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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?

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You require countability of I in order to guarantee this. For example, suppose X = I = R and C_i = {i}. – Qiaochu Yuan Nov 29 '10 at 19:25
your statement is dangling: there is no $i$ in the statement before "where $i\in I$"; you only have $i_n$. And for the statement to hold, you would require there to be a countable subcover; this will hold trivially if $I$ itself is countable. Otherwise, no, you cannot. – Arturo Magidin Nov 29 '10 at 19:33
what is a locally finite basis? – Qiaochu Yuan Nov 29 '10 at 20:09
@Qiaochu Yuan: a base (in the standard sense of topology) which is locally finite, i.e a collection of sets is said to be locally finite if for every point in the ambient space , i.e $x \in X$ there exists a neighborhood of $x$ which intersects only finitely many members of the collection. – student Nov 29 '10 at 20:19

We cannot say it. It depends on the specific family you consider. This is possible even if $I$ is uncountable.

For example, say $X$ has size $\aleph_1$, the first uncountable cardinal. Let $I=X$, and suppose $i\in C_i$ and $X\setminus C_i$ is countable for each $i\in X$. A natural example is to let $X=\omega_1$, the first uncountable ordinal, and $C_i=[i,\infty)$ with the usual order of ordinals. Then Only countably many $C_i$ are needed to cover $X$, so we can pick for each $x$ an $i_n$ as you require.

However, if $I=X$ and $C_i=\{i\}$ for each $i$, then of course all $C_i$ are needed.

Note that we can easily put a topology in each case so the $C_i$ are open sets, so that additional condition would not change the answer.

(So, to guarantee that it is always possible to pick such $i_n$, we need $I$ to be countable, or some topological condition if we want the $C_i$ to be open, such as $X$ being Lindelöf, which is a weakening of being compact.)

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just modified the question, can you please have a look? – student Nov 29 '10 at 19:39
@student : I think Qiaochu's answer addresses the modified version. But let me know if something still needs clarifying. – Andrés E. Caicedo Nov 29 '10 at 22:36

You are misunderstanding the author's argument. Suppose $x \in B_1$. By local finiteness, $B_1$ only contains finitely many elements of $X$, say $n$. If $y \in B_1$ is any element other than $x$, then by the $T_1$ assumption $B_1 - \{ y \} = B_2$ is also an open set containing $x$. Continuing in this way we obtain a sequence $B_1 \supset B_2 \supset ... $ of strictly decreasing open sets containing $x$ which stabilizes after at most $n$ steps; this gives an open set $B_n = \{ x \}$, hence the topology is discrete.

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