# If $f_n(x)=x^n$ converges to $f$, why is $f$ not continuous?

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?

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Also, is the (analysis) tag good or is there a more accurate tag for this question? – Wheepy Feb 21 '13 at 10:21
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

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