Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$X$ is a compact metric space and $( C(X), ||.||_\infty )$ the space of the continous functions on X with the maximum norm.

If it holds that every sequence converges in $X$ : $$\lim _{n\rightarrow \infty} x_n = x $$ in $X$ and every sequence of functions converges in $C(X)$: $$\lim_{n\rightarrow \infty} f_n = f$$ in $C(X)$

how can one conclude by choosing ($\frac{\epsilon}{3}$) as boundary for $||f_n-f||$ and $|f(x_n)-f(x)|$ :

$$|f_{n}(x_n)-f(x)| = |f_n(x_n)+f(x_n)-f(x_n)+f_n(x)-f_n(x) - f(x)| \le 2 || f_n-f|| + |f(x_n)-f(x)| = \epsilon$$

that $$ \lim_{n\rightarrow \infty} f_n(x_n)=f(x)$$ ?

share|cite|improve this question
Your "every"'s in the second sentence need to "a"'s... That is, $(x_n)$ and $(f_n)$ are given convergent sequences. – David Mitra Jan 18 '13 at 14:58
How does one find these boundaries or rather what is the reasoning behind choosing $\epsilon / 3$ ? And would this method of choosing $\epsilon /3$ also work if X were not compact? – bakabakabaka Jan 18 '13 at 15:45
up vote 2 down vote accepted

First, it seems the statement you want to prove is "if $(x_n)$ is a sequence in $X$ with $\lim x_n=x$ and $(f_n)$ is a convergent sequence in $C(X)$ with limit $f$, then $\lim\limits_{n\rightarrow\infty}f_n(x_n)=f(x)$".

In your proof, it seems you've written $$ |f_n(x_n)-f(x)| \le |f_n(x_n)-f(x_n)|+|f(x_n)-f_n(x) |+| f(x_n)-f(x)|. $$ You do know that $|f_n(x_n)-f(x_n)|\le\Vert f_n-f\Vert$, but how do you conclude the same for $|f(x_n)-f_n(x) |$? This is true for sufficiently large $n$, but would require an additional argument.

An alternative method would be to write, for any fixed $N$ and arbitrary $n$: $$\eqalign{ |f_n(x_n)-f(x)| &=| f_n(x_n)-f_N(x_n) +f_N(x_n)-f_N(x)+f_N(x) -f(x) | \cr &\le \color{maroon}{|f_n(x_n)-f_N(x_n)|} +\color{darkgreen}{|f_N(x_n)-f_N(x)|}+\color{darkblue}{|f_N(x)-f(x)|}.} $$

Then, given $\epsilon>0$, you can select $N$ so large that:

$\ \ \ 1)\ \ \color{darkblue}{|f_N(x)-f(x)|}<\epsilon/3$, since $\Vert f_n-f\Vert\rightarrow 0$.

$\ \ \ 2)\ \ \color{darkgreen}{|f_N(x_n)-f_N(x)|}<\epsilon/3$ for all $n\ge N$, since $f_N$ is continuous and since $x_n\rightarrow x$,


$\ \ \ 3)\ \ \color{maroon}{|f_n(x_n)-f_N(x_n)|}<\epsilon/3$ for all $n\ge N$, since $\Vert f_n-f\Vert\rightarrow 0$.

It would follow then that for every $\epsilon>0$, there is an $N$ so that for all $n\ge N$, $$ |f_n(x_n)-f(x)|<\epsilon; $$ which shows that $\lim\limits_{n\rightarrow\infty}f_n(x_n)=f(x)$.

Incidentally, I don't think the compactness of $X$ is required.

share|cite|improve this answer
Compactness of $X$ is only needed to make $C(X)$ a normed space. But it won't hurt to use only the bounded continuous functions. – Hagen von Eitzen Jan 18 '13 at 15:57
Thank you very much. – bakabakabaka Jan 18 '13 at 15:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.