I'm having a lot of difficulty getting past one small step of the proof of the Continuous Extension Theorem.

Theorem (one direction): Given a function $f$ that is uniformly continuous on the interval $(a, b)$, show that $f$ can be defined at the endpoints $a$ and $b$ such that the extended function is continuous on $[a,b]$.

What I know thus far: If $(x_n)$ is a sequence in $(a,b)$ with $ \lim_{n \to \infty} ( x_n ) = a $, then it is a Cauchy sequence (since convergent sequences are Cauchy sequences). Since $f$ is uniformly continuous on $(a,b)$ and $(x_n)$ is in (a,b), then $(f(x_n))$ is also a Cauchy sequence. Since $(f(x_n))$ is a Cauchy sequence, it converges in $\mathbb{R}$. Thus, the limit $ \lim_{n \to \infty} ( f(x_n) ) = L $ exists. If $(y_n)$ is any other sequence in $(a,b)$ that converges to $a$, then $ \lim_{n \to \infty} ( x_n - y_n ) = a - a = 0 $.

Where I'm stuck (third equality): Thus, by the uniform continuity of $f$, we have $$ \lim_{n \to \infty} ( f(y_n) ) = \lim_{n \to \infty} ( f(y_n) - f(x_n) + f(x_n)) = \lim_{n \to \infty} ( f(y_n) - f(x_n) ) + \lim_{n \to \infty} ( f(x_n) ) = 0 + L = L $$

Question: Why does $\lim_{n \to \infty} ( f(x_n)-f(y_n) ) = 0 $ in the third equality? I think that maybe $ \lim_{n \to \infty} | x_n-y_n | = 0 \implies \lim_{n \to \infty} | f(x_n)-f(y_n) | = 0 $. Is that true? I don't know why...

I understand the rest of the proof and the proof of the converse of the theorem, but not this one step.

  • 1
    $\begingroup$ You should contemplate the definition "uniformly continuous". $\endgroup$
    – GEdgar
    Dec 13 '16 at 2:15
  • $\begingroup$ Alternative approach: Consider the Cauchy sequence given by $x_1, y_1, x_2, y_2, x_3, y_3, \ldots$. $\endgroup$ Dec 13 '16 at 2:18
  • $\begingroup$ I understand that the key to solving this is something to do with the uniform continuity of $f$. I haven't been able to figure out the missing piece of the puzzle though. I know that $f(y_n)$ is also a Cauchy sequence and must, therefore, converge. $\endgroup$
    – hgil
    Dec 13 '16 at 2:24
  • $\begingroup$ @GEdgar From the Definition of Uniform Continuity, I understand that I can find a delta to make $(f(x_n))$ and $(f(y_n))$ arbitrarily close to each other, but I guess I haven't seen that directly imply that $\lim_{n \to \infty} ( f(x_n)-f(y_n) ) = 0$ before. Intuitively, I believe I get it; but I'm having trouble seeing this "algebraically". $\endgroup$
    – hgil
    Dec 13 '16 at 3:19
  • $\begingroup$ @GEdgar Oh so I think that the Definition of Uniform Continuity implies (rather directly) that the $\lim_{n \to \infty} ( f(x_n)-f(y_n) ) = 0$. That is because if you were to view the definition of uniform continuity from a limit perspective, you would have $| f(x)-f(y) - L| < \epsilon, \forall \epsilon > 0$. But $L=0$ in the definition. So from this you can infer that $\lim_{n \to \infty} ( f(x_n)-f(y_n) ) = 0$. Right? $\endgroup$
    – hgil
    Dec 13 '16 at 3:31

I suspect you're overthinking it - writing down the definition of uniform continuity gives you all you need.

By the definition of uniform continuity of $f$, for any $\epsilon > 0$ there is a $\delta > 0$ so that if $|x - y| < \delta$ then $|f(x) - f(y)| < \epsilon$. Since $(x_n)$ and $(y_n)$ both converge to $a$, we can find $n$ sufficiently large that $|x_n - a| < \delta/2$ and $|y_n - a| < \delta/2$, so $|x_n - y_n| < \delta$. Therefore $|f(x_n) - f(y_n)| < \epsilon$. Hence, for any $\epsilon > 0$, there is $n$ sufficiently large that $|f(x_n) - f(y_n)| < \epsilon$; by definition, this means that $\lim_{n\to\infty}|f(x_n)-f(y_n)| = 0$.


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