From introductory real analysis, I know that the sequence of functions $f_n(x)=x^n$ converges pointwise in that $f_n \rightarrow 0$ for $0 \leq x < 1$ and $f_n \rightarrow 1$ whenever $x=1$. Thus, the limit function $f(x)$ is not continuous, and hence it does not converge uniformly. This is clear, but how can one prove this topologically? I believe we have to use some sort of metric, but I'm not sure how to proceed in this manner. Well, in terms of metrics, say we let $f_n: X \rightarrow Y$ from the set $X$ to the metric space $Y$ and let $d$ be the metric for $Y$. Then the sequence $(f_n)$ is said to converge uniformly to the function $f: X \rightarrow Y$ if given $\epsilon > 0$, there exists $N$ such that $d(f_n(x), f(x)) < \epsilon, \forall n > N, \forall x \in X$. How would this apply here? I would greatly appreciate some input on this.

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    $\begingroup$ Do you have some guidelines for what you'd consider a "topologic" proof? $\endgroup$ – Henning Makholm Oct 9 '11 at 21:50

The metric that corresponds to "uniform convergence" for functions is the "sup norm": $$\lVert f\rVert_{\infty} = \sup\{ |f(x)|\},$$ and the corresponding "uniform metric": $$d(f,g) = \lVert f-g\rVert_{\infty}.$$

Theorem. A sequence $(f_n)$ of functions converges uniformly to $f$ if and only if it converges to $f$ in the uniform metric.

This is clear once you write out what everything means: if $f_n$ converges to $f$ uniformly, then for every $\epsilon\gt 0$ you can find $N\gt 0$ such that for all $x$ and all $n\geq N$, $|f(x)-f_n(x)|\lt\epsilon$; this implies that $\lVert f-f_n\rVert_{\infty}\leq \epsilon$, so uniform convergence implies convergence in the uniform metric. Conversely, if $\lVert f-f_n\rVert_{\infty}\to 0$ as $n\to\infty$, then for every $\epsilon\gt 0$ we can find $N$ such that for all $n\geq N$, $\sup\{|f(x)-f_n(x)|\}\lt \epsilon$, so $f_n$ converges uniformly to $f$.

Of course, you need to check that $\lVert\cdot\rVert_{\infty}$ is a norm on the set of functions (viewed as a vector space over $\mathbb{R}$), so that you can conclude that the uniform metric is indeed a matric and so gives you a topology on the space of functions.

Moreover, if $f_n\to f$ pointwise, and $f_n\to g$ in the uniform metric, then $f=g$. So it would suffice to show that $f_n$ does not converge to $f(x) = \chi_{\{1\}}$ uniformly to show that the sequence does not converge uniformly.

Alternatively, you could show that the sequence is not Cauchy in the uniform metric.


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