April 3, 2014

If X is a set, a *filter* on X is a set \mathcal{F} of subsets of X such that \emptyset \not \in \mathcal{F}; if A,B \in \mathcal{F} then A \cap B \in \mathcal{F}; if A \subseteq X and there is some B \in \mathcal{F} such that B \subseteq A, then A \in \mathcal{F}. For example, if x \in X then the set of all subsets of X that include x is a filter on X.^{1} A *basis* for the filter \mathcal{F} is a subset \mathcal{B} \subseteq \mathcal{F} such that if A \in \mathcal{F} then there is some B \in \mathcal{B} such that B \subseteq A.

If X is a set, a *topology* on X is a set \mathcal{O} of subsets of X such that: \emptyset, X \in \mathcal{O}; if U_\alpha \in \mathcal{O} for all \alpha \in I, then \bigcup_{\alpha \in I} U_\alpha \in \mathcal{O}; if I is finite and U_\alpha \in \mathcal{O} for all \alpha \in I, then \bigcap_{\alpha \in I} U_\alpha \in \mathcal{O}. If N \subseteq X and x \in X, we say that N is a *neighborhood* of x if there is some U \in \mathcal{O} such that x \in U \subseteq N. In particular, an open set is a neighborhood of every element of itself. A *basis* for a topology \mathcal{O} is a subset \mathcal{B} of \mathcal{O} such that if x \in X then there is some B \in \mathcal{B} such that x \in B, and such that if B_1,B_2 \in \mathcal{B} and x \in B_1 \cap B_2, then there is some B_3 \in \mathcal{B} such that x \in B_3 \subseteq B_1 \cap B_2.^{2}

On the one hand, suppose that X is a topological space with topology \mathcal{O}. For each x \in X, let \mathcal{F}_x be the set of neighborhoods of x; we call \mathcal{F}_x the *neighborhood filter* of x. It is straightforward to verify that \mathcal{F}_x is a filter for each x \in X. If N \in \mathcal{F}_x, there is some U \in \mathcal{F}_x that is open, and for each y \in U we have N \in \mathcal{F}_y.

On the other hand, suppose X is a set, for each x \in X there is some filter \mathcal{F}_x, and: if N \in \mathcal{F}_x then x \in N; if N \in \mathcal{F}_x then there is some U \in \mathcal{F}_x such that if y \in U then N \in \mathcal{F}_y. We define \mathcal{O} in the following way: The elements U of \mathcal{O} are those subsets of X such that if x \in U then U \in \mathcal{F}_x. Vacuously, \emptyset \in \mathcal{O}, and it is immediate that X \in \mathcal{O}. If U_\alpha \in \mathcal{O}, \alpha \in I and x \in U=\bigcup_{\alpha \in I} U_\alpha then there is at least one \alpha \in I such that x \in U_\alpha and so U_\alpha \in \mathcal{F}_x. As x \in U_\alpha \subseteq U and \mathcal{F}_x is a filter, we get U \in \mathcal{F}_x. If I is finite and U_\alpha \in I, \alpha \in I, let U=\bigcap_{\alpha \in I} U_\alpha. If x \in U, then for each \alpha \in I, x \in U_\alpha, and hence for each \alpha \in I, U_\alpha \in \mathcal{F}_x. As \mathcal{F}_x is a filter, the intersection of any two elements of it is an element of it, and thus the intersection of finitely many elements of it is an element of it, so U \in \mathcal{F}_x, showing that U \in \mathcal{O}. This shows that \mathcal{O} is a topology. We will show that a set N is a neighborhood of a point x if and only if N \in \mathcal{F}_x.

If N \in \mathcal{F}_x, then let V=\{y \in N: N \in \mathcal{F}_y\}. There is some U_0 \in \mathcal{F}_x such that if y \in U_0 then N \in \mathcal{F}_y. If y \in U_0 then N \in \mathcal{F}_y, which implies that y \in N, and hence U_0 \subseteq V. U_0 \subseteq V and U_0 \in \mathcal{F}_x imply that V \in \mathcal{F}_x, which implies that x \in V. If y \in V then N \in \mathcal{F}_y, and hence there is some U \in \mathcal{F}_y such that if z \in U then N \in \mathcal{F}_z. If z \in U then N \in \mathcal{F}_z, which implies that z \in N, and hence U \subseteq V. U \subseteq V and U \in \mathcal{F}_y imply that V \in \mathcal{F}_y. Thus, if y \in V then V \in \mathcal{F}_y, which means that V is open, x \in V \subseteq N tells us that N is a neighborhood of x.

If a set N is a neighborhood of a point x, then there is some open set U with x \in U \subseteq N. U being open means that if y \in U then U \in \mathcal{F}_y. As x \in U we get U \in \mathcal{F}_x, and as U \subset N we get N \in \mathcal{F}_x. Therefore a set N is a neighborhood of a point x if and only if N \in \mathcal{F}_x.

In conclusion: If X is a topological space and for each x \in X we define \mathcal{F}_x to be the neighborhood filter of x, then these filters satisfy the two conditions that if N \in \mathcal{F}_x then x \in N and that if N \in \mathcal{F}_x there is some U \in \mathcal{F}_x such that if y \in U then N \in \mathcal{F}_y. In the other direction, if X is a set and for each point x \in X there is a filter \mathcal{F}_x and the filters satisfy these two conditions, then there is a topology on X such that these filters are precisely the neighborhood filters of each point.