Difference between revisions of "Bases of Open Sets"

From Department of Mathematics at UTSA
Jump to navigation Jump to search
(Created page with "A set <math>S \subseteq \mathbb{R}^n</math> is said to be open if <math>S =\mathrm{int} (S)</math>, that is, for every point <math>\mathbf{a} \in S</math> we have that there e...")
 
 
(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
A set <math>S \subseteq \mathbb{R}^n</math> is said to be open if <math>S =\mathrm{int} (S)</math>, that is, for every point <math>\mathbf{a} \in S</math> we have that there exists a positive real number <span class="math-inline"><math>r > 0</math></span> such that the ball centered at <math>\mathbf{a}</math> with radius <math>r</math> is contained in <math>S</math>, i.e., <math>B(\mathbf{a}, r) \subseteq S</math>.
+
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.
  
Furthermore, we said that <math>S \subseteq \mathbb{R}^n</math> is closed if <math>S^c</math> is open.
+
<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>
  
For any general metric space <math>(M, d)</math>, we define open and closeds subsets <math>S</math> of <math>M</math> in a similar manner.
+
''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>.''
  
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
+
Let's look at some examples.
:'''Definition:''' If <math>(M, d)</math> is a metric space and <math>S \subseteq M</math> then <math>S</math> is said to be '''Open''' if <math>S = \mathrm{int} (S)</math> and <math>S</math> is said to be '''Closed<''' if <math>S^c</math> is open. Moreover, <math>S</math> is said to be '''Clopen''' if it is both open and closed.
+
 
</blockquote>
+
=== 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>.
  
It is important to note that the definitions above are somewhat of a poor choice of words. A set <math>S</math> may just be open, just closed, open and closed (clopen), or even neither. Unfortunately these definitions are standard and we should note that saying a set is "not open" does not mean it is closed and likewise, saying a set is "not closed" does not mean it is open.
+
=== 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>
  
Now consider the whole set <math>M</math>. Is <math>M</math> open or closed? Well by definition, for every <math>a \in M</math> there exists a positive real number <math>r > 0</math> such that <math>B(a, r) \subseteq M</math> since the ball centered at <math>a</math> with radius <math>r</math> is defined to be the set of all points <em>IN <math>M</math></em> that are of a distance less than <math>r</math> of <math>a</math>. Therefore <math>M</math> is an open set.
+
(The collection of bounded open intervals) is a base for the Euclidean topology.
  
So then the complement of <math>M</math> is <math>M^c = M \setminus M = \emptyset</math> is a closed set. However, it is vacuously true that for all <math>a \in \emptyset</math> there exists a ball centered at <math>a</math> fully contained in <math>\emptyset</math> since <math>\emptyset</math> contains no points to begin with. Therefore <math>\emptyset</math> is also an open set and so <math>M</math> is also a closed set.
+
=== Example 4 ===
  
This is the case for all metric spaces <math>(M, d)</math>. The whole set <math>M</math> and empty set <math>\emptyset</math> are trivially clopen sets!
+
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==
+
== Licensing ==  
 
Content obtained and/or adapted from:
 
Content obtained and/or adapted from:
* [http://mathonline.wikidot.com/open-and-closed-sets-in-metric-spaces Open and Closed Sets in Metric Spaces, mathonline.wikidot.com] under a CC BY-SA license
+
* [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: