Lindelöf Theorem - Revision history

Lila: Lila moved page Lindeloff Theorem to Lindelöf Theorem: Misspelled title

2021-10-26T19:13:57Z

Lila moved page Lindeloff Theorem to Lindelöf Theorem: Misspelled title

Lila at 19:13, 26 October 2021

2021-10-26T19:13:06Z

Lila at 19:11, 26 October 2021

2021-10-26T19:11:47Z

Lila: /* The Lindelöf Lemma */

2021-10-26T19:10:16Z

The Lindelöf Lemma

Lila: /* The Lindelöf Lemma */

2021-10-26T19:09:44Z

The Lindelöf Lemma

Lila at 19:08, 26 October 2021

2021-10-26T19:08:51Z

Lila at 19:08, 26 October 2021

2021-10-26T19:08:20Z

Lila at 19:04, 26 October 2021

2021-10-26T19:04:57Z

Lila at 18:57, 26 October 2021

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Lila at 18:56, 26 October 2021

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@@ Line 25: / Line 25: @@
 <p>A space <math>X</math> is second countable if there exists a countable basis for the topology on <math>X</math>.</p>
 <p>We will now look at a very important connection between Lindelöf spaces and second countable spaces which we state and prove below.</p>
-<table class="wiki-content-table">
 <tr>
-<td><strong>Lemma 1 (The Lindelöf Lemma):</strong> If <math>X</math> is a second countable topological space then <math>X</math> is Lindelöf.</td>
+<blockquote style="background: white; border: 1px solid black; padding: 1em;"><td><strong>Lemma 1 (The Lindelöf Lemma):</strong> If <math>X</math> is a second countable topological space then <math>X</math> is Lindelöf.</td>
 </tr>
-</table>
 <ul>
 <li><strong>Proof:</strong> Let <math>X</math> be a second countable topology space. Then there exists a countable basis <math>\mathcal B</math> of the topology <math>\tau</math> on <math>X</math>.</li>

@@ Line 59: / Line 59: @@
 ==Licensing==
 Content obtained and/or adapted from:
-* [http://mathonline.wikidot.com/lindeloef-and-countably-compact-topological-spaces Lindelof and Countably Compact Topological Spaces, mathonline.wikidot.com] under a CC BY-SA license
+* [http://mathonline.wikidot.com/lindeloef-and-countably-compact-topological-spaces Lindelöf and Countably Compact Topological Spaces, mathonline.wikidot.com] under a CC BY-SA license
-* [http://mathonline.wikidot.com/the-lindeloef-lemma The Lindelof Lemma, mathonline.wikidot.com] under a CC BY-SA license
+* [http://mathonline.wikidot.com/the-lindeloef-lemma The Lindelöf Lemma, mathonline.wikidot.com] under a CC BY-SA license

@@ Line 23: / Line 23: @@
 <p>A topological space <math>X</math> is said to be Lindelöf if every open cover of <math>X</math> has a countable subcover.</p>
 <p>Furthermore, we said that <math>X</math> is countably compact if every countable open cover of <math>X</math> has a finite subcover.</p>
-<p>Awhile back, on the <a href="/second-countable-topological-spaces">Second Countable Topological Spaces</a> page we said that a space <math>X</math> is second countable if there exists a countable basis for the topology on <math>X</math>.</p>
+<p>A space <math>X</math> is second countable if there exists a countable basis for the topology on <math>X</math>.</p>
 <p>We will now look at a very important connection between Lindelöf spaces and second countable spaces which we state and prove below.</p>
 <table class="wiki-content-table">

@@ Line 45: / Line 45: @@
 <li>Define an open cover <math>\mathcal C</math> of <math>X</math> as follows:</li>
 </ul>
-<math>\begin{align} \quad \mathcal C = \left \{ B \in \mathcal B : A = \bigcup_{B \in \mathcal B_A} B, \: \mathrm{for \: some \:} A \in \mathcal F \right \} \end{align}</math>
+<math>\begin{align} \quad \mathcal C = \left \{ B \in \mathcal B : A = \bigcup_{B \in \mathcal B_A} B,  \mathrm{for  some } A \in \mathcal F \right \} \end{align}</math>
 <ul>
 <li>Then <math>\mathcal C</math> is countable since <math>\mathcal C \subseteq \mathcal B</math> and <math>\mathcal B</math> is countable.</li>
@@ Line 56: / Line 56: @@
 <li>Hence every open cover <math>\mathcal F</math> of <math>X</math> has a countable subcover <math>\mathcal F^*</math>. Therefore, <math>X</math> is Lindelöf. <math>\blacksquare</math></li>
 </ul>
 ==Licensing==

@@ Line 36: / Line 36: @@
 <li>Now, let <math>\mathcal F</math> be any open cover of <math>X</math> so that:</li>
 </ul>
-<div class="math-equation" id="equation-1">\begin{align} \quad X = \bigcup_{A \in \mathcal F} A \end{align}</div>
+<math>\begin{align} \quad X = \bigcup_{A \in \mathcal F} A \end{align}</math>
 <ul>
 <li>Note that each <math>A \in \mathcal F</math> is an open set since <math>\mathcal F</math> is an open cover of <math>X</math>. Therefore, for each <math>A</math> there exists a subcollection <math>\mathcal B_A \subseteq \mathcal B</math> such that:</li>

← Older revision	Revision as of 19:13, 26 October 2021
(No difference)

@@ Line 1: / Line 1: @@
-<h1 id="toc0">Lindelöf and Countably Compact Topological Spaces</h1>
+==Lindelöf and Countably Compact Topological Spaces==
 <p>If <math>X</math> is a topological space and <math>A \subseteq X</math>, then <math>A</math> is said to be compact in <math>X</math> if every open cover of <math>A</math> has a finite subcover.</p>
 <p>Moreover, we said that <math>X</math> is a compact topological space if every open cover of <math>X</math> has a finite subcover.</p>
@@ Line 20: / Line 20: @@
 <p>So as we can see, the concept of compactness, Lindelöfness, and countable compactness are different properties.</p>
-<h1 id="toc0">The Lindelöf Lemma</h1>
+==The Lindelöf Lemma==
 <p>A topological space <math>X</math> is said to be Lindelöf if every open cover of <math>X</math> has a countable subcover.</p>
 <p>Furthermore, we said that <math>X</math> is countably compact if every countable open cover of <math>X</math> has a finite subcover.</p>
@@ Line 41: / Line 41: @@
 </ul>
-<div class="math-equation" id="equation-2">\begin{align} \quad A = \bigcup_{B \in \mathcal B_A} B \end{align}</div>
+<math>\begin{align} \quad A = \bigcup_{B \in \mathcal B_A} B \end{align}</math>
 <ul>
 <li>Define an open cover <math>\mathcal C</math> of <math>X</math> as follows:</li>
 </ul>
-<div class="math-equation" id="equation-3">\begin{align} \quad \mathcal C = \left \{ B \in \mathcal B : A = \bigcup_{B \in \mathcal B_A} B, \: \mathrm{for \: some \:} A \in \mathcal F \right \} \end{align}</div>
+<math>\begin{align} \quad \mathcal C = \left \{ B \in \mathcal B : A = \bigcup_{B \in \mathcal B_A} B, \: \mathrm{for \: some \:} A \in \mathcal F \right \} \end{align}</math>
 <ul>
 <li>Then <math>\mathcal C</math> is countable since <math>\mathcal C \subseteq \mathcal B</math> and <math>\mathcal B</math> is countable.</li>
@@ Line 52: / Line 52: @@
 <li>We now define an open subcover of <math>\mathcal F</math>. For each element <math>C \in \mathcal C</math>, take an element <math>A \in \mathcal F</math> such that <math>C \subseteq A</math>. This is possible since <math>\mathcal C</math> is a subcollection of the countable basis <math>\mathcal B</math> and every element <math>A \in \mathcal F</math> is the union of some collection of basis elements. Then <math>\mathcal F^*</math> is countable, and moreover:</li>
 </ul>
-<div class="math-equation" id="equation-4">\begin{align} \quad X = \bigcup_{A \in \mathcal F^*} A \end{align}</div>
+<math>\begin{align} \quad X = \bigcup_{A \in \mathcal F^*} A \end{align}</math>
 <ul>
 <li>Hence every open cover <math>\mathcal F</math> of <math>X</math> has a countable subcover <math>\mathcal F^*</math>. Therefore, <math>X</math> is Lindelöf. <math>\blacksquare</math></li>

← Older revision		Revision as of 19:04, 26 October 2021
Line 19:		Line 19:
	<p>Unfortunately, <math>\mathbb{N}</math> is not countably compact if we use the example open cover which showed that <math>\mathbb{N}</math> was not compact.</p>		<p>Unfortunately, <math>\mathbb{N}</math> is not countably compact if we use the example open cover which showed that <math>\mathbb{N}</math> was not compact.</p>
	<p>So as we can see, the concept of compactness, Lindelöfness, and countable compactness are different properties.</p>		<p>So as we can see, the concept of compactness, Lindelöfness, and countable compactness are different properties.</p>
		+
		+	<h1 id="toc0">The Lindelöf Lemma</h1>
		+	<p>A topological space <math>X</math> is said to be Lindelöf if every open cover of <math>X</math> has a countable subcover.</p>
		+	<p>Furthermore, we said that <math>X</math> is countably compact if every countable open cover of <math>X</math> has a finite subcover.</p>
		+	<p>Awhile back, on the <a href="/second-countable-topological-spaces">Second Countable Topological Spaces</a> page we said that a space <math>X</math> is second countable if there exists a countable basis for the topology on <math>X</math>.</p>
		+	<p>We will now look at a very important connection between Lindelöf spaces and second countable spaces which we state and prove below.</p>
		+	<table class="wiki-content-table">
		+	<tr>
		+	<td><strong>Lemma 1 (The Lindelöf Lemma):</strong> If <math>X</math> is a second countable topological space then <math>X</math> is Lindelöf.</td>
		+	</tr>
		+	</table>
		+	<ul>
		+	<li><strong>Proof:</strong> Let <math>X</math> be a second countable topology space. Then there exists a countable basis <math>\mathcal B</math> of the topology <math>\tau</math> on <math>X</math>.</li>
		+	</ul>
		+	<ul>
		+	<li>Now, let <math>\mathcal F</math> be any open cover of <math>X</math> so that:</li>
		+	</ul>
		+	<div class="math-equation" id="equation-1">\begin{align} \quad X = \bigcup_{A \in \mathcal F} A \end{align}</div>
		+	<ul>
		+	<li>Note that each <math>A \in \mathcal F</math> is an open set since <math>\mathcal F</math> is an open cover of <math>X</math>. Therefore, for each <math>A</math> there exists a subcollection <math>\mathcal B_A \subseteq \mathcal B</math> such that:</li>
		+	</ul>
		+
		+	<div class="math-equation" id="equation-2">\begin{align} \quad A = \bigcup_{B \in \mathcal B_A} B \end{align}</div>
		+	<ul>
		+	<li>Define an open cover <math>\mathcal C</math> of <math>X</math> as follows:</li>
		+	</ul>
		+	<div class="math-equation" id="equation-3">\begin{align} \quad \mathcal C = \left \{ B \in \mathcal B : A = \bigcup_{B \in \mathcal B_A} B, \: \mathrm{for \: some \:} A \in \mathcal F \right \} \end{align}</div>
		+	<ul>
		+	<li>Then <math>\mathcal C</math> is countable since <math>\mathcal C \subseteq \mathcal B</math> and <math>\mathcal B</math> is countable.</li>
		+	</ul>
		+	<ul>
		+	<li>We now define an open subcover of <math>\mathcal F</math>. For each element <math>C \in \mathcal C</math>, take an element <math>A \in \mathcal F</math> such that <math>C \subseteq A</math>. This is possible since <math>\mathcal C</math> is a subcollection of the countable basis <math>\mathcal B</math> and every element <math>A \in \mathcal F</math> is the union of some collection of basis elements. Then <math>\mathcal F^*</math> is countable, and moreover:</li>
		+	</ul>
		+	<div class="math-equation" id="equation-4">\begin{align} \quad X = \bigcup_{A \in \mathcal F^*} A \end{align}</div>
		+	<ul>
		+	<li>Hence every open cover <math>\mathcal F</math> of <math>X</math> has a countable subcover <math>\mathcal F^*</math>. Therefore, <math>X</math> is Lindelöf. <math>\blacksquare</math></li>
		+	</ul>
		+
		+
		+	==Licensing==
		+	Content obtained and/or adapted from:
		+	* [http://mathonline.wikidot.com/lindeloef-and-countably-compact-topological-spaces Lindelof and Countably Compact Topological Spaces, mathonline.wikidot.com] under a CC BY-SA license
		+	* [http://mathonline.wikidot.com/the-lindeloef-lemma The Lindelof Lemma, mathonline.wikidot.com] under a CC BY-SA license

@@ Line 13: / Line 13: @@
 <p>For another example, consider the set of natural numbers <math>\mathbb{N}</math> with the discrete topology, i.e., every subset of <math>\mathbb{N}</math> is open. Then <math>\mathbb{N}</math> is not compact, because of the following open cover of <math>\mathbb{N}</math>:</p>
-<math>\begin{align} \qqquad \mathcal F = \{ \{n \} : n \in \mathbb{N} \} = \{ \{ 1 \}, \{ 2 \}, ..., \{ n \}, ... \} \end{align}</math>
+<div class="center" style="width: auto; margin-left: auto; margin-right: auto;"> <math> \mathcal{F} = \{ \{n \} : n \in \mathbb{N} \} = \{ \{ 1 \}, \{ 2 \}, ..., \{ n \}, ... \} </math> </div>
 <p>Clearly there does not exist any subcollection <math>\mathcal F^* \subseteq \mathcal F</math> that is finite and still covers <math>\mathbb{N}</math>! So <math>\mathbb{N}</math> is not compact.</p>