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I started reading about the topology of pointwise convergence. So far I do not feel quite comfortable with this theory. Maybe one can help me out in a more concrete example case.

Let's consider sequences, so $a : \mathbb{N} \to \mathbb{R},x \mapsto a(x)$.

In the topology of pointwise convergence it holds now that

$$(a)_n \to (A) \iff (a(x))_n \to A(x),\forall x \in \mathbb{N} \quad (1)$$

I hope so far I somehow boiled it down correctly to that concrete case.

Now I wonder what it means for a set $\mathcal{S} \subseteq \mathbb{R}^\mathbb{N}$ of sequences to be compact in the topology of pointwise convergence.

First, I need to get a feeling about "open sets" in that topology and how (1) defines those open sets -- I do not quite get the link.

After that I would need to show, that every open cover of $\mathcal{S}$ has a finite subcover, right? What would that mean precisely, I mean in doing..

ADDED STUFF

Consider we want to show, that $\mathcal{S} \subseteq \mathbb{R}^\mathbb{N}$ is compact in the topology of pointwise convergence.

By, $S_n$ we denote the projections. Assume we know, that all $S_n$ are bounded.

Further assume, that for every sequence in $\mathcal{S}$ we know, that the resulting projected sequences have a convergent subsequence of which the limit lies in the corresponding $S_n$. Does this suffice to show that $\mathcal{S}$ is compact?

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  • $\begingroup$ Use that the topology of pointwise convergence is the product topology $\mathbb{R}^\mathbb{N}$. In that topology, a set is open exactly iff if it's a product of open subsets $(U_i)_{i\in\mathbb{N}}$ of $\mathbb{R}$ and only for only finitely many of these $U_i \neq \mathbb{R}$. $\endgroup$ – fgp Apr 1 '14 at 15:52
  • $\begingroup$ @topoload: Such subsets are indeed compact, but there are many more compact subsets which are not of this form. Think of compact subsets of $R^2$: Are all of them products of compacts in $R$? $\endgroup$ – Moishe Kohan Apr 1 '14 at 16:34
  • $\begingroup$ @topoload Btw, I my first comment I made a mistake - the cylinder sets, i.e. the products of open subsets $(U_i)_{i\in \mathbb{N}}$ where only for finitely many $i$ you have $U_i \neq \mathbb{R}$, aren't all the open sets of the product topology. In general, each open set is a union of such cylinder sets, not directly equal to a cylinder set. $\endgroup$ – fgp Apr 1 '14 at 23:41
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For a $M \subset \mathbb{R}^\mathbb{N}$, where $\mathbb{R}^\mathbb{N}$ endowed with the topology of pointwise convergence, i.e. the product topology, let $$ M_n = \{x_n \,: \, (x_i)_{i\in\mathbb{N}} \in \mathbb{R}^\mathbb{N}\} $$ be the projections of $M$ onto the factors of the product space $\mathbb{R}^\mathbb{N}$. Then you have that

  1. If $M$ is compact, all the projections $M_n$ are compact. Thus, all the $M_n$ being compact is a necessary condition for $M$ being compact.

  2. It is possible that all $M_n$ are compact but $M$ isn't. Thus, all the $M_n$ being compact is not, in general, a sufficient condition for $M$ being compact.

  3. However, if $M = \prod_{i=1}^\infty M_n$, i.e. if $M$ is the product of all it's projections, rather than just a subset of that product, and if all the $M_n$ are compact, then $M$ is compact. Thus, for products all the $M_n$ being compact is a sufficient condition for $M$ being compact.


Proof of (1):

The product topology on $X^Y$ is the coarsest topology such that all the projections ($y \in Y$) $$ p_y \,:\, X^Y \to X \,:\, (x_i)_{i \in Y} \to x_y $$ are continuous. In the language of sequences of real numbers, that means that the topology of pointwise convergence is the coarsest topology on $\mathbb{R}^\mathbb{N}$ such that all the mappings $$ p_n \,:, \mathbb{R}^\mathbb{N} \to \mathbb{R} \,:\, (x_i)_{i \in \mathbb{N}} \to x_n \text{,} $$ i.e. the mappings which map a sequence to it's $n$-th value, are continuous.

Thus, since the image of compact sets is compact under continuous mappings, if $M \subset \mathbb{R}^\mathbb{N}$ is compact in the topology of pointwise convergence, then all the sets $$ M_n := p_n(M) = \{x_n \,:\, (x_i)_{i\in\mathbb{N}} \in M\} $$ are compact in the usual topology of $\mathbb{R}$. In other words, for every compact set of sequences (in the topology of pointwise convergence) the set of $n$-th values is compact (in $\mathbb{R}$).

That leaves the question of whether the $M_n$ being all compact (in $\mathbb{R}$) is a sufficient condition open, though - it only proves the compactness of the $M_n$ (in $\mathbb{R}$) to be necessary for the compactness of $M$ (in the topology of pointwise convergence).

Proof of (2):

Let $$ M = \{s^k \,|\, k \in \mathbb{N}\} \text{ where } s^k = (\underbrace{0,\ldots,0}_{k-1\textrm{ times}},k,0,\ldots) \in \mathbb{R}^\mathbb{N} \text{,} $$ i.e. $M$ is the set of sequences $s^k$ with $s^k_n = k\delta_{n,k}$ ($\delta$ is the Kronecker delta here). Then $M_n = \{0,k\}$ which are all clearly compact. Now let $$ O_n = p^{-1}_n\left((n-\tfrac{1}{2},n+\tfrac{1}{2})\right) = \left\{ (x_i)_{i\in\mathbb{N}} \,\big|\, x_n \in (n-\tfrac{1}{2},n+\tfrac{1}{2})\right\} $$ and observe that $s^k \in O_n$ exactly if $n = k$, since only then is $s^k_n = k\delta_{n,k} \in (n-\tfrac{1}{2},n+\tfrac{1}{2})$. Also note that the $O_n$ are, as preimages under a continuous function of an open set, open. Thus, $$ \bigcup_{n=1}^\infty O_i \supset M \text{,} $$ is an covering of $M$ with open sets. But since $s^k \in O_n$ only if $k=n$, no finite number of sets $O_n$ suffices to cover $M$. We've thus found a covering of $M$ with open sets without a finite subcover, and $M$ is therefore not compact, even though all the projections $M_n$ are compact (even finite).


There are two intuitive reasons for (2)

  1. The projections lose quite a lot of information about $M$. Start with a sequence of compact $(M_n)_{n\in\mathbb{N}}$, and set $M = \prod_{n\in\mathbb{N}} M_n$. Now, this $M$ will (I think) be compact (basically by tychonov's theorem). But there are a lot of subsets $\hat M \subset M$ which have the same projections $M_n$. If the condition were sufficient, all of these would have to be compact.

  2. To show that the condition is sufficient, one would need to somehow find a finit subcover for every cover $(O_i)_{i\in I}$ of an $M$ whose projections $M_n$ are compact. But the only starting point would be the projections of the open sets $O_i$. Now, one can of course, for each $n$ seperately, find a finite subcover of $M_n$ within the $(p_n(O_i))_{i\in I}$. But that leaves you with countably many finite subcovers, which doesn't really help with finding one finite subcover with the product space.

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  • $\begingroup$ @topoload Well, in $\mathbb{R}$ compact means bounded and closed... $\endgroup$ – fgp Apr 1 '14 at 18:26
  • $\begingroup$ Exactly. I understood "Is there a way to obtain a necessary (sic!) condition for compactness in the topology of pointwise convergence which uses that all $M_n$ are bounded" as asking for a version of my necessary condition which, instead of saying $M_n$ must be compact, says $M_n$ must be bounded. My answer is yes, but you must say $M_n$ must be bounded and closed because that means compact in $\mathbb{R}$. But the condition is still only necessary, i.e. if $M$ is compact than all the $M_n$ must be bounded and closed. $\endgroup$ – fgp Apr 1 '14 at 18:36
  • $\begingroup$ Re-reading your question, I'm not sure we're looking at the space you're really interested in. Are you aware that $\mathbb{R}^\mathbb{N}$ is the space of sequences of real numbers, not the space of function $\mathbb{R} \to \mathbb{R}$? In particular - pointwise convergence for $\mathbb{R}^\mathbb{N}$ means pointwise convergence of sequences of sequences to a particular sequence. $\endgroup$ – fgp Apr 1 '14 at 19:15
  • $\begingroup$ @topoload I.e., say you have a sequence $(s_n)$ where each $s_n$ is itself a sequence $s_n = (a^n_i) \in \mathbb{R}^\mathbb{N}$ ($a^n$ is NOT a power here!). Pointwise convergence of $s_n$ to $s = (a_i)$ then means that for every $i$, the sequnce of $i$-th values of the $s_n$ converges to $a_i$, i.e. that for every $i$, $a^n_i \to a_i$ as $n \to \infty$. $\endgroup$ – fgp Apr 1 '14 at 19:19
  • $\begingroup$ @topoload However, the usual use-case of pointwise convergence is for sequences of functions $\mathbb{R}\to\mathbb{R}$, i.e. you're looking at sequences $(f_n)$ where $f \in \mathbb{R}^\mathbb{R}$, and say $f_n \to f$ if $f_n(x) \to f(x)$ for all $x$. The correct space to look at for this situation is $\mathbb{R}^\mathbb{R}$, not $\mathbb{R}^\mathbb{N}$. You still use the corresponding topology of pointwise convergence though (just on a larger space), so my answer stays essentially correct, since it doesn't use that $\mathbb{N}$ is countable.. $\endgroup$ – fgp Apr 1 '14 at 19:24

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