Let f be continuous on [a, b) and define: f(x) = max a less than or equal to t l
ID: 1891499 • Letter: L
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Let f be continuous on [a, b) and define: f(x) = max a less than or equal to t less than or equal to x f(t), a less than or equal to x less than b. (How do we know that f is well defined?) Show that f is continuous on [a, b)Explanation / Answer
Such a function does exist. I will describe here how to construct one such function. (The description will be fairly complicated, but this should not be unexpected, as it is certainly a rather bizarre function that is strictly increasing but never assumes a rational value.) First I'll describe the function, and then I'll show that it has the two properties you want (strictly increasing and never rational). (I will be more verbose than usual since you have stated that you have not had an analysis course. I apologize in advance to other interested readers who are familiar with more advanced notation or more formal arguments.) --- I'll start by defining f(x) just for numbers *between* 0 and 1 (not 0 and 1 themselves), and then show at the end how I can extend to these numbers. First, make a list of all the rational numbers between 0 and 1, in the following way: 1/2, 1/3, 2/3, 1/4, 2/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 2/6, 3/6, 4/6, 5/6, ... (First list the fractions with denominator 2; then those with denominator 3, in order; and so on. Some numbers are listed more than once--for example, 1/2 and 2/4 are listed separately--but this will not matter.) Call the nth fraction in the list r_n. (So r_1 = 1/2; r_2 = 1/3; and so on.) Now, I will define f(x) for when x is between 0 and 1. I will define the value of f(x) to always be between 0 and 1, and I'll define each of its digits after the decimal point separately, as follows: If x > r_n, then the nth decimal place of f(x) should be 1. Otherwise, the nth decimal place of f(x) should be 0. As an example, let's find the first few decimal places of f(1/2). The first decimal place should be 0 (since 1/2 is not greater than 1/2, the first fraction in the list above). The second decimal place should be 1 (since 1/2 is greater than 1/3, the second fraction). The third decimal place should be 0 (since 1/2 is not greater than 2/3, the third fraction). The fourth decimal place should be 1 (since 1/2 is greater than 1/4). The fifth decimal place should be 0 (since 1/2 is not greater than 2/4). ...and so on. So f(1/2) starts out as 0.01010... --- I will show that f is a strictly increasing function, by showing that if a > b, then f(a) > f(b). So suppose a > b. I'll show that each decimal place in f(a) will be greater than or equal to the corresponding decimal place in f(b). If a and b are both greater than one of the fractions in the list, both corresponding decimal places will be 1; if a and b are both less than one of the fractions in the list, both corresponding decimal places will be 0; and if a is greater than the fraction but b is less than it, then the corresponding decimal place will be 1 in f(a) but 0 in f(b). (It can't happen that a is larger than the fraction but b isn't, because we know a > b.) Since each decimal place in f(a) is at least as large as each decimal place in f(b), then f(a) >= f(b); but they can't be equal, because there is *some* rational number in between a and b. So they differ at some decimal place, and f(a) must be larger then f(b) at that decimal place. So f(a) > f(b). --- Now I'll show that f(x) is irrational for every x between 0 and 1. Recall that a number is rational just when its decimal expansion is either terminating or repeating. I'll show that the decimal expansion of f(x) can be neither terminating nor repeating. First, notice that f(x) has infinitely many 0's in it, because x is certainly less than *some* fraction, and if it's greater than, say, 1/10, it's certainly going to be greater than 1/11, 1/12, and so on, which occur later in the list of fractions. Next, notice that f(x) has strings of 1's as long as you like it in. (It has a string of at least ten zeros somewhere; a string of at least a hundred zeros somewhere; and so on.) This is true because x is bigger than *some* fraction. Say x is bigger than 1/100. Then when we get to the part in the sequence of fractions that goes 1/1000, 2/1000, ..., 9/1000, we have a string of nine 1's in the decimal expansion of f(x). When we get to the part that goes 1/10000, 2/10000, ..., 99/10000, we get a sequence of ninety-nine 1's. And so on. Since f(x) has infinitely many zeros, but arbitrarily long strings of 1's in it, there's no way it can be repeating. (Can it start repeating after 10,000 decimal places? No, because we haven't found our string of 10,001 1's in a row yet!) So f(x) is always irrational. --- Thus, f is a function with the properties you seek. So far it's only defined for numbers *between* 0 and 1, but we can easily extend it by defining f(0) = -square root of 2, f(1) = square root of 2. Then the function is still strictly increasing, because f(0) is the smallest value in the range and f(1) is the biggest value, and its range is still entirely irrational numbers. --- (A postscript note to those familiar with analysis: The function I've described is strictly increasing, so it has at most countably many discontinuities. I'm not sure where these are, but my hunch is that it's probably discontinuous at every rational number and continuous at every irrational number. Thoughts?)
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