# Basis

basis for a set

a basis for a set \( X \) is a collection \( \mathcal{B} \) of subsets of \( X \) called basis elements such that

- for each \( x \in X \), there is at least one basis element \( B \) containing \( x \)
- if \( x \) belongs in the intersection of two basis elements \( B_{1} \) and \( B_{2} \), then there is a basis element \( B_{3} \) containing \( x \) such that \( B_{3} \subseteq B_{1} \cap B_{2} \)

collection generated by a basis

Given a basis \( \mathcal{B} \) we define the set \( G_{\mathcal{B}} \) by \( U \in G_{\mathcal{B}} \) iff for each \( x \in U \) there is a \( B \in \mathcal{B} \) such that \( x \in B \) and \( B \subseteq U \)

\( G_{\mathcal{B}} \) is a topology

The collection \( G_{\mathcal{B}} \) is a topology

We will verify it's a topology from the definition

\( \emptyset \) is in \( G_{\mathcal{B}} \) because it vacuously satisfies the condition. \( X \) is in \( G_{\mathcal{B}} \) since given any point \( x \in X \), we have a \( B \in \mathcal{B} \) such that \( x \in B \subseteq X \) by the definition of a basis

Now suppose we have an arbitrary union of elements from \( G_{\mathcal{B}} \), say \( S \subseteq G_{\mathcal{B}} , M = \bigcup_{U \in S} U \). We need to verify that \( M \in G_{\mathcal{B}} \), so let \( x \in M \), therefore \( x \in \overline{U} \) for some \( \overline{U} \in S \) and since \( \overline{U} \) was assumed to be in \( G_{\mathcal{B}} \) we know that there is some \( B \in \mathcal{B} \) such that \( x \in B \subseteq \overline{U} \) but also \( \overline{U} \subseteq M \) since \( M \) includes \( \overline{U} \) in the union so then \( x \in B \subseteq M \) meaning by the definition that \( M \in G_{\mathcal{B}} \)

Now if we are given a finite intersection of elements from \( G_{\mathcal{B}} \) we will prove by induction that it is also an element of \( G_{\mathcal{B}} \)

- base case
if \( n = 1 \) then the union of one element from \( G_{\mathcal{B}} \) (namely just itself) is also an element of \( G_{\mathcal{B}} \) trivially.

Before moving to the inductive step, notice that it also holds for \( n = 2 \), because given \( U_{1} , U_{2} \in G_{\mathcal{B}} \), we can see that for any \( x \in U_{1} \cap U_{2} \), we know that we have a basis elements \( B_{1} , B_{2} \in \mathcal{B} \) such that \( x \in B_{1} \subseteq U_{1} \) and \( x \in B_{2} \subseteq U_{2} \) and thus \( x \) also belongs to the intersection of \( B_{1} \) and \( B_{2} \), so by the definition of a basis, we get a \( B_{3} \subseteq B_{1} \cap B_{2} \) that contains \( x \) since \( B_{1} \cap B_{2} \subseteq U_{1} \cap U_{2} \), we have \( x \in B_{3} \subseteq U_{1} \cap U_{2} \), which means that \( U_{1} \cap U_{2} \in G_{\mathcal{B}} \)

- inductive step
Let \( k \in \mathbb{N}_{1} \) and assume that the intersection of \( k \) elements from \( G_{\mathcal{B}} \) is also in \( G_{\mathcal{B}} \) now we'll show it's true for \( k + 1 \). So consider the intersection \( U_{1} \cap \ldots \cap U_{k + 1} \), then we can re-write it as \( \left ( U_{1} \cap \ldots \cap U_{k} \right ) \cap U_{k + 1} \) so by our inductive hypothesis \( \left ( U_{1} \cap \ldots \cap U_{k} \right ) \in G_{\mathcal{B}} \) and therefore this \( k + 1 \) union is a union of two elements already in \( G_{\mathcal{B}} \) so by our \( n = 2 \) argument \( U_{1} \cap \ldots \cap U_{k + 1} \in G_{\mathcal{B}} \), as needed.

topology generated by a basis

Since a collection generated by a basis turns out to be a topology we define \( \mathcal{T}_{\mathcal{B}} := G_{\mathcal{B}} \) and call it the topology generated by \( \mathcal{B} \)

basis for a topology

We say that \( \mathcal{B} \) is a basis for a topology \( \mathcal{T} \) when \( \mathcal{T}_{\mathcal{B}} \) \( = \mathcal{T} \). In this case we may also say that \( \mathcal{B} \) generates \( \mathcal{T} \)

basis elements are open in a generated topology

Suppose we have the topology \( \mathcal{T}_{\mathcal{B}} \), then given any \( B \in \mathcal{B} \), \( B \in T_{\mathcal{b}} \), in other words \( \mathcal{B} \subseteq T_{\mathcal{B}} \)

Considering a topology generated by a basis \( \mathcal{B} \) we can see that for any \( B \in \mathcal{B} \), we can use \( B \) itself such that for each \( x \in B \), we have \( B \) so that \( B \subseteq B \), to show that \( B \in \mathcal{T}_{\mathcal{B}} \)

topology generated by a basis consists of unions

\( \mathcal{T}_{\mathcal{B}} \) equals the collection of all unions of elements of \( \mathcal{B} \)

Symbolically we can write this collection of unions as \( \mathcal{U} = \left \lbrace \bigcup_{B \in \mathcal{S}} B : \mathcal{S} \subseteq \mathcal{C} \right \rbrace \)

Suppose we have a subset of elements from \( \mathcal{B} \) called \( \mathcal{S} \), then for each \( B \in \mathcal{S} \) it is open in \( X \) because it is a basis element, and therefore by the definition of a topology, as \( \bigcup_{B \subseteq \mathcal{S}} B \) is an arbitrary union of open sets it is also open in \( X \), so we've shown that \( \mathcal{U} \subseteq \mathcal{T}_{\mathcal{B}} \)

We'll show the converse now, so suppose that \( U \in \mathcal{T}_{\mathcal{B}} \), we want to prove that it's equal to an arbitrary union of elements of \( \mathcal{B} \).

Recall the definition of a topology generated by a basis so since \( U \in \mathcal{T}_{\mathcal{B}} \) then for each \( x \in U \) we have some \( B_{x} \in \mathcal{B} \) with \( x \in B_{x} \subseteq U \), therefore \( U = \bigcup_{x \in U} B_{x} \), and as a union of of elements of \( \mathcal{ B } \) thus \( U \) is an element of \( \mathcal{U} \) so \( \mathcal{T}_{\mathcal{B}} \subseteq \mathcal{U} \).

basis criterion

Let \( \left ( X , \mathcal{T} \right ) \) be a topological space. Suppose that \( \mathcal{C} \) is a collection of open sets of \( X \) such that for each open set \( U \) of \( X \) and each \( x \in U \), there is an element \( C \) of \( \mathcal{C} \) such that \( x \in C \subseteq U \). Then \( \mathcal{C} \) is a basis and \( \mathcal{T}_{\mathcal{C}} = \mathcal{T} \)

We'll first show that \( \mathcal{C} \) is a basis, so let \( x \in X \), but \( X \) is open and so we can use \( X \) as the element of \( \mathcal{C} \) satisfying \( x \in X \subseteq X \).

Now let \( C_{1} , C_{2} \) be elements of \( \mathcal{C} \) and now consider \( C_{1} \cap C_{2} \), since it's a finite intersection, then we know it's also open in \( X \) from the definition of a topology.

Since \( C_{1} \cap C_{2} \) is open, let \( x \in C_{1} \cap C_{2} \), then by the definition of \( \mathcal{C} \), there is some \( C \in \mathcal{C} \) such that \( x \in C \subseteq C_{1} \cap C_{2} \), so we can take \( B_{3} = C \) in the definition of basis, to see that the second condition is satisfied, and thus we know that \( \mathcal{C} \) is a basis

Now we want to show that \( \mathcal{T}_{\mathcal{C}} = \mathcal{T} \). So suppose \( U \in \mathcal{T}_{\mathcal{C}} \), then it can be written as a union as elements from \( \mathcal{C} \), but since \( \mathcal{C} \) was defined to consist of open sets of \( X \), then this union is also open with respect to \( X \) by the definition of a topology.

Now suppose that \( U \in \mathcal{T} \) and consider some \( x \in U \) then by the definition of \( \mathcal{C} \), there is some \( C \in \mathcal{C} \) such that \( x \in C \subseteq U \), looking back at the definition for a topology generated by a basis, we can see this means \( U \in \mathcal{T}_{\mathcal{C}} \), as needed.

basis finer equivalence

Let \( \mathcal{B} \) and \( \mathcal{B} ' \) be bases for the topologies \( \mathcal{T} \) and \( \mathcal{T} ' \), respectively on \( X \). Then the following are equivalent:

- \( \mathcal{T} ' \) is finer than \( \mathcal{T} \)
- for each \( x \in X \) and each basis element \( B \in \mathcal{B} \) containing \( x \) there is a basis element \( B ' \in \mathcal{B} ' \) such that \( x \in B ' \subseteq B \)

TODO

topology equality with basis

Given two bases \( \mathcal{B} \) and \( \mathcal{B}^{'} \) for the toplogies \( \mathcal{T} \) and \( \mathcal{T}^{'} \) on \( X \), then \( \mathcal{T} = \mathcal{T}^{'} \) if and only if all of the following hold:

- for each \( x \in X \) and each basis element \( B \in \mathcal{B} \) containing \( x \) there is a basis element \( B ' \in \mathcal{B} ' \) such that \( x \in B ' \subseteq B \)
- for each \( x \in X \) and each basis element \( B^{'} \in \mathcal{B}^{'} \) containing \( x \) there is a basis element \( B \in \mathcal{B} \) such that \( x \in B \subseteq B^{'} \)

TODO

standard topology on \( \mathbb{R}^{n} \)

Let \( \mathcal{B} \) be the collection of all sets of the form \( \left \lbrace y \in \mathbb{R}^{n} : \left \| x - y \right \| < \epsilon \right \rbrace \) for any \( x \in \mathbb{R}^{n} \) and \( \epsilon \in \mathbb{R}^{\gt 0} \), then the topology generated by \( \mathcal{B} \) is called the standard topology on the \( \mathbb{R}^{n} \). Whenever we consider \( \mathbb{R}^{n} \), we shall suppose it is given this topology unless we specifically state otherwise

lower limit topology on R

If B is the collection of all half-open intervals of the form
[a, b) = {x | a ≤ x < b}, where a < b, the topology generated by B is called the lower limit topology on R.
When R is given the lower limit topology, we denote it by Rl

K topology

Finally let K denote the
set of all numbers of the form 1/n, for n ∈ Z+, and let B be the collection of all open
intervals (a, b), along with all sets of the form (a, b) − K. The topology generated
by B will be called the K-topology on R. When R is given this topology, we denote
it by RK

subbasis

A subbasis \( S \) for a topology on \( X \) is a collection of subsets of \( X \) whose union equals \( X \). The topology generated by the subbasis \( S \) is defined to be the collection \( \mathcal{T} \) of all unions of finite intersections of elements of \( S \)

topology generated by a subbasis

Given a subbasis \( S \), we know that this induces a basis \( \mathcal{B}_{S} \), and therefore a topology \( \mathcal{T}_{\mathcal{B}_{S}} \) which we call the topology generated by the subbasis \( S \).

subbasis for a topology

We say that \( S \) is a subbasis for the topology \( \mathcal{T} \) when \( \mathcal{T}_{\mathcal{B}_{S}} \)\( = \mathcal{T} \)