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Let $\{f_{n}\}$ be a sequence of pointwise bounded continuous functions on a separable metric space $X$. There is a common diagonalization theorem (see Baby Rudin, Theorem 7.23) which states that if $E$ is a countable subset of $X$, then we can find a susequence $\{f_{n_{k}}\}$ which converges on every point of $E$ as $k\to \infty$.

My question is that if $E$ is also dense, must $\{f_{n_{k}}\}$ converge for every point in $X$? The Arzela theorem states that if equicontinuity and compactness of $X$ is also assumed then we have uniform convergence. What results are there if one or both of these assumptions are removed?

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Let $f_n(x) = \max\{1-n\operatorname{dist}(x,\mathbb{Z}),0\}$ for $n \geq 4$. The graph of $f_n\colon \mathbb{R} \to [0,1]$ has triangular spikes of height $1$ and with base of length $\frac{2}{n}$ centered around the integers.

With this picture in mind, you can see that $$ \lim_{n\to\infty} f_n(x) = \begin{cases} 0, & \text{if }x \notin\mathbb{Z}, \\ 1, & \text{if }x \in \mathbb{Z}. \end{cases} $$ If you put $g_{2n}(x) = f_n(x)$ and $g_{2n+1}(x) = f_{n}(x-\frac{1}{2})$ then you obtain a sequence of continuous and bounded functions $g_n$ that converges if and only if $x \notin \frac{1}{2}\mathbb{Z}$.

Nothing prevents you from obtaining a sequence like $(g_n)_{n\in\mathbb{N}}$ from Rudin's theorem you describe (for example taking the sequence $g_n$ and taking a dense set $E$ in the complement of $\frac{1}{2}\mathbb{Z}$).

Restricting the $g_n$'s to the interval $[-100,100]$ also shows that compactness alone doesn't help.

However, the sequence of functions in this example is not equicontinuous.

You can show that if the $f_n$ are equicontinuous and converge pointwise on a dense subset $E$ of $X$ then their limit $\tilde{f}(e) = \lim_{n\to\infty} f_n(e)$ is uniformly continuous on $E$ and hence extends uniquely to a continuous function $f$ on $X$ (this is proved as in the argument for Ascoli's theorem). Then using equicontinuity one can even show that $f_n|_K \to f|_{K}$ uniformly for each compact $K \subset X$.

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Thanks for the response. –  Alex Lapanowski Oct 26 '12 at 15:03

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