Let $f:[a,b]\to \Bbb R$. We say that $f(x)$ is uniformly continuous if for every two sequences $x_n,y_n$ in $[a,b]$ satisfying $x_n-y_n\to 0$, we have $f(x_n)-f(y_n)\to 0$. It is well known that this is equivalent to the usual epsilon-delta definition of uniform continuity.
I wish to prove the Heine-Cantor theorem, that if $f(x)$ is continuous in $[a,b]$ then it is uniformly continuous there, but I want to prove this directly using the sequential definition above (without resorting to the equivalence with the epsilon-delta definition).
I tried the following, which is inspired by the standard epsilon-delta proof of the theorem, but it doesn't quite work: let $x_n,y_n$ be such that $x_n-y_n\to 0$. Since $x_n\in[a,b]$ then $x_n$ is bounded, so it has a convergent subsequence $x_{n_j}\to L$. It follows that $y_{n_j}\to L$, and by continuity $f(x_{n_j})\to f(L)$ and $f(y_{n_j})\to f(L)$ so that $f(x_{n_j}) – f(y_{n_j}) \to 0$. However, this isn't enough; I wanted to conclude that $f(x_n) – f(y_n) \to 0$. Is there an extra step I'm missing that will allow me to easily conclude that $f(x_n)-f(y_n)\to 0$? Is there a different way to approach this proof?
Best Answer
You can appeal to the following theorem: Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.. You have already shown that there is a subsequence of $\{x_n-y_n\}$ that converges to $0$, but of course you could have taken a subsequence instead of $\{x_n-y_n\}$, so indeed you know that every subsequence has a subsubsequence that converges to $0$, and you are done by the theorem.