Confusion regarding the proof to "every Lindelöf metric space is second countable".

Question

The proof given in my book for "every Lindelöf metric space is second countable" is:

Let there exist an open covering $\{B(x,\epsilon)\}, \forall x\in X$, where $X$ is a Lindelöf metric space. There consequently exists a countable subcover $\{B(x_{n},\epsilon)\}$. Hence every point $y\in X$ is a part of some $B(x_{i},\epsilon)$. Let $B(y,d)$ be an arbitrary open set containing $y$, where $d\in\Bbb{R}$. If $d(x_{i},y)+\epsilon\leq d$ along with the condition $\epsilon\geq d(x_{i},y)$ then we have $B(x_{i},\epsilon)\subseteq B(y,d)$ and $y\in B(x_{i},\epsilon)$. Hence, the set of all such $\{B(x_{i},\epsilon)\}$ will form a basis, of which there is a countable number. Hence, the space is second countable.

My doubt is this: For every arbitrary $B(y,d)$ of a point $y\in X$, you will have a separate countable cover $\{B(x_{n},\epsilon)\}$, where $\epsilon$ satisfies

1. $d(x_{i},y)+\epsilon\leq d$

2. $\epsilon\geq d(x_{i},y)$

Note that the value of $\epsilon$ will remain the same for all $\{B(x_{n},\epsilon)\}$.

To form a basis, you will have to take the collection of all such countable subcovers $\{B(x_{n},\epsilon)\}$. For the particular point $y$, there might be an uncountable number of different points $x_{i}$ such that $B(x_{i},\epsilon)\subseteq B(y,d)$, where $d$ varies over $\Bbb{R}$. Why can't the collection of all such sets be uncountable, making the space not second countable?

Answer 1

Part of the argument is simply wrong: the base constructed is not countable, because it uses all $\epsilon>0$ instead of a countable set of $\epsilon>0$ containing arbitrarily small members, like $\{2^{-n}:n\in\Bbb N\}$ or $\left\{\frac1n:n\in\Bbb Z^+\right\}$. The intended argument could be made much more clearly. Let me simply do a more intelligible version, rather than comment specifically on this one.

For each $n\in\Bbb N$, $\{B(x,2^{-n}:x\in X\}$ is an open cover of the Lindelöf space $X$, so it has a countable subcover $\mathscr{B}_n$. Let $\mathscr{B}=\bigcup_{n\in\Bbb N}\mathscr{B}_n$; $\mathscr{B}$ is a countable family of open subsets of $X$, and I claim that it’s a base for the topology of $X$.

To see this, let $U$ be any non-empty open set in $X$, and fix $x\in U$; we must show that there is some $B\in\mathscr{B}$ such that $x\in B\subseteq U$. Since $x\in U$, and $U$ is open, we know that there is an $\epsilon>0$ such that $B(x,\epsilon)\subseteq U$. Choose $n\in\Bbb N$ large enough so that $2^{-n}<\frac{\epsilon}2$. Now $\mathscr{B}_n$ covers $X$, so there is a $B(y,2^{-n})\in\mathscr{B}_n$ such that $x\in B(y,2^{-n})$; I claim that $B(y,2^{-n})\subseteq U$.

Suppose that $z\in B(y,2^{-n})$; then $d(z,y)<2^{-n}<\frac{\epsilon}2$. We also know that $d(x,y)<2^{-n}<\frac{\epsilon}2$, so by the triangle inequality we have $d(x,z)<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$, $z\in B(x,\epsilon)\subseteq U$, and hence $B(y,2^{-n})\subseteq U$, as claimed. It follows that $\mathscr{B}$ is indeed a base for the topology on $X$.

Added: Getting back to your specific questions about the original form of the argument, I think that what you’ve not realized is that for each $\epsilon>0$ there is only one family $\{B(x_{\epsilon,n},\epsilon:n\in\Bbb N\}$ of $\epsilon$ balls being considered; there is not a separate one for each $B(y,d)$. (The double subscript on $x_{\epsilon,n}$ is necessary, because for each $\epsilon>0$ there is potentially a different $x_n$.) We use the Lindelöf property to find these countable covers of $X$ at the beginning; we don’t find new ones for each $B(y,d)$. Part of the problem, I think, is that the argument given does not adequately explain why we can be certain that there is a $B(x_{\epsilon,n},\epsilon)$ such that $d(x_n,y)+\epsilon<d$ and $\epsilon\ge d(x_{\epsilon,n},y)$.