While reading Michael Field's $Essential\,Real\,Analysis$ I hit a wall in trying to understand an essential proof (pages 335-336). I write here only the part that I don't understand. We have sets
$Z=K_1\cup K_2\cup\,...\,\cup K_{N-1}\cup\overline{B(K_N,1)}\,$ and
$L=\bigcap_{n\geq 1}\overline{\cup_{m\geq n}}K_m$.
$K_n\subset\overline{B(K_N,1)}$ with all $n\geq N$ and $B(K_N,1)$ is an open ball with radius of $1$, centered at $K_N$. All the sets $K_n$ are sequentially compact so $Z$ and $L$ are also sequentially compact and $L$ is not empty. All the sets $K_n$ are in space $\mathbb{R}^n$ so sequential compactness is equivalent with compactness. $(K_n)$ is a Cauchy sequence.
There's a sequence $(x_n)$ in $Z$ such that $x_n\in K_n$. $Z$ is sequentially compact so $(x_n)$ has a subsequence converging to a point $z\in Z$.
I understand the proof except for the part where Field writes that because $z\in Z$, then $z\in L$. I see that $z\in L$ implies $z\in Z$ so Field is essentially saying that $Z=L$. Could someone point out where I got lost? Or how does Field's claim $z\in Z\Rightarrow z\in L$ hold?
First, as per Angina Seng's comment, $B(x,1), x \in \Bbb R^n$ is the open ball of radius $1$ about the point $x$. But $K_N$ is a set, not a point. Surely Field would not have made the horrendously bad choice of using $K_N$ to represent both a point and a set, so presumably $B(K_N, 1)$ would be $$B(K_N, 1) = \{y \in \Bbb R^n \mid \exists x \in K_N, \|x - y\| < 1\} = \bigcup_{x\in K_N} B(x,1)$$ the set of all points within a distance of $1$ of the set $K_N$, which is not in general going to be a ball.
From the information reproduced here, $Z$ need not be a subset of $L$, so it is not strictly true that $z \in L$ because $z \in Z$. For example, in $\Bbb R, K_n = \left[0, 2-\frac 1n\right]$ meets all the conditions, but $Z = \left[-1, 3 - \frac 1N\right]$ while $L = [0,2]$.
But $z \in L$ anyway. If $\{z_k\}$ is the sequence converging to $z$, then by how $z$ was constructed, there is an increasing sequence $n_k$ such that $z_k \in K_{n_k}$ for each $k$. That means for each $m$, eventually $n_k$ will rise above $m$, so from then on $z_k \in \bigcup_{n\ge m} K_n$, and therefore $z \in \overline{\bigcup_{n\ge m} K_n}$. Hence $z$ is also in the intersection $L$.