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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. enter image description here

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

enter image description here

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.

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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 $(\xi,\xi')_o = \lim_{n \to \infty} G(n,i_n)$:

  1. Choose $i_1$ so that the row 1 entry $G(1,i_1)$ is within distance $2^{-1}$ of the liminf of the row $1$;
  2. Choose $i_2$ so that the row 2 entry $G(2,i_2)$ is within distance $2^{-1}$ of the liminf of row $2$;
  3. Choose $i_3$ so that the row $3$ entry $G(3,i_3)$ is within distance $2^{-3}$ of the liminf of row $3$;

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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 $(\xi,\xi')_o$.

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  • $\begingroup$ How does one garantee the convergence of the sequence? And even after showing the sequence we have to show the sequences are points of the boundary . $\endgroup$ Jan 12 at 11:19
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    $\begingroup$ I wrote why the sequence converges. You're right that there's more to be done to obtain sequences $(x^n_{i_n})$ and $(x'{}^n_{i_n})$ that still represent the same points $\xi,\xi'$ respectively, but that can be arranged by further diagonalization. I can add that later if you like. $\endgroup$
    – Lee Mosher
    Jan 12 at 13:51
  • $\begingroup$ Thanks , I have figured it out myself . $\endgroup$ Jan 12 at 15:18

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