Compact metric space: distance from any given point.

Question

$(N, p)$ is compact metric space. Pick $x_1 \in N$. Define $x \mapsto p(x, x_1)$, then $\sup \limits_{x \in N} p(x, x_1) < \infty$. Why?

Argue by contradiction: Suppose $\sup \limits_{x \in N} p(x, x_1) \nless \infty$, then:

...............

$\exists$ a sequence $\left(z_n \in N \right)_{n \in \mathbb{N}}$ such that $p(z_n, x_1) \geq n$ for all $n$.

...............

But $N$ is compact, so $\exists$ a subsequence $z_{n_k}$ such that $\lim \limits_{k \to \infty} z_{n_k} = z \in N$, i.e. $p(z_{n_k}, z) \to 0$. But then $p(z_{n_k}, x_1) \to p(z, x_1) < \infty$. Hence, a contradiction.

I just don't understand the stuff surrounded by .......... and ............ .

Answer 1

To clarify what's going on in the line you mentioned, you could expand it like this:

Suppose $\sup_{x\in N} p(x, x_1) \not < \infty$. This means that the set of real numbers $S=\{p(x, x_1) : x\in N\}$ is not bounded above; if it were, then the supremum, or least upper bound, would be finite. Then since $1$ is not an upper bound for $S$, there must exist an element $x\in N$ such that $p(x,x_1)>1$. Call that element $z_1$, so $p(z_1, x_1) > 1$. Next, since $2$ is not an upper bound for $S$, there must exist an $x\in N$ such that $p(x, x_1) > 2$; call it $z_2$. In fact, for each natural number $n$, we know that $n$ cannot be an upper bound for $S$, so there must exist some element $x\in N$ such that $p(x, x_1)> n$, and we can let $z_n$ be such an element. Thus there exists a sequence $z_1, z_2, z_3, \dots$ of elements of $N$, where each $z_n$ has the property that $p(z_n, x_1) > n$.

Answer 2

If $K$ is a compact metric space and $f\colon K\longrightarrow\mathbb R$ is a continuous map, then $f(K)$ is bounded (since it's a compact subset of $\mathbb R$). Since $f(K)$ is a bounded subset of $\mathbb R$, $\sup f(K)<\infty$. Now, apply this to your map $x\mapsto p(x,x_1)$.