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The following is copied verbatim from a book (I. Protasov, Combinatorics of numbers, p. 14):

Suppose that to each point $x$ of a set $X$ a collection $\mathcal{B}(x)$ of subsets of $X$, which are called neighborhoods of $x$, is assigned so that the following conditions are satisfied:

(B1) $x\in U$ for every neighborhood $U \in \mathcal{B}(x)$;
(B2) if $U \subseteq V, U \in \mathcal{B}(x)$, then $V\in \mathcal{B}(x)$;
(B3) if $U_1, \dots, U_n \in \mathcal{B}(x)$, then $U_1 \cap \dots \cap U_n\in \mathcal{B}(x)$;
(B4) if $U\in \mathcal{B}(x)$, then there is a neighborhood $V\in\mathcal{B}(x)$ such that $U\in \mathcal{B}(y)$ for every $y\in V$.

A subset $A\subseteq X$ is defined to be open, if $A$ is a neighborhood of each of its points, i.e. $A\in\mathcal{B}(x)$ for every $x\in A$. Evidently, open sets satisfy the following properties:

(O1) $X, \varnothing$ are open sets;
(O2) if $U_1, \dots, U_n$ are open sets, then $U_1 \cap \dots \cap U_n$ is an open set;
(O3) if $U_\alpha, \alpha\in J$, is a collection of open sets, then $\bigcup\{U_\alpha : \alpha \in J\}$ is an open set.

I don't see why (O1) is true. More generally, I don't see what guarantees that $\mathcal{B}(x) \neq \varnothing$ for any $x \in X$. In fact, if we set $\mathcal{B}(x) = \varnothing$ for all $x \in X$, conditions (B1)-(B4) are all satisfied vacuously. In this case there could be no open sets, so (O1) could not hold.

Am I missing something?

If not, there must be some error in the book, and I'd like to know how to fix it such that what results agrees with the standard way of defining open sets in terms of neighborhoods.

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2 Answers

Your observation appears to be correct, and you don't appear to be missing anything.

Note that simply adding the condition that $\mathcal{B}(x) \neq \emptyset$ for all $x \in X$ (or, equivalently, $X \in \mathcal{B}(x)$ for all $x \in X$) will yield the usual definition of a topology. (The sets in $\mathcal{B}(x)$ will be those sets whose interior contains $x$.)

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up vote 3 down vote accepted

I think I finally got it. All we need to do is replace B3 with

(B3) each collection $\mathcal{B}(x)$ is closed with respect to finite (including empty) intersections;

Then the empty intersection (namely $X$) will belong to every $\mathcal{B}(x)$.

(It could be argued that the original (B3) already implies this.)

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This could work, but one would have to explicitly mention that the intersection of a/the empty family of subsets of $X$ is taken to equal $X$. This is just a convention, as the empty set is the empty family of subsets of any set, be it $\mathbb{N}$ or $\mathbb{R}$, or any other set. –  Arthur Fischer May 28 '13 at 16:32
@ArthurFischer: I don't think it's a mere convention: there's a consistent logic to it; the empty intersection should be the identity with respect to intersection, just like the empty union is the identity with respect to unions. This is a widespread idea in mathematics. E.g. the empty sum (e.g. $\sum_{i=1}^0 x_i$) is $0$ (the identity wrt addition), the empty product (e.g. $\prod_{i=1}^0 x_i$) is $1$ (the identity wrt multiplication), the empty power, namely $x^0$, is (as a special case of the empty product) also $1$, etc., etc. –  kjo May 28 '13 at 17:45
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