Definition: A subbasis S for a topology on X is a collection of subsets of X whose union equals X.

Lemma: If S is a subbasis for a topology on X, then B = {∩_{i=1}^{n}S_{i} | S_{i} ∈ S} is a basis for a topology.

Note that B is a collection of subsets of X (S_{i} ⊂ X implies ∩_{i=1}^{n}S_{i} ⊂ X).

(1) Show for each x ∈ X, there is at least one basis element B containing x.

Proof: Take x ∈ X. Find B ∈ B such that x ∈ B.

We will first show that S ⊂ B: Suppose that S_{1} ∈ S. Then S_{1} = ∩_{n=1}^{1}S_{1}. Thus S_{1} is a finite intersection of elements of S since 1 is a finite number. Hence S_{1} ∈ B.

Since S is a collection of subsets of X whose union equals X, there exists an S ∈ S ⊂ B such that x ∈ S. Hence, there is at least one basis element B containing x.

(2) Show that if x ∈ B_{1} ∩ B_{2} where B_{1}, B_{2} ∈ B, then there exists B_{3} ∈ B such that

x ∈ B_{3} ⊂ B_{1} ∩ B_{2}.

The topology T generated by a basis B is defined as follows: U is open if and only if for all x ∈ U, there exists B ∈ B such that x ∈ B ⊂ U.

Negation: U is NOT open in T iff the following is false: for all x ∈ U, there exists B ∈ B such that x ∈ B ⊂ U.

U is NOT open in T iff there exists an x ∈ U, such that the following is false: there exists B ∈ B such that x ∈ B ⊂ U.

U is NOT open in T iff there exists an x ∈ U, such that for all B ∈ B the following is false: x ∈ B ⊂ U.

U is NOT open in T iff there exists an x ∈ U, such that for all B ∈ B the following is false: x ∈ B and B ⊂ U.

U is NOT open in T iff there exists an x ∈ U, such that for all B ∈ B either x ∉ B OR B ⊄ U.

U is NOT open in T iff there exists an x ∈ U, such that for all B ∈ B, x ∈ B implies B ⊄ U.