proof of Thm 15.1

Thm 15.1
For $X(C,d)$ a Metric Space

proof 1 → 2

Suppose $C$ is closed, we want to prove $X$ is complete.
Consider $(x_n)\subseteq C$ as an arbitrary Cauchy sequence in $C$, according to $C$ is closed, which imply $C$ is sequentially Closed by Prop 15.2, we can say

$$\underset{n}{\lim }(x_n)=x^{\prime }⟹x^{\prime }\in (x_n)$$

is true. Hence it’s safe to say $X$ is complete. $\square$

proof 2 → 1

Suppose $X$ is complete, we want to prove $C$ is closed.