Lemma 2

There exists a sequence $(x_m)_{m \geq 0}$ consisting of distinct elements of A, which is bounded in E.

We shall construct a sequence $(x_m)_{m \geq 0}$ of points of A by induction; and a decreasing sequence $(C_m)_{m \geq 0}$ of subsets of A satisfying the following conditions : a) None of the sets $C_m$ is countable. b) For every $n \geq 0$, the set $\mathrm{pr}_k(C_m)$ is bounded in $F_k$ for $0 \leq k \leq m$. c) For every $m \geq 0$, we have $x_m \in C_m - C_{m+1}$.

We put $C_0 = A$. Suppose the sets $C_m$ for $0 \leq m \leq n$ have been defined, so as to satisfy a) and b) for $0 \leq m \leq n$, and also the points $x_m$ in $C_m - C_{m+1}$ for $0 \leq m < n$. For every integer $r \geq 1$, let $C_{n,r}$ be the set of all $x \in C_n$ such that

Since $C_n$ is not countable, there exists an integer $r \geq 1$ such that $C_{n,r}$ is not countable. We choose a point $x_n$ in $C_{n,r}$ and put $C_{n+1} = C_{n,r} - \{ x_n \}$. Evidently $C_{n+1} \subset C_n$ and $x_n \in C_n - C_{n+1}$, the set $C_{n+1}$ is not countable and $\mathrm{pr}_k(C_{n+1})$ is bounded in $F_n$ for $0 \leq k \leq n + 1$.

We have $x_m \in C_m$, and so $x_m \in C_n$ where $m \geq n$. The projection of the sequence $(x_m)_{m \geq 0}$ on $F_n$ is therefore bounded for all $n \geq 0$; in other words, the sequence $(x_m)_{m \geq 0}$ is bounded in E, and this establishes lemma 2.

Q.E.D.

With the notations of lemma 2, the bounded sequence $(x_m)_{m \geq 0}$ has a limit point $y$ in E. Therefore the sequence $(\mathrm{pr}_0(x_m))_{m \geq 0}$ has a limit point $\mathrm{pr}_0(y)$ in $F_0$, but this contradicts the construction of $A_0$.