# comparing compactness theorems: bolzano-weierstrass and arzela-ascoli

Rudin says In [BW] we saw that every bounded sequence of complex numbers contains a convergent subsequence, and the question arises whether something similar is true for sequences of functions. I'm trying to follow his statement and understand the role of equicontinuity in Rudin's Arzela-Ascoli theorem in the framework of Bolzano-Weierstrass.

For reference, Rudin's AA theorem states:

If $K$ is compact, if $f_n \in \mathcal{C}(K)$ for $n=1,2,\ldots$, and if $\{f_n\}$ is pointwise bounded and equicontinuous on $K$, then

1. $\{f_n\}$ is uniformly bounded on $K$
2. $\{f_n\}$ contains a uniformly convergent subsequence

and his BW states:

1. If $\{p_n\}$ is a sequence in a compact metric space $X$, then some subsequence of $\{p_n\}$ converges to a point of $X$.
2. Every bounded sequence in $\mathbb{R}^k$ contains a convergent subsequence.

For BW, I have a simple mental picture of two points in a closed/bounded interval of $\mathbb{R}$, a sequence that alternates between the two points, and a convergent subsequence constructed by taking a sequence of one of the points.

I'm looking for another simple view of AA similar to this so that I can develop some more tangible idea of what equicontinuity provides. Does anyone have a nice, simple interpretation?

• Your view of BW should be more general. The point is that if you have any infinite sequence in this interval, there is some subsequence which converges. The points can be spread out quite sporadically. Somehow the compactness says that there is 'finite space' and you are trying to put infinitely many points in this space. So somehow they accumulate. – Alfred Yerger Jun 2 '17 at 16:16
• On the other hand, you can think of $C(K)$ as a metric space itself, with metric induced by say, the sup norm. This space is not itself compact, but if it is bounded and equicontinuous, then it becomes (relatively) compact. The general form of BW says that a sequence of points in a compact metric space has a subsequence, so really Arzela-Ascoli gives conditions on when $C(X)$ is a compcat metric space - when the functions are bounded and equicontinuous. – Alfred Yerger Jun 2 '17 at 16:19
• Great, thinking in terms of "finite space" helps. Is there anything in the BW proof for points that is analagous to using equicontinuity for functions? Or is it that we have that "points are always equicontinuous"? I feel I'm trying to force this reinterpretation, but I also feel there might be some connection between the topogy of points and the topology of a family of equicontinuous functions? Is there a way to think about it like this? – jjjjjj Jun 2 '17 at 17:24
• I think that's a good question, and I haven't thought about it all that closely. Perhaps you can think of points as constant functions? These are trivially equicontinuous. – Alfred Yerger Jun 2 '17 at 17:30
• I can copy paste my comments, and then we can close this question. It would probably be good for you to ask a new question with a link to the old one. That will also attract people who see the new question to this one, and that might help you reach a good answer. – Alfred Yerger Jun 2 '17 at 18:06

Your view of BW should be more general. The point is that if you have any infinite sequence in this interval, there is some subsequence which converges. The points can be spread out quite sporadically. Somehow the compactness says that there is 'finite space' and you are trying to put infinitely many points in this space. So somehow they accumulate. On the other hand, you can think of $C(K)$ as a metric space itself, with metric induced by say, the sup norm. This space is not itself compact, but if it is bounded and equicontinuous, then it becomes (relatively) compact. The general form of BW says that a sequence of points in a compact metric space has a subsequence, so really Arzela-Ascoli gives conditions on when $C(X)$ is a compact metric space - when the functions are bounded and equicontinuous.