Explanations for a "diagonal process" construction of a sequence I am reading Peter Lax's "Functional analysis".
Let $y_n$ be a bounded sequence of vectors in a Banach reflexive space, $X, Y$ their closed linear span. Take a countable set  $m_j$ of applications belonging to the dual $Y^*$ of $Y$ ($Y^*$ is separable).
The author then says that we can apply the "classical diagonal process" to obtain a subsequence $z_n$ of vectors such that $\lim_n m_j(z_n)$ exists for every $j$. 
I do not understand how he builds this subsequence.
 A: The key ingredient is the observation that if a sequence is convergent, then any subsequence is also convergent.


*

*First, you construct recursively a sequence of subsequences of $y_n$ as follows.

*To begin, put $z_{n,0}:=y_n$.

*Given $(z_{n,j})_n$, you can find (by compactness) a subsequence $(z_{n,j+1})_n$ such that $\lim_{n}m_j(z_{n,j+1})$ exists.

*Note that then all $\lim_n m_{j'}(z_{n,j+1})$ exist, for $j'\leq j$ (because in this case $(z_{n,j+1})_n$ is a subsequence of $(z_{n,j'+1})_n$).

*Finally, put $z_n:=z_{n,n}$ (i.e. $z_n$ is the diagonal of $z_{n,j}$, hence the name). To see that $\lim_n m_j(z_n)$ exists, note that except for finitely many elements, $z_n$ is a subsequence of $z_{n,j}$.


You don't need $X$ to be reflexive, or even a Banach space for this argument. All that really matters is that all the functions $m_j$ map the sequence $y_n$ into compact metric spaces (so that we can use compactness in step 3.) -- the topology on $X$ (or its linear structure) is completely irrelevant beyond this, guaranteed by the fact that bounded functionals by definition map bounded sequences to bounded subsets of the real line.
