Compactness in Metric Spaces - Revision history

Lila: /* Basic Theorems Regarding Compact Sets in a Metric Space */

2021-11-09T19:56:10Z

Basic Theorems Regarding Compact Sets in a Metric Space

Lila: /* Licensing */

2021-11-09T19:55:16Z

Licensing

Lila: /* Licensing */

2021-11-08T21:56:32Z

Licensing

Lila: /* Boundedness of Compact Sets in a Metric Space */

2021-11-08T21:55:36Z

Boundedness of Compact Sets in a Metric Space

Lila at 21:55, 8 November 2021

2021-11-08T21:55:08Z

Lila at 21:53, 8 November 2021

2021-11-08T21:53:52Z

Lila: /* Licensing */

2021-11-08T21:44:50Z

Licensing

Lila at 21:42, 8 November 2021

2021-11-08T21:42:47Z

Lila at 21:41, 8 November 2021

2021-11-08T21:41:52Z

Lila at 21:41, 8 November 2021

2021-11-08T21:41:24Z

@@ Line 66: / Line 66: @@
 <div style="text-align: center;"><math>\begin{align} \quad S \cap T \subseteq T \end{align}</math></div>
 <ul>
-<li>Furthermore, <span class="math-inline"><math>S \cap T</math></span> is closed. This is because <span class="math-inline"><math>S</math></span> is given as closed, and since <span class="math-inline"><math>T</math></span> is compact we know that <span class="math-inline"><math>T</math></span> is closed (and bounded). So the finite intersection <span class="math-inline"><math>S \cap T</math></span> is closed. But any closed subset of a compact set is also compact as we proved on the <a href="/closed-subsets-of-compact-sets-in-metric-spaces">Closed Subsets of Compact Sets in Metric Spaces</a> page, so <span class="math-inline"><math>S \cap T</math></span> is compact in <span class="math-inline"><math>M</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
+<li>Furthermore, <span class="math-inline"><math>S \cap T</math></span> is closed. This is because <span class="math-inline"><math>S</math></span> is given as closed, and since <span class="math-inline"><math>T</math></span> is compact we know that <span class="math-inline"><math>T</math></span> is closed (and bounded). So the finite intersection <span class="math-inline"><math>S \cap T</math></span> is closed. But any closed subset of a compact set is also compact, so <span class="math-inline"><math>S \cap T</math></span> is compact in <span class="math-inline"><math>M</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
 </ul>
 <blockquote style="background: white; border: 1px solid black; padding: 1em;">
@@ Line 96: / Line 96: @@
 <li>Furthermore, since each <span class="math-inline"><math>C \in \mathcal C</math></span> is compact, then each <span class="math-inline"><math>C</math></span> is closed (and bounded). An arbitrary intersection of closed sets is closed, and so <span class="math-inline"><math>\displaystyle{\bigcap_{C \in \mathcal C}}</math></span> is a closed subset of the compact set <span class="math-inline"><math>C</math></span>. Therefore by the theorem referenced earlier, <span class="math-inline"><math>\displaystyle{\bigcap_{C \in \mathcal C} C}</math></span> is compact in <span class="math-inline"><math>M</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
 </ul>
 ==Licensing==

@@ Line 54: / Line 54: @@
 <li>Hence <span class="math-inline"><math>S</math></span> is bounded. <span class="math-inline"><math>\blacksquare</math></span></li>
 </ul>
 ==Licensing==
@@ Line 59: / Line 102: @@
 * [http://mathonline.wikidot.com/compact-sets-in-a-metric-space Compact Sets in a Metric Space, mathonline.wikidot.com] under a CC BY-SA license
 * [http://mathonline.wikidot.com/boundedness-of-compact-sets-in-a-metric-space Boundedness of Compact Sets in a Metric Space, mathonline.wikidot.com] under a CC BY-SA license

← Older revision		Revision as of 21:56, 8 November 2021
Line 58:		Line 58:
	Content obtained and/or adapted from:		Content obtained and/or adapted from:
	* [http://mathonline.wikidot.com/compact-sets-in-a-metric-space Compact Sets in a Metric Space, mathonline.wikidot.com] under a CC BY-SA license		* [http://mathonline.wikidot.com/compact-sets-in-a-metric-space Compact Sets in a Metric Space, mathonline.wikidot.com] under a CC BY-SA license
		+	* [http://mathonline.wikidot.com/boundedness-of-compact-sets-in-a-metric-space Boundedness of Compact Sets in a Metric Space, mathonline.wikidot.com] under a CC BY-SA license

@@ Line 27: / Line 27: @@
 <td><strong>Theorem 1:</strong> If <span class="math-inline"><math>(M, d)</math></span> be a metric space and <span class="math-inline"><math>S \subseteq M</math></span> is a compact subset of <span class="math-inline"><math>M</math></span> then <span class="math-inline"><math>S</math></span> is bounded.</td>
 </blockquote>
-<div class="image-container aligncenter"><img src="http://mathonline.wdfiles.com/local--files/boundedness-of-compact-sets-in-a-metric-space/Screen%20Shot%202015-10-05%20at%209.56.39%20PM.png" alt="Screen%20Shot%202015-10-05%20at%209.56.39%20PM.png" class="image" /></div>
 <ul>
 <li><strong>Proof:</strong> For a fixed <span class="math-inline"><math>x_0 \in S</math></span> and for <span class="math-inline"><math>r > 0</math></span>, consider the ball centered at <span class="math-inline"><math>x_0</math></span> with radius <span class="math-inline"><math>r</math></span>, i.e., <span class="math-inline"><math>B(x_0, r)</math></span>. Let <span class="math-inline"><math>\mathcal F</math></span> denote the collection of balls centered at <span class="math-inline"><math>x_0</math></span> with varying radii <span class="math-inline"><math>r > 0</math></span>:</li>
@@ Line 54: / Line 54: @@
 <li>Hence <span class="math-inline"><math>S</math></span> is bounded. <span class="math-inline"><math>\blacksquare</math></span></li>
 </ul>
 ==Licensing==
 Content obtained and/or adapted from:
 * [http://mathonline.wikidot.com/compact-sets-in-a-metric-space Compact Sets in a Metric Space, mathonline.wikidot.com] under a CC BY-SA license

@@ Line 29: / Line 29: @@
 <div class="image-container aligncenter"><img src="http://mathonline.wdfiles.com/local--files/boundedness-of-compact-sets-in-a-metric-space/Screen%20Shot%202015-10-05%20at%209.56.39%20PM.png" alt="Screen%20Shot%202015-10-05%20at%209.56.39%20PM.png" class="image" /></div>
 <ul>
-<li><strong>Proof:</strong> For a fixed <span class="math-inline"><math>x_0 \in S</math></span> and for <span class="math-inline"><math>r &gt; 0</math></span>, consider the ball centered at <span class="math-inline"><math>x_0</math></span> with radius <span class="math-inline"><math>r</math></span>, i.e., <span class="math-inline"><math>B(x_0, r)</math></span>. Let <span class="math-inline"><math>\mathcal F</math></span> denote the collection of balls centered at <span class="math-inline"><math>x_0</math></span> with varying radii <span class="math-inline"><math>r &gt; 0</math></span>:</li>
+<li><strong>Proof:</strong> For a fixed <span class="math-inline"><math>x_0 \in S</math></span> and for <span class="math-inline"><math>r > 0</math></span>, consider the ball centered at <span class="math-inline"><math>x_0</math></span> with radius <span class="math-inline"><math>r</math></span>, i.e., <span class="math-inline"><math>B(x_0, r)</math></span>. Let <span class="math-inline"><math>\mathcal F</math></span> denote the collection of balls centered at <span class="math-inline"><math>x_0</math></span> with varying radii <span class="math-inline"><math>r > 0</math></span>:</li>
 </ul>
-<div style="text-align: center;"><math>\begin{align} \quad \mathcal F = \{ B(x_0, r) : r &gt; 0 \} \end{align}</math></div>
+<div style="text-align: center;"><math>\begin{align} \quad \mathcal F = \{ B(x_0, r) : r > 0 \} \end{align}</math></div>
 <ul>
-<li>It should not be hard to see that <span class="math-inline"><math>\mathcal F</math></span> is an open covering of <span class="math-inline"><math>S</math></span>, since for all <span class="math-inline"><math>s \in S</math></span> we have that <span class="math-inline"><math>d(x_0, s) = r_s &gt; 0</math></span>, so <span class="math-inline"><math>s \in B(x_0, r_s) \in \mathcal F</math></span>.</li>
+<li>It should not be hard to see that <span class="math-inline"><math>\mathcal F</math></span> is an open covering of <span class="math-inline"><math>S</math></span>, since for all <span class="math-inline"><math>s \in S</math></span> we have that <span class="math-inline"><math>d(x_0, s) = r_s > 0</math></span>, so <span class="math-inline"><math>s \in B(x_0, r_s) \in \mathcal F</math></span>.</li>
 </ul>
 <ul>
@@ Line 44: / Line 44: @@
 <div style="text-align: center;"><math>\begin{align} \quad S \subseteq \bigcup_{k=1}^{p} B(x_0, r_k) \end{align}</math></div>
 <ul>
-<li>Each of the open balls in the open subcovering <span class="math-inline"><math>\mathcal F^*</math></span> is centered at <span class="math-inline"><math>x_0</math></span> with <span class="math-inline"><math>r_1, r_2, ..., r_p &gt; 0</math></span>. Since the set <span class="math-inline"><math>\{ r_1, r_2, ..., r_p \}</math></span> is a finite set, there exists a maximum value. Let:</li>
+<li>Each of the open balls in the open subcovering <span class="math-inline"><math>\mathcal F^*</math></span> is centered at <span class="math-inline"><math>x_0</math></span> with <span class="math-inline"><math>r_1, r_2, ..., r_p > 0</math></span>. Since the set <span class="math-inline"><math>\{ r_1, r_2, ..., r_p \}</math></span> is a finite set, there exists a maximum value. Let:</li>
 </ul>
 <div style="text-align: center;"><math>\begin{align} \quad r_{\mathrm{max}} = \max \{ r_1, r_2, ..., r_p \} \end{align}</math></div>
@@ Line 54: / Line 54: @@
 <li>Hence <span class="math-inline"><math>S</math></span> is bounded. <span class="math-inline"><math>\blacksquare</math></span></li>
 </ul>
 ==Licensing==
 Content obtained and/or adapted from:
 * [http://mathonline.wikidot.com/compact-sets-in-a-metric-space Compact Sets in a Metric Space, mathonline.wikidot.com] under a CC BY-SA license

← Older revision		Revision as of 21:42, 8 November 2021
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	<div style="text-align: center;"><math>\begin{align} \quad \bigcup_{k=1}^{p} \left ( 0, 1 - \frac{1}{n_p} \right ) = \left ( 0, 1 - \frac{1}{n^*} \right ) \end{align}</math></div>		<div style="text-align: center;"><math>\begin{align} \quad \bigcup_{k=1}^{p} \left ( 0, 1 - \frac{1}{n_p} \right ) = \left ( 0, 1 - \frac{1}{n^*} \right ) \end{align}</math></div>
	<p>But for <span class="math-inline"><math>(0, 1) \subseteq \left ( 0, 1 - \frac{1}{n^} \right )</math></span> we need <span class="math-inline"><math>1 \leq 1 - \frac{1}{n^}</math></span>. But <span class="math-inline"><math>n^* \in \mathbb{N}</math></span>, so <span class="math-inline"><math>n^* > 0</math></span> and <span class="math-inline"><math>\frac{1}{n^} > 0</math></span>, so <span class="math-inline"><math>1 - \frac{1}{n^} < 1</math></span>. Therefore any finite subset <span class="math-inline"><math>\mathcal F^*</math></span> of <span class="math-inline"><math>\mathcal F</math></span> cannot cover <span class="math-inline"><math>S = (0, 1)</math></span>. Hence, <span class="math-inline"><math>(0, 1)</math></span> is not compact.</p>		<p>But for <span class="math-inline"><math>(0, 1) \subseteq \left ( 0, 1 - \frac{1}{n^} \right )</math></span> we need <span class="math-inline"><math>1 \leq 1 - \frac{1}{n^}</math></span>. But <span class="math-inline"><math>n^* \in \mathbb{N}</math></span>, so <span class="math-inline"><math>n^* > 0</math></span> and <span class="math-inline"><math>\frac{1}{n^} > 0</math></span>, so <span class="math-inline"><math>1 - \frac{1}{n^} < 1</math></span>. Therefore any finite subset <span class="math-inline"><math>\mathcal F^*</math></span> of <span class="math-inline"><math>\mathcal F</math></span> cannot cover <span class="math-inline"><math>S = (0, 1)</math></span>. Hence, <span class="math-inline"><math>(0, 1)</math></span> is not compact.</p>
		+
		+	==Licensing==
		+	Content obtained and/or adapted from:
		+	* [http://mathonline.wikidot.com/compact-sets-in-a-metric-space Compact Sets in a Metric Space] under a CC BY-SA license

@@ Line 5: / Line 5: @@
 <p>We also said that a subset <span class="math-inline"><math>\mathcal S \subseteq \mathcal F</math></span> is a subcover/subcovering (or open subcover/subcovering if <span class="math-inline"><math>\mathcal F</math></span> is an open covering) if <span class="math-inline"><math>\mathcal S</math></span> is also a cover of <span class="math-inline"><math>S</math></span>, that is:</p>
-<div style="text-align: center;"><math>\begin{align} \quad S \subseteq \bigcup_{A \in \mathcal{S}} A \quad \text{where} \: \mathcal S \subseteq \mathcal F \end{align}</math></div>
+<div style="text-align: center;"><math>\begin{align} \quad S \subseteq \bigcup_{A \in \mathcal{S}} A \quad \text{ where }  \mathcal S \subseteq \mathcal F \end{align}</math></div>
 <p>We can now define the concept of a compact set using the definitions above.</p>
 <blockquote style="background: white; border: 1px solid black; padding: 1em;">

@@ Line 4: / Line 4: @@
 <p>Furthermore, we said that an open cover (or open covering) is simply a cover that contains only open sets.</p>
 <p>We also said that a subset <span class="math-inline"><math>\mathcal S \subseteq \mathcal F</math></span> is a subcover/subcovering (or open subcover/subcovering if <span class="math-inline"><math>\mathcal F</math></span> is an open covering) if <span class="math-inline"><math>\mathcal S</math></span> is also a cover of <span class="math-inline"><math>S</math></span>, that is:</p>
-<span class="equation-number">(2)</span>
-<div style="text-align: center;"><math>\begin{align} \quad S \subseteq \bigcup_{A \in \mathcal S} A \quad \text{ where }  \mathcal S \subseteq \mathcal F \end{align}</math></div>
+<div style="text-align: center;"><math>\begin{align} \quad S \subseteq \bigcup_{A \in \mathcal{S}} A \quad \text{where} \: \mathcal S \subseteq \mathcal F \end{align}</math></div>
 <p>We can now define the concept of a compact set using the definitions above.</p>
 <blockquote style="background: white; border: 1px solid black; padding: 1em;">
@@ Line 19: / Line 19: @@
 <p>Let <span class="math-inline"><math>n^* = \max \{ n_1, n_2, ..., n_p \}</math></span>. Then due to the nesting of the open covering <span class="math-inline"><math>\mathcal F</math></span>, we see that:</p>
 <div style="text-align: center;"><math>\begin{align} \quad \bigcup_{k=1}^{p} \left ( 0, 1 - \frac{1}{n_p} \right ) = \left ( 0, 1 - \frac{1}{n^*} \right ) \end{align}</math></div>
-<p>But for <span class="math-inline"><math>(0, 1) \subseteq \left ( 0, 1 - \frac{1}{n^*} \right )</math></span> we need <span class="math-inline"><math>1 \leq 1 - \frac{1}{n^*}</math></span>. But <span class="math-inline"><math>n^* \in \mathbb{N}</math></span>, so <span class="math-inline"><math>n^* &gt; 0</math></span> and <span class="math-inline"><math>\frac{1}{n^*} &gt; 0</math></span>, so <span class="math-inline"><math>1 - \frac{1}{n^*} &lt; 1</math></span>. Therefore any finite subset <span class="math-inline"><math>\mathcal F^*</math></span> of <span class="math-inline"><math>\mathcal F</math></span> cannot cover <span class="math-inline"><math>S = (0, 1)</math></span>. Hence, <span class="math-inline"><math>(0, 1)</math></span> is not compact.</p>
+<p>But for <span class="math-inline"><math>(0, 1) \subseteq \left ( 0, 1 - \frac{1}{n^*} \right )</math></span> we need <span class="math-inline"><math>1 \leq 1 - \frac{1}{n^*}</math></span>. But <span class="math-inline"><math>n^* \in \mathbb{N}</math></span>, so <span class="math-inline"><math>n^* > 0</math></span> and <span class="math-inline"><math>\frac{1}{n^*} > 0</math></span>, so <span class="math-inline"><math>1 - \frac{1}{n^*} < 1</math></span>. Therefore any finite subset <span class="math-inline"><math>\mathcal F^*</math></span> of <span class="math-inline"><math>\mathcal F</math></span> cannot cover <span class="math-inline"><math>S = (0, 1)</math></span>. Hence, <span class="math-inline"><math>(0, 1)</math></span> is not compact.</p>