I am starting to studying Functional Analysis through Kreyszig's book. Right now I am studying this proposition:
Theorem (Complete subspace). A subspace $M$ of a complete metric space $X$ is itself complete if and only if the set $M$ is closed in $X$.
I am having problems understanding the part of the proof which proves $M$ is complete. The argument below is exactly as written in the book.
Proof:
Conversely, let $M$ be closed and $\left(x_{n}\right)$ Cauchy in $M$. Then $x_{n} \longrightarrow x \in X$, which implies $x \in \bar{M}$ by 1.4-6(a), and $x \in M$ since $M=\bar{M}$ by assumption. Hence the arbitrary Cauchy sequence $\left(x_{n}\right)$ converges in $M$, which proves completeness of $M$.
What I don't get is why exactly $x_{n} \longrightarrow x \in X$. Why does $x_{n}$ converges and why $x \in X$?
What I am getting stuck is that $x$ should be such that $x \in M$, since $M$ is closed. Also, why $x_{n}$ must be convergent?
Important definitions and results he uses in the proof
1.4-6 Theorem (Closure, closed set).
Let $M$ be a nonempty subset of a metric space $(X, d)$ and $\bar{M}$ its closure as defined in the previous section. Then:
(a) $x \in \bar{M}$ if and only if there is a sequence $\left(x_{n}\right)$ in $M$ such that $x_{n} \longrightarrow x$.
(b) $M$ is closed if and only if the situation $x_{n} \in M, x_{n} \longrightarrow x$ implies that $x \in M$.
Thanks in advance, Lucas
Obs: Also, I noticed that I would not know how to explain why $x_{n} \longrightarrow x \in X$, implies $x \in \bar{M}$
To show that $M$ is complete, you need to produce an $x \in M$ so that $x_n\longrightarrow x$. While your hypothesis is that $X$ is complete, the sequence $x_n$ being Cauchy in $M$ is also Cauchy in $X$, hence there is a $y\in X$ so that $x_n\longrightarrow y$. So if you can show that the element $y$ belongs to $M$ then you can take the desired $x$ to be $y$ which completes the proof.