Construction of an uncountable set using well ordering principle. I was trying to prove that a metric space in which every uncountable set has a limit point is separable.Now I was trying to show that it is Lindelof.I proceed like this,If not lindelof,then there is an uncountable set with no limit points.
I want to do it in a similar way we prove: If a metric space is not compact,then there is an infinite set with no limit points. 
What do we do in this case.Since there is an open cover with no finite subcover.So,there is a countable open cover with no finite subcover(as countable compactness $\iff $ compactness in metric spaces).Now we start with a countable open cover $\{U_n\}$ which has no finite subcover and construct an infinite set as follows $x_n \in X-\bigcup\limits_{k=1}^{n} U_k$.Then $\{x_n:n\in \mathbb N\}$ is an infinite set with no limit points.
Now if I want to do the proof of the original problem in a similar way.Then first I have to consider that we have an open cover $\{G_\alpha\}$ with no countable subcover.Then we must choose $x_\beta \in X-\bigcup\limits_{countable} G_\alpha$,i.e. the union runs over a countable subcollection of $G_\alpha$.But we must go sequentially as in the previous problem.So,we define an well order on $I$,then index set and go one by one.Now my question is we can go one by one in case of the previous problem because in each stage it was a finite union.But my question is can we transcend finiteness and go to countably many times i.e. choosing $1$ st element,then $2$ nd element,and so on and we keep doing it till the union remains countable.Is this possible logically?Please can someone tell me if I can transcend finiteness and extend the same idea for $\aleph_0$ cardinality and generalize the same process.
 A: Yes, you can construct the points $x_\beta$: let $\le$ be a well-order on $I$, and for each $\beta\in I$ choose $x_\beta\in X\setminus\bigcup_{\alpha<\beta}G_\alpha$. However, this won’t be enough to prove the result, because in general the set $\{x_\beta:\beta\in I\}$ can have limit points; you’ll need to use the fact that $X$ is a metric space.
I think that it’s probably at least as easy to prove separability directly. For each $n\in\Bbb N$ let $A_n$ be a maximal subset of $X$ with the property that $d(x,y)\ge 2^{-n}$ whenever $x,y\in A_n$ and $x\ne y$. (You can use Zorn’s Lemma to get each $A_n$, or you can build each $A_n$ by transfinite recursion much as you were constructing your points $x_\beta$.) Let $A=\bigcup_{n\in\Bbb N}A_n$; it’s not hard to show that $A$ is dense in $X$, so we need only show that $A$ is countable. For this it suffices to show that each $A_n$ is countable.
Fix $n\in\Bbb N$. The open balls $B(x,2^{-(n+1)})$ for $x\in A_n$ are pairwise disjoint, so $A_n$ is a discrete set, though it need not be closed. It is, however, a relatively open subset of $\operatorname{cl}A_n$, since we have
$$\bigcup_{x\in A_n}B\left(x,2^{-(n+1)}\right)\cap\operatorname{cl}A_n=A_n\;.$$
Every open set in a metric space is an $F_\sigma$-set, i.e, the union of countably many closed sets, so there is a family $\{F_k:k\in\Bbb N\}$ of closed subsets of $X$ such that
$$\bigcup_{x\in A_n}B\left(x,2^{-(n+1)}\right)=\bigcup_{k\in\Bbb N}F_k\;.$$
For each $k\in\Bbb N$ let $D_k=F_k\cap A_n$; $F_k\cap A_n=F_k\cap\operatorname{cl}A_n$, so $D_k$ is closed, and as a subset of $A_n$ it is also discrete, so by hypothesis it is countable. Finally, $A_n=\bigcup_{k\in\Bbb N}D_k$, so $A_n$ is the union of countably many countable sets and is therefore itself countable.
A: Let $(X,d)$ be a metric space.


*

*If $(X,d)$ is $not$ separable then $X$ has an uncountable subset with no limit points.
Proof: For each $a\in \omega_1$ let $g(a)\in X\setminus \overline {\{g(b):b\in a\}}$.... Note that $g(a)$ exists because $\{g(a):b\in a\}$ is countable and hence not dense in $X.$ Note that $g:\omega_1\to X$ is injective. 
For each $a\in \omega_1$ let $Q(a)\in \Bbb Q^+$ such that $\forall b\in a\,(d(g(a),g(b))>Q(a)\,).$  Now $Q:\omega_1\to \Bbb Q^+$ but $\Bbb Q^+$ is countable, so take some $q\in \Bbb Q^+$ such that the set $S= Q^{-1}\{q\}$ is uncountable. Now $Y=\{g(a):a\in S\}$ is also uncountable because $g$ is injective. And if $a,b\in S$ with $a\ne b$ then $d(g(a),g(b))>q.$
In any metric space $(X,d),$ if $Y\subset X$ and $\inf \{d(y,y'): y,y'\in Y\land y\ne y'\}>0$ then $Y$ has no limit points.


*If $E$ is a dense subset of $X$ then $\{B_d(e,q):q\in \Bbb Q^+\}$ is a base (basis) for $(X,d).$
Proof: If $x\in U\subset X$ where $U$ is open, take $r>0$ such that $B_d(x,r)\subset U.$ Take $q\in \Bbb Q\cap (0,r/2)$ and take $e\in E$ with $d(x,e)<q.$ Then $x\in B_d(e,q)\subset B_d(x,r)\subset U.$ 
Corollary 2a. If $(X,d)$ is separable then it is 2nd-countable.
Any 2nd-countable space is Lindelof. Hence: 
Corollary 2b. If $(X,d)$ is separable then it is Lindelof.


*Suppose $(X,d)$ is not compact.
(i). If it is not separable then by 1. there is an uncountable $Y\subset X$ with no limit points.
(ii). If it is separable, let $C$ be an open cover of $X$ with no finite sub-cover. By Corollary 2b. let $\{C_n:n\in \Bbb N\}$ be a sub-cover of $C.$ For $n\in \Bbb N$ let $H(n)\in X\setminus \cup_{j<n}C_j,$ and let $Y=\{H(n): n\in \Bbb N\}.$ Note that $H:\Bbb N\to X$ is injective, so $Y$ is infinite.
For any $x\in X$ there exists $n_x\in \Bbb N$ such that $x\in C_{n_x}.$ Now $C_{n_x}$ is a nbhd of $x$ which contains only finitely many members of $Y,$ because $n>n_x\implies H(n)\not \in \cup_{j<n}C_j\supset C_{n_x}.$ So $x$ is not a limit point of $Y.$ 
