In Royden's text the Arzela-Ascoli Theorem states:

Let X be a compact metric space and $f_n$ a uniformly bounded, equicontinuous sequence of real valued functions on X. Then $f_n$ has a subsequence that converges uniformly on X, to a continuous function f on X.

However I cannot seem to see where the hypothesis of uniform boundedness is used in the proof - it seems that only pointwise boundedness of the sequence is required. My question: is uniform boundedness actually required, or could we replace "uniformly bounded" with "pointwise bounded", in the statement of the theorem? And if we cannot: what is an example of a pointwise but not uniformly bounded sequence for which the theorem fails?

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    $\begingroup$ @SangchulLee thank you for your clear explanation. Seems as though the theorem would be more clearly stated in Royden with pointwise boundedness, seeing as this is what is used in the proof, not uniform boundedness (even though they are equivalent).. Thank you for your help. $\endgroup$ – qwert4321 Jul 3 '16 at 4:06

(Migrated from the comment) Recall that uniform continuity on a totally bounded set (for instance, a compact set) implies that a function is bounded.

Similarly, for a family $\mathcal{F}$ of functions on a metric space $X$,

  1. Equicontinuity of $\mathcal{F}$,
  2. Totally-boundedness of $X$, and
  3. Pointwise boundedness of $\mathcal{F}$

implies that the family of function is uniformly bounded. So if you assume that $X$ is compact and $(f_n)$ is equicontinuous, then $(f_n)$ is pointwise bounded iff it is uniformly bounded.

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    $\begingroup$ @Masacroso, One of the naturally occurring examples is the family of Taylor polynomials of an entire function. For instance, take $f_n(x) = \sum_{k=0}^{n} \frac{(-1)^k}{(2k)!} x^{2k}$ then $(f_n)$ is pointwise bounded since $f_n(x) \to \cos x$ and the limit is uniformly bounded, but still $(f_n)$ is not uniformly bounded. $\endgroup$ – Sangchul Lee Jul 3 '16 at 4:14

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