Suppose I is an index set (not necessarily countable). Say we have a family of subsets $\{C_{i}: i \in I\}$ and suppose that this family is a cover of a space $X$, that is, for each $x$ in $X$ we have $x \in C_{j}$ for some $j \in I$.
Can we say the following? for each $x \in X$ there exists some natural number $n$ such that $x \in C_{i_{n}}$ where $i \in I$ ? or do we require countability of $I$ in order to guarantee this?
Let me state the question more precisely. I'm trying to understand the proof of the following fact: Let $X$ be a $T_{1}$ space with locally-finite basis. Then $X$ must be discrete. Proof of the author: say $B$ is a locally finite basis. Now pick $x \in B$ then by definition there is some element $B_{1}$ of $B$ such that $x \in B_{1}$. Using recursion and the assumption of $T_{1}$ we can find $B_{n}$ in B such that $x$ in all $B_{n}$ and for all n we have the inclusion $B_{n+1} \subset B_{n}$. Thus $B$ cannot be locally finite, contradiction.
Isn't the above proof assuming I is countable? Can you please ellaborate more?