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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}.2010-11-29
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    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.2010-11-29
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    what is a locally finite basis?2010-11-29
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    @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.2010-11-29

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