Helly's Selection Theorem

[l'Hôpital]

Homework Statement

It's from Rudin, Chapter 7, #13:

Suppose {f_n} is a sequence of monotonically-increasing functions on R with 0 < f_n(x) < 1 for all x and n.

a) Show there is a function f and a sequence {n_k} such that
$$f(x) = \lim_{k \rightarrow \infty} f_{n_k} (x)$$
b) If, moreover, if f is continuous, show the convergence is uniform.

Homework Equations

The Attempt at a Solution

So I've proven it's true for rational x and for the points at which f and the f_n are discontinuous. I don't know how to figure out when they are all continuous at a point x. Every use of the triangle inequality requires some sort of equicontinuity requirement or something of the sort which I don't have. Help?

 

[mighty2000]

I would first like to commend you for your efforts in proving this statement for rational x and points of discontinuity. However, I believe you may have overlooked a key piece of information in the problem statement - specifically, the fact that the f_n functions are monotonically-increasing. This means that for any given x, the sequence {f_n(x)} is also monotonically-increasing and therefore converges to a limit, which we can call f(x). In other words, we can define f(x) as the supremum of the sequence {f_n(x)}, and since each f_n(x) is bounded between 0 and 1, f(x) must also be bounded between 0 and 1.

Now, for part (a), we can choose n_k to be any subsequence of the natural numbers (such as the prime numbers, for example), and the limit of f_{n_k}(x) as k approaches infinity will still be f(x). This is because, as mentioned before, the sequence {f_n(x)} is monotonically-increasing and bounded, so it must converge to a limit regardless of the specific subsequence chosen.

For part (b), since f is defined as the supremum of the sequence {f_n(x)}, it is clear that f is continuous. Therefore, for any given epsilon > 0, we can find a delta > 0 such that for all x' in the interval (x-delta, x+delta), we have |f(x') - f(x)| < epsilon. We can also find a value K such that for all k > K, we have |f_{n_k}(x') - f(x')| < epsilon for all x' in the same interval. Therefore, for all x' in this interval, we have |f_{n_k}(x') - f_{n_k}(x)| < 2epsilon. Combining these two inequalities, we have |f(x') - f_{n_k}(x)| < 3epsilon for all x' in the interval. Since epsilon was arbitrary, this shows that the convergence is uniform.

I hope this helps in your understanding of the problem. Keep up the good work!