Basis for a Topology
If X is a set, a basis on X is a collection \mathcal{B} of subsets B \subset X such that
For all x \in X, there exists some basis element B containing X.
If x belongs to two separate basis elements, B_1,B_2, then there exist some B_3 such that x is in B_3 and {B_3 \subset B_1 \cap B_2}.
We can use a basis on a set to generate a topology on that set.
Let X be a set and \mathcal{B} a basis on X. Let \mathcal{T} be the collection of sets U \subset X such that for each x \in U. there exists B \in \mathcal{B} such x \in B and B \subset U.
Then the collection \mathcal{T} is a topology, and is called the topology generated by \mathcal{B}.
Despite calling \mathcal{T} a topology, we have not yet proven that fact.
Given a set X and a basis \mathcal{B} on X, the collection generated by \mathcal{B} is a topology.
We have to show that the collection \mathcal{T} generated by \mathcal{B} is a topology, meaning we have to show it abides the three properties of a topology:
- The empty set \varnothing and the set itself X are in \mathcal{T}.
If U is the empty set, then U is open vacously. If U is X, then for each x \in X, there exists some basis element containing x and a subset of X by definition.
- Any union of elements in \mathcal{T} is in \mathcal{T}.
We now must show than any indexed family \{U_\alpha\}_{\alpha \in J} of elements in \mathcal{T} has a union such that
U = \bigcup_{\alpha \in J}U_{\alpha}
is an element of \mathcal{T}. Given any x \in U, there is some index \alpha such that x \in U_\alpha. Since U_\alpha is open, there is a basis element B such that x \in B and B \subset U_a \subset U. Therefore U is an element of \mathcal{T}.
- Any finite intersection of elements in \mathcal{T} is in \mathcal{T}.
Now, given any U_1,U_2 we must show U_1 \cap U_2 is an element of \mathcal{T}. Given x \in U_1 \cap U_2, there exist B_1 \subset U_1 and B_2 \subset U_2 both containing x. By the definition of a basis, there exist some B_3 \subset B_1 \cap B_2 containing x. Then x \in B_3 and B_3 \subset U_1 \cap U_2, and therefore U_1 \cap U_2 is an element of \mathcal{T}.
By induction, one can clearly see that this holds for any finite amount of intersections.
There is another way to construct a topology from a given basis.
Let X be a set and \mathcal{B} be a basis for a topology \mathcal{T} on X. Then \mathcal{T} equals the collection of all unions of elements of \mathcal{B}.
Any subset B of \mathcal{B} is an element of \mathcal{T}. Because \mathcal{T} is a topology, any arbitrary union of elements in \mathcal{B} are in \mathcal{T}.
We now must show that each element of \mathcal{T} is a union of basis elements. Given any U \in \mathcal{T}, for each element x \in U, there exists B_x \in \mathcal{B} such that x \in B_x \subset U by the definition of a basis. Then U = \bigcup_{x \in U}B_x, so U is a union of elements of \mathcal{B}.
Now we have two ways to create a topology given a basis. Now we examine creating a basis from a topology.
Let X be a set and \mathcal{T} be a topology on X. Suppose that \mathcal{C} is a collection of open sets of X such that for each U in \mathcal{T} and each x \in U, there exists some C \in \mathcal{C} such that {x \in C \subset U}. Then \mathcal{C} is a basis for \mathcal{T}.
We must first show \mathcal{C} is a basis.
- For each x \in X, there is at least one basis element B containing x.
We know X is an open set in \mathcal{T} by the definition of a topology on a set. Thus, for each x \in X, by our hypothesis there exists some C \in \mathcal{C} such that {x \in C}.
- If x belongs to two separate basis elements, B_1,B_2, then there exist some B_3 such that x is in B_3 and {B_3 \subset B_1 \cap B_2}.
Let x be an element of C_1 \cap C_2 where C_1,C_2 are basis elements of \mathcal{C}. Since C_1 \cap C_2 is an open set in \mathcal{T}, ny our hypothesis there exists C_3 such that x \in C_3 \subset C_1 \cap C_2.
Now we must show that the topology generated by \mathcal{C}, denoted \mathcal{T}^\prime, is the same as the topology \mathcal{T}. We do this by subset inclusion. This means we must show
- \mathcal{T} \subset \mathcal{T}^\prime
For each U \in \mathcal{T}, and each x \in U, there exists some C \in \mathcal{C} by hypothesis such that x \in C \subset U. Then it follows that U \in \mathcal{T}^\prime.
- \mathcal{T}^\prime \subset \mathcal{T}
For each W \in \mathcal{T}^\prime, we have that W is a union of elements of \mathcal{C}. Since every element in \mathcal{C} is also an element of \mathcal{T}, it follows that W \in \mathcal{T}.
Thus we have that \mathcal{T} = \mathcal{T}^\prime, and \mathcal{C} forms a basis for \mathcal{T}.
When topologies are given by bases, we use the following lemma to determine whether one topology is finer than another
Let \mathcal{B} and \mathcal{B}^\prime be bases for the topologies \mathcal{T} and \mathcal{T}^\prime respectively, on a set X. Then the following are equivalent.
(1) \mathcal{T}^\prime is finer than \mathcal{T}.
(2) For each x \in X and each basis element B \in \mathcal{B}, there exist B^\prime \in \mathcal{B}^\prime such that x \in B^\prime \subset B.
To show the two are equivalent, we must show that (1) \implies (2), and that (2) \implies (1).
- (2) \implies (1)
Given U \in \mathcal{T}, we want to show that U \in \mathcal{T}^\prime. By hypothesis we have that for all x \in U, there exists some B \in \mathcal{B} such that x \in B \subset U. By condition (2) we have that there exists some B^\prime \in \mathcal{B}^\prime such that x \in B^\prime \subset B \subset U. Therefore U \in \mathcal{T}^\prime.
- (1) \implies (2)
Let x be an element of X and B \in \mathcal{B} such that x \in B. We have by (1) that \mathcal{T} \subset \mathcal{T}^\prime, and thus B \in \mathcal{T}^\prime. Then since \mathcal{T}^\prime is generated by \mathcal{B}^\prime, there exists some B^\prime \in \mathcal{B}^\prime such that x \in B^\prime \subset B.
And therefore the two statements are equivalent.
We can now use basies to generate some interesting topologies on the real line.
If \mathcal{B} is a subset of \mathbb{R} consisting of all open intervals
(a,b) = \{x \in a < x < b\}
the topology generated by \mathcal{B} is called the standard topology on the real line.
We may also create a topology on some set by partitioning it, and then taking finite intersections and arbitrary unions of that set.
A subbasis \mathcal{S} for a topology on X is a collection of subsets of X whose union equals X.
The topology generated by a subbasis \mathcal{S} is the collection \mathcal{T} of all unions of finite intersections of elements of \mathcal{S}.