Gromov product on the boundary I was reading the book Elements of Asymptotic Geometry by Sergei Buyalo and Viktor Schroeder. There in the second chapter Gromov boundary is defined as the equivalence classes of sequences (in the usual way). And then they try to extend the Gromov product to the boundary as given in the image below.

Now I am having problem proof of Lemma 2.2.2.(1).

I couldn't understand what they mean by 'standard diagonal procedure' in the second line of the proof.I was able to follow rest of the argument except how one can produce a pair of sequences with the given property
Any help shall be highly appreciated.
 A: The definition of $(\xi,\xi')_o$ is an infimum taken over all pairs $x$, $x'$, such that $x=(x_i)$ is a sequence representing $\xi$, and $x'=(x'_i)$ is a sequence representing $\xi'$. The quantity being infimized is $\liminf_{i \to \infty} (x_i,x'_i)_o$.
For any nonempty set of real numbers that has an infimum, there is a sequence in that set such that the limit of the sequence equals the infimum of the set.
Therefore, there exists a sequence of pairs $x^n$, $x'^n$ such that each $x^n=(x^n_i)$ is a sequence representing $\xi$, each $x'{}^n = (x'{}^n_i)$ is a sequence representing $\xi'$, and
$$(\xi,\xi')_o = \lim_{n \to \infty} \left( \liminf_{i \to \infty} \underbrace{(x^n_i,x'{}^n_i)_o}_{G(n,i)} \right)
$$
Here's where the "diagonalization" procedure kicks in. Imagine the doubly indexed sequence of real numbers $G(n,i)=(x^n_i,x'{}^n_i)_o$ arranged in a table, with $n$ indexing the rows and $i$ indexing the columns. Knowing the above equation, you carefully choose a sequence of entries $G(n,i_n)$ out of the table so that $(\xi,\xi')_o = \lim_{n \to \infty} G(n,i_n)$:

*

*Choose $i_1$ so that the row 1 entry $G(1,i_1)$ is within distance $2^{-1}$ of the liminf of the row $1$;

*Choose $i_2$ so that the row 2 entry $G(2,i_2)$ is within distance $2^{-1}$ of the liminf of row $2$;

*Choose $i_3$ so that the row $3$ entry $G(3,i_3)$ is within distance $2^{-3}$ of the liminf of row $3$;

.
.
.
The sequence $G(n,i_n)$ converges to $(\xi,\xi')_0$ by the sandwich lemma:
$$\liminf_{i \to \infty} (x^n_i,x'{}^n_i) - 2^{-n} \le G(n,i_n) \le \liminf_{i \to \infty} (x^n_i,x'{}^n_i) + 2^{-n}
$$
and since $\liminf_{i \to \infty} (x^n_i,x'{}^n_i)$ converges to $(\xi,\xi')_o$ and $2^{-n}$ converges to $0$, it follows that  $G(n,i_n)$ converges to $(\xi,\xi')_o$.
