# understand proof of compactness in product topology

I am trying to understand the following reasoning.

Call $\mathcal{F_\lambda}$ the set of functions $a:\mathbb{N} \to \mathbb{R}$ for which $Na(i) := \sum_{j \in \mathbb{N}} n_{ij} a(j)\leq \lambda a(i) \forall i \in \mathbb{N}$ where $N = n_{ij}$ is an irreducible stochastic matrix. Show that $\mathcal{B_\lambda} = \{a \in \mathcal{F_\lambda} | a(0) = 1, a \geq 0\}$ is compact in the set of pointwise convergence.

In the given proof there are two things shown:

(1) Using that $N$ is irreducible it is shown, that $f(i) \leq \lambda^kn^{(k)}_{0i}$ for a $k$ and for all $a \in \mathcal{B_\lambda}, i \in \mathbb{N}$, where $n^{(k)}_{0i}$ denotes the entry at $0i$ of $N^k$.

(2) It says, that closedness of $\mathcal{B_\lambda}$ can be derived using Fatou's lemma.

The proof finishes, saying, that from (1) and (2) the compactness follows.

How (1) is done is clear to me. My questions:

(Updated the questions to be more precise and specific.)

Question (a): Is the following meant by (2): considering a convergent sequence of sequences $(a_n) \to a$ of $a_n \in \mathcal{B_\lambda}$, which hence all suffice the given condition, one can derive for every $i \in \mathbb{N}$ that $$Na(i) = \sum_{j\in \mathbb{N}}\lim n_{ij}a_n(j) \leq \liminf \sum_{j\in \mathbb{N}} n_{ij}a_n(j) \leq \liminf \lambda a_n(i) = a(i)$$ and hence $a \in \mathcal{B_\lambda}$.

Question (b): Can someone explain why from (1) and (2) compactness follows? Considering what is mentioned here Compactness in $\mathbb{R}^{X}$ my guess is the following: (1) shows pointwise boundedness and (2) shows closedness, hence $\mathcal{B_\lambda}$ is compact in the product topology, i.e. in the topology of pointwise convergence. Is that right?

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Be carefull! The product of compact sets is compact! But there are many more sets with the same projections, and not all of them are compact. I haven't read this through carefully, so I don't know if $\mathcal{B}_{\lambda}$ really is a product - but if it is, and if they factors are compact, then it's compact. You can show this, I think, by using tychonov's theorem, which tells you that the product of compact spaces is compact, and apply it to the trace topologies induced by the projections. –  fgp Apr 2 '14 at 12:29
@topload Yes, you are - see my updated answer to the other question and my comment there. –  fgp Apr 2 '14 at 12:54
@fgp I found a question on compactness in the product topology, which might answer the second part of my question. Wanted to let you know this equivalent condition of compactness in the product topology (if it is true). –  topoload Apr 4 '14 at 9:25
@topload Yeah, this answers it. The idea is that, as mentioned, products of compact sets are compact in the product topology. Thus, if you can establish that (a) some set $M$ is a subset of such a product and you can establish that (b) $M$ is closed, then $M$ is compact since all closed subsets of compact sets are compact. You get (a) from pointwise boundedness, because you can then find an interval $I_n = [a_n,b_n] \subset \mathbb{R}$ for each $n$ such that $M \subset \prod_{n\in\mathbb{N}} I_n$, which as a product of compact sets is compact. –  fgp Apr 6 '14 at 12:32