compactness in topology of pointwise convergence 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?
 A: 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


*

*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.

*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.

*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)


*

*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.

*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.
