I was reading, continuous extension theorem for uniformly continuous function Theorem 5.4.8 in "Introduction to real analysis by Robert G. Bartle" fourth edition In the proof, i didn't get one point.

Author has written in end of first para of proof that, if $(u_n)$ is any sequence in $(a,b)$ that converges to $a$ then $lim(u_n-x_n)=0$. So "By uniform Continuity of $f$ we have,


How uniform Continuity of $f$ is used here?

My opinion: as $lim(u_n-x_n)=0$ hence we have $lim(u_n)=lim(x_n)= a$ (since $(x_n)$, $(u_n)$ are sequences in $(a,b)$ both converging to $a$)

So as $f$ is uniformly continuous on $(a,b)$ and hence it is continuous on $(a,b)$ hence by "Sequential criterion of continuity" we have,

$lim(f(u_n))=lim(f(x_n))=f(a)=L$ (say)

Hence, $lim(f(u_n)-f(x_n))=0$.

So that, $lim(f(u_n))=lim(f(u_n)-f(x_n))+lim(f(x_n))=0+L=L$ as required.

Is my opinion (reason) is correct? Please help.....

  • $\begingroup$ Is the hypothesis since $(x_n)$, $(u_n)$ are sequences in $(a,b)$ both converging to $a$ really one of the author? $\endgroup$ Nov 21, 2020 at 16:14
  • $\begingroup$ @mathcounterexamples.net author consider, if $(x_n)$ is any sequence in $(a,b)$ that converges to $a$ then,...... $\endgroup$ Nov 21, 2020 at 16:17
  • 1
    $\begingroup$ FYI - \lim provides a very nice $\lim f(u_n)$ instead of the funky looking $lim(f(u_n))$. Similar shortcut formatting is available for many other common mathematical keywords. $\endgroup$ Nov 21, 2020 at 23:38
  • $\begingroup$ @PaulSinclair thanks sir. I will definitely use it from next post. $\endgroup$ Nov 22, 2020 at 4:10

1 Answer 1


$f$ being continuous on $(a,b)$ does not mean that $f$ is continuous at $a$, because $a \notin (a,b)$. $f$ is not even defined at $a$, so there is no "$f(a)$" for those two limits to converge to.

The whole point of this theorem is that when $f$ is uniformly continuous on $(a,b)$ you can extend the domain of $f$ to include $a$ in such a way that $f$ will be continuous at $a$. You do not get to assume that the continuous extension has already happened.

Because $x_n$ and $u_n$ converge to point outside the domain of $f$, you need uniform continuity to assure $$\lim_{n\to\infty} f(u_n) - f(x_n) = 0$$

Since you cannot assume a common limit for $f(u_n)$ and $f(x_n)$, you will need to examine the definition of uniform continuity to see why else it indicates they converge to the same value.

To see that plain continuity is not sufficient, consider $(a,b) = (0,1), f(x) = \sin 1/x$.

  • $\begingroup$ Sir, Thanks for point out my mistake. Yes $f$ is continuous on $(a,b)$ does not mean that $f$ is continuous at $a$. By your answer i think this will be correct: As $\lim (u_n-x_n)=0$ so that, for any $ε_0>0$ there exists $n_0\in\mathbb{N}$ such that, $|x_n-u_n|<ε_0$ for all $n≥n_0$. Also by defi. of UC of $f$ on $(a,b)$ for each $ε>0$ there exists $δ>0$ such that, if $x,u\in(a,b)$ such that, $|x-u|<δ$ then $|f(x)-f(u)|<ε$. Now as $ε_0$ is arbitrary we can let $ε_0<δ$ so that, $|x_n-u_n|<δ$ hence $|f(x_n)-f(u_n)|<ε$ for all $n≥n_0$. Hence $\lim (f(x_n)-f(u_n))=0$ as required. Am i correct sir? $\endgroup$ Nov 22, 2020 at 4:06
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    $\begingroup$ Yes, that is how uniform continuity comes into the theorem. $\endgroup$ Nov 22, 2020 at 14:52

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