Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was reading my Analysis course notes and had some trouble. I hope you can help me.

Let $C(X)=\{ f | f:X \longrightarrow \mathbb{R} \text{ is a continuous function}\}$.

It was already stated and proved in the notes that, whenever $X$ is a compact metric space, then $C(X)$ is a complete metric space regarding the uniform metric $d_\infty(f,g)= \displaystyle \sup_{x \in X} \lvert f(x)-g(x) \rvert$.

What I thought: if $C(X)$ is complete, then every Cauchy sequence of functions $f_n$ in $C(X)$ must converge to a function $f \in C(X)$.

But then it came to me that the sequence $f_n(x)=x^n$ is such that $f_n \in C([0,1])$ for all $n \in \mathbb{N}$, but $f_n$ converges to $f \in l^\infty(X)$, where $$ f(x) = \begin{cases} 0, \text{ if } x \in [0,1)\\ 1, \text{ if } x = 1 \end{cases} $$ and $l^\infty(X)$ is the metric space of all the limited real functions defined in $X$ with the metric above.

If $f_n \longrightarrow f \in l^\infty(X)$, then $f_n$ is a Cauchy sequence in $C(X)$. As $C(X)$ is complete, why is it that $f \notin C(X) \subset l^\infty(X)$?

I'm sure my mistake is as silly as a wrong assumption, but I can't spot where it is. Could someone give me a clue?

Thanks in advance.

share|improve this question
    
Also, is the (analysis) tag good or is there a more accurate tag for this question? –  Wheepy Feb 21 '13 at 10:21
1  
What is $d_\infty(f_n,f)$? The convergence is not with respect to the metric $d_\infty$. –  P.. Feb 21 '13 at 10:21
    
Ends up it really was something silly. Sorry for that and thank you all for helping. –  Wheepy Feb 21 '13 at 10:55
    
@Pambos, your answer was the closest to being a clue, which is what I wanted, but as it's not posted as an answer, I couldn't mark it as the correct one. –  Wheepy Feb 21 '13 at 10:58
add comment

2 Answers

up vote 0 down vote accepted

$f_n \to f$, but only pointwise. Convergence in $C(X)$ is the same as uniform convergence.

share|improve this answer
    
Both answers were helpful, but yours is more succinct and made me realize my mistake instantly. Thanks! –  Wheepy Feb 21 '13 at 10:59
add comment

The sequence does not converge to that function in the uniform norm. The function $x^n$ will always take values near 1 for $x\in(0,1)$, regardless of the $n$ (they're just closer to $1$) so $\|f-f_n\|_u$ is always 1.

share|improve this answer
add comment

Your Answer

 
discard

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.