Difference between revisions of "Bases of Open Sets"

From Department of Mathematics at UTSA
Jump to navigation Jump to search
(Blanked the page)
Tag: Blanking
 
(One intermediate revision by the same user not shown)
Line 1: Line 1:
 +
We will now focus our attention at a special type of subset of a topology called a base for <math>\tau</math> which we define below.
  
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
:'''Definition:''' Let <math>(X, \tau)</math> be a topological space. A <strong>Base</strong> (sometimes <strong>Basis</strong>) for the topology <math>\tau</math> is a collection <math>\mathcal B</math> of subsets from <math>\tau</math> such that every <math>U \in \tau</math> is the union of some collection of sets in <math>\mathcal B</math>.</td>
 +
</blockquote>
 +
 +
''Note that by definition, <math>\mathcal B = \tau</math> is a base of <math>\tau</math> - albeit a rather trivial one! The emptyset is also obtained by an empty union of sets from <math>\mathcal B</math>.''
 +
 +
Let's look at some examples.
 +
 +
=== Example 1 ===
 +
 +
Consider any nonempty set <math>X</math> with the discrete topology <math>\tau = \mathcal P (X)</math>. Consider the collection:
 +
 +
<div style="text-align: center;"><math>\begin{align} \quad \mathcal B = \{ \{ x \} : x \in X \} \end{align}</math></div>
 +
 +
We claim that <math>\mathcal B</math> is a base of the discrete topology <math>\tau</math>. Let's verify this. First, since <math>\tau</math> is the discrete topology we see that every subset of <math>X</math> is contained in <math>\tau</math>. For each <math>B = \{ x \} \in \mathcal B</math> we therefore have that:
 +
 +
<div style="text-align: center;"><math>\begin{align} \quad B = \{ x \} \in \mathcal B \in \tau \end{align}</math></div>
 +
 +
For the second condition, let <math>U \in \tau</math>. Then since <math>\tau</math> is the discrete topology, we have that <math>U \subseteq X</math>. For all <math>x \in U</math>, we have that <math>U</math> can be expressed as the union of some collection of sets in <math>\mathcal B</math>. In particular, for each <math>U \in \tau</math> we have that:
 +
 +
<div style="text-align: center;"><math>\begin{align} \quad U = \bigcup_{x \in U} \{ x \} \end{align}</math></div>
 +
 +
Therefore <math>\mathcal B = \{ \{ x \} : x \in X \}</math> is a base of the discrete topology.
 +
 +
=== Example 2 ===
 +
 +
For another example, consider the set <math>X = \{ a, b, c, d \}</math> and the following topology on <math>X</math>:
 +
 +
<div style="text-align: center;"><math>\begin{align} \quad \tau = \{ \emptyset, \{ a \}, \{d \}, \{a, d \}, \{ b, c\}, \{a, b, c \}, \{ b, c, d \}, X \} \end{align}</math></div>
 +
 +
Consider the collection of open sets <math>\mathcal B = \{ \{ a \}, \{ d \}, \{b, c \} \}</math>. We claim that <math>\mathcal B</math> is a base of <math>\tau</math>. Clearly all of the sets in <math>\mathcal B</math> are contained in <math>\tau</math>, so every set in <math>\mathcal B</math> is open.
 +
 +
For the second condition, we only need to show that the remaining open sets in <math>\tau</math> that are not in <math>\mathcal B</math> can be obtained by taking unions of elements in <math>\mathcal B</math>. The <math>\emptyset \in \tau</math> can be obtained by taking the empty union of elements in <math>\mathcal B</math>. Furthermore:
 +
 +
<div style="text-align: center;"><math>\begin{align} \quad & \{ a \} \cup \{ d \} = \{ a, d \} \\ \quad & \{ a \} \cup \{ b, c \} = \{ a, b, c \} \\ \quad & \{ a \} \cup \{ b, c \} \cup \{ d \} = X \end{align}</math></div>
 +
 +
Therefore every <math>U \in \tau</math> is the union of some collection of sets from <math>\mathcal B</math>, so <math>\mathcal B</math> is a base of <math>\tau</math>.
 +
 +
=== Example 3 ===
 +
If <math>\mathbb{R}</math> has the usual Euclidean topology, then the collection:
 +
<div style="text-align: center;"><math>\begin{align} \quad \mathcal B = \{ (a, b) : a, b \in \mathbb{R}, a &lt; b \} \end{align}</math></div>
 +
 +
(The collection of bounded open intervals) is a base for the Euclidean topology.
 +
 +
=== Example 4 ===
 +
 +
If <math>X</math> is any metric space, then the collection:
 +
<div style="text-align: center;"><math>\begin{align} \quad \mathcal B = \{ B(x, \epsilon) : x \in X, \epsilon &gt; 0 \} \end{align}</math></div>
 +
 +
(The collection of open balls relative to the metric defined on <math>X</math>) is a base for the topology resulting from the metric on <math>X</math>.
 +
 +
== Licensing ==
 +
Content obtained and/or adapted from:
 +
* [http://mathonline.wikidot.com/bases-of-a-topology Bases of a topology, mathonline.wikidot.com] under a CC BY-SA license

Latest revision as of 17:49, 13 November 2021

We will now focus our attention at a special type of subset of a topology called a base for which we define below.

Definition: Let be a topological space. A Base (sometimes Basis) for the topology is a collection of subsets from such that every is the union of some collection of sets in .

Note that by definition, is a base of - albeit a rather trivial one! The emptyset is also obtained by an empty union of sets from .

Let's look at some examples.

Example 1

Consider any nonempty set with the discrete topology . Consider the collection:

We claim that is a base of the discrete topology . Let's verify this. First, since is the discrete topology we see that every subset of is contained in . For each we therefore have that:

For the second condition, let . Then since is the discrete topology, we have that . For all , we have that can be expressed as the union of some collection of sets in . In particular, for each we have that:

Therefore is a base of the discrete topology.

Example 2

For another example, consider the set and the following topology on :

Consider the collection of open sets . We claim that is a base of . Clearly all of the sets in are contained in , so every set in is open.

For the second condition, we only need to show that the remaining open sets in that are not in can be obtained by taking unions of elements in . The can be obtained by taking the empty union of elements in . Furthermore:

Therefore every is the union of some collection of sets from , so is a base of .

Example 3

If has the usual Euclidean topology, then the collection:

(The collection of bounded open intervals) is a base for the Euclidean topology.

Example 4

If is any metric space, then the collection:

(The collection of open balls relative to the metric defined on ) is a base for the topology resulting from the metric on .

Licensing

Content obtained and/or adapted from: