Topological spaces and neighborhood filters Jordan Bell [email protected] Department of Mathematics, University of Toronto April 3, 2014 If X is a set, a filter on X is a set F of subsets of X such that ; 62 F; if A; B 2 F then A \ B 2 F; if A ⊆ X and there is some B 2 F such that B ⊆ A, then A 2 F. For example, if x 2 X then the set of all subsets of X that include x is a filter on X.1 A basis for the filter F is a subset B ⊆ F such that if A 2 F then there is some B 2 B such that B ⊆ A. If X is a set, a topology on X is a set O of subsets of X such that: ;;X 2 O; S if Uα 2 O for all α 2 I, then Uα 2 O; if I is finite and Uα 2 O for all α 2 I, T α2I then α2I Uα 2 O. If N ⊆ X and x 2 X, we say that N is a neighborhood of x if there is some U 2 O such that x 2 U ⊆ N. In particular, an open set is a neighborhood of every element of itself. A basis for a topology O is a subset B of O such that if x 2 X then there is some B 2 B such that x 2 B, and such that if B1;B2 2 B and x 2 B1 \ B2, then there is some B3 2 B such that 2 x 2 B3 ⊆ B1 \ B2. On the one hand, suppose that X is a topological space with topology O. For each x 2 X, let Fx be the set of neighborhoods of x; we call Fx the neighborhood filter of x. It is straightforward to verify that Fx is a filter for each x 2 X. If N 2 Fx, there is some U 2 Fx that is open, and for each y 2 U we have N 2 Fy. On the other hand, suppose X is a set, for each x 2 X there is some filter Fx, and: if N 2 Fx then x 2 N; if N 2 Fx then there is some U 2 Fx such that if y 2 U then N 2 Fy. We define O in the following way: The elements U of O are those subsets of X such that if x 2 U then U 2 Fx. Vacuously, ; 2 O, and S it is immediate that X 2 O. If Uα 2 O, α 2 I and x 2 U = α2I Uα then there is at least one α 2 I such that x 2 Uα and so Uα 2 Fx. As x 2 Uα ⊆ U and Fx T is a filter, we get U 2 Fx. If I is finite and Uα 2 I, α 2 I, let U = α2I Uα. If x 2 U, then for each α 2 I, x 2 Uα, and hence for each α 2 I, Uα 2 Fx. As Fx 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 2 Fx, showing that U 2 O. This shows that O is a topology. We will show that a set N is a neighborhood of a point x if and only if N 2 Fx. If N 2 Fx, then let V = fy 2 N : N 2 Fyg. There is some U0 2 Fx such 1cf. Fran¸coisTr`eves, Topological Vector Spaces, Distributions and Kernels, p. 6. 2cf. James R. Munkres, Topology, second ed., p. 78. 1 that if y 2 U0 then N 2 Fy. If y 2 U0 then N 2 Fy, which implies that y 2 N, and hence U0 ⊆ V . U0 ⊆ V and U0 2 Fx imply that V 2 Fx, which implies that x 2 V . If y 2 V then N 2 Fy, and hence there is some U 2 Fy such that if z 2 U then N 2 Fz. If z 2 U then N 2 Fz, which implies that z 2 N, and hence U ⊆ V . U ⊆ V and U 2 Fy imply that V 2 Fy. Thus, if y 2 V then V 2 Fy, which means that V is open, x 2 V ⊆ 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 2 U ⊆ N. U being open means that if y 2 U then U 2 Fy. As x 2 U we get U 2 Fx, and as U ⊂ N we get N 2 Fx. Therefore a set N is a neighborhood of a point x if and only if N 2 Fx. In conclusion: If X is a topological space and for each x 2 X we define Fx to be the neighborhood filter of x, then these filters satisfy the two conditions that if N 2 Fx then x 2 N and that if N 2 Fx there is some U 2 Fx such that if y 2 U then N 2 Fy. In the other direction, if X is a set and for each point x 2 X there is a filter Fx 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. 2.