Understanding completness of a metric space I have a question about completeness of a metric space.
The lecture notes that I am reading on my own says the following.
PROPOSITION
 1. Let X be a complete metric space. Let Y be a closed subset of X. Then Y
is complete (as a metric space in its own right).
 2. Let X be a metric space. Let Y be a subset of X which is complete (as a
metric space in its own right). Then Y is closed in X.
Then the author says that
"This Proposition gives the correct impression that completeness is a kind of
global closedness property."
What does he mean by global closedness property?
I see that closedness and compactness are related, but I do not see how the Proposition is related to "global closedness property". 
 A: You should be comfortable with the notion that whether $X$ is closed as a subset of $Y$ depends on $Y$. For example if $X = \mathbb{Q}$ then it is closed as a subset of $Y = \mathbb{Q}$, but not as a subset of $Y = \mathbb{R}$. Completeness, on the other hand, is a property which a metric space has that is invariant under embeddings into larger metric spaces. Similar statements hold for compactness and connectedness (although these are purely topological, while completeness depends on metric)
Propositions (i) and (ii) from your post establish a close relationship between completeness and closedness, I shall attempt to take this a bit farther. Suppose $X \subset Y$ are metric spaces. $X$ is closed if every sequence $\{x_n\}$ of elements of $x$, that converges in $Y$ has a limit in $x$. Such a sequence will certainly be Cauchy, so if $X$ is complete, it most be closed in $Y$ for any $Y$ it isometrically embeds into. This is Prop. (ii). Proposition (i) will be explained later. To make the analogy a bit clearer: to check if something is closed we check if sequences that converge in some bigger space converge to something in $X$. To check if something is complete we check if every sequence that ever could converge in any bigger space converges in $X$. A little later when i discuss the completion of a space we shall see why Cauchy sequences are precisely those sequences which converge in some bigger space. 
The idea of completeness as a form of global closedness can be formulated in the following way. We call a metric space $X$ "always closed" if it satisfies Proposition (ii). If a closed subspace is one that contains all it's limits in some particular larger space, an "always closed" space is one that will always contain all it's possible limits. We now claim that $X$ is "always closed" if and only if it is complete. We already know complete spaces are "always closed", so now we simply need to show that an "always closed" space is complete.
The meat of this proof lies in the idea of the completion of a metric space. The claim is that every metric space $X$ can be realized as a dense subspace of a complete space $X'$. To construct $X'$ we do something similar to one construction of the real numbers. Let $X'$ be the set of all Cauchy sequences in $X$, mod an equivalence relation $\{x_n\} \sim \{y_n\}$ if $\lim\limits_{n\to\infty} d(x_n,y_n) =0$. We give $X'$ a metric $d'(\{x_n\},\{y_n\})=\lim\limits_{n\to\infty} d(x_n,y_n)$. Then one simply has to prove $X'$ is complete, and that if $i:X \to X'$ is the map $x \mapsto \{x,x,x,...\}$ then this identifies $X$ with a dense subset of $X'$.
Once this is done, simply observe that if $X$ is "always closed" then $X = X'$ is its own completion, as $X$ is a dense, closed subspace of $X'$. Thus $X$ is complete.
As a final remark: Why do we want Prop. (i). Well, basically, if we want to with any authority say that "complete sets have a global closedness property" then closed subsets had better be globally closed. After all, a sequence in a subset is a sequence in the whole space. 
