Prohorov theorem I had a doubt on one "diagonlization argument" used to show Prohorov's Theorem. I am simplifying the discussion, so lets say that we consider the space $\mathbb{R}^{\infty}$
, and a set Π of probability measures on $\mathbb{R}^{\infty}$. Let us assume that $\Pi$ is tight. Now we want to show that it is relatively compact, i.e. every sequence $P_n \in \Pi$ has a subsequence $P_{n^{\prime}}$ which has a weak limit.
This statement below is easily proved:
Statement: Assume that we know that if the replace in the above statement $\mathbb{R}^{\infty}$ by $\mathbb{R}^{k}$
, then it is true, i.e. for a tight set in the set of measures on $\mathbb{R}^{k}$
, every sub-sequence has a weak limit.
Now we note that the projections $π_k$ of elements of $\mathbb{R}^{\infty}$ to $\mathbb{R}^{k}$
 are continuous, which implies that the sets ${Pπ^{−1}_k:P∈\Pi}$ are tight. Hence for every sequence $P_n$, by passing onto subsequence, $π^{−1}_k P_{n^{\prime}}$ has a weak limit. Now there is a "diagonalization" argument, which I had a problem absorbing. It essentially says that "prune" the above $P_{n^{\prime}}$ inductively to obtain a sequence $P_{\tilde{n}}$ which is such that the projection sequence of measures $π^{−1}_l P_{\tilde{n}}$ have a weak limit for all $l=1,2,\ldots$.
My problem is that if the "pruning" of the sequence $P_{n^{\prime}}$ was to be done only finitely many times, I know that after "pruning" the left over sequence is non-empty. But the problem here is that I have to prune the sequence infinitely many times, so how come in the end I get a non-empty set? I have seen this diagonalization argument used at many places but no one talks of the non-empty limit set.
I posted this question at some other site and was told this is connected to Tychonoff's Theorem. I get it that if at each step of pruning if I take the closure of the the subsequence, then they are compact, and hence their intersection is compact. But then who guarantees me that the intersection is non-empty??
 A: You don't need to have anything "in the end"; that's not how the argument works.
The construction produces a sequence of sequences, each of which is a subsequence of the previous one.  That is, for each $k$, we get a sequence $\{P_n^k\}_{n=1}^\infty$ such that the pushforwards $\{\pi_k^{-1} P_n^k\}$ converges weakly, and such that $\{P_n^k\}$ is a subsequence of $\{P_n^{k-1}\}$.  (I'll stick with your notation for the pushforwards, but $(\pi_k)_* P_n$ would probably be better.)
The "diagonalization" step is now to consider the sequence $\{P_i^i\}_{i=1}^\infty$, which is again a subsequence of our original sequence $\{P_n\}$.  Note carefully this is not the intersection of all the sequences $\{P_n^k\}, k=1,2,\dots$.   It's entirely possible that, for instance, $P_1^1$ is not a term in the sequence $\{P^8_n\}$.  But this is fine because, except for the seven terms $P^1_1, \dots, P^7_7$, the sequence $\{P^i_i\}$ is still a subsequence of $\{P^8_n\}$, and as such, we know that $\{\pi_8^{-1} P^i_i\}$ converges weakly. 
Of course, the same argument shows that for every $k$, $\{\pi_k^{-1} P^i_i\}$ converges weakly.  And this is what you need to get weak convergence of $\{P_i^i\}$ as measures on $\mathbb{R}^\infty$.
My reference to "Tychonoff" was to suggest that this is exactly the same argument that one can use to prove that, for instance, $[0,1]^\infty$ is compact in the product topology.  If we view $[0,1]^\infty$ as the set of all functions from $\mathbb{N}$ to $[0,1]$, then given a sequence $\{x_n\}_{n=1}^\infty$ in $[0,1]^\infty$, we can produce a sequence of subsequences $\{x_n^k\}, k=1, 2,\dots$ such that for each $k$, $\{x_n^k(k)\}_n$ converges (using the compactness of $[0,1]^k$ for each $k$).  Now the diagonal subsequence $\{x^i_i\}_{i=1}^\infty$ has the property that for each $k$, $\{x_i^i(k)\}$ converges, which is exactly what it means for $\{x_i^i\}$ to converge in the product topology.
Your last paragraph doesn't quite make sense because in order to conclude that the closure of a sequence of elements from $\Pi$ was compact, we'd have to know that $\Pi$ itself was relatively compact!  So that becomes circular.
