [2] S. P. Glasby, Enumerating the rationals from left to right, Amer. Math. Monthly 118 (9) (2011) 830-835. [3] G. Hamel, Eine Basis aller Zalen und die unstetigen Losungen der Funktionalgleichung: f (x + y) = f (x) + f (y). Math. Ann., vol. 60, pp. 459462, 1905.
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[4] M. France and G. Tenenbaum, The Prime Numbers and Their Distribution, Student Mathematical Library, Volume 6, AMS (2000). [5] A. B. Kharazishvili, Strange Functions in Real Analysis, Pure and Applied Mathematics, Chapman & Hall, 2006.
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