View source for Compactness in Metric Spaces

=== Compact Sets in a Metric Space ===
<p> If <span class="math-inline"><math>(M, d)</math></span> is a metric space and <span class="math-inline"><math>S \subseteq M</math></span> then a cover or covering of <span class="math-inline"><math>S</math></span> is a collection of subsets <span class="math-inline"><math>\mathcal F</math></span> in <span class="math-inline"><math>M</math></span> such that:</p>
<div style="text-align: center;"><math>\begin{align} \quad S \subseteq \bigcup_{A \in \mathcal F} A \end{align}</math></div>
<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>

<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;">
<td><strong>Definition:</strong> Let <span class="math-inline"><math>(M, d)</math></span> be a metric space. The subset <span class="math-inline"><math>S \subseteq M</math></span> is said to be <strong>Compact</strong> if every open covering <span class="math-inline"><math>\mathcal F</math></span> of <span class="math-inline"><math>S</math></span> has a finite subcovering of <span class="math-inline"><math>S</math></span>.</td>
</blockquote>
<p>In general, it may be more difficult to show that a subset of a metric space is compact than to show a subset of a metric space is not compact. So, let's look at an example of a subset of a metric space that is not compact.</p>
<p>Consider the metric space <span class="math-inline"><math>(\mathbb{R}, d)</math></span> where <span class="math-inline"><math>d</math></span> is the Euclidean metric and consider the set <span class="math-inline"><math>S = [0, 1) \subseteq \mathbb{R}</math></span>. We claim that this set is not compact. To show that <span class="math-inline"><math>S</math></span> is not compact, we need to find an open covering <span class="math-inline"><math>\mathcal F</math></span> of <span class="math-inline"><math>S</math></span> that does not have a finite subcovering. Consider the following open covering:</p>
<div style="text-align: center;"><math>\begin{align} \quad \mathcal F = \left \{ \left( (0, 1 - \frac{1}{n} \right ) : n \in \mathbb{Z}, n \geq 1 \right \} = \left \{ \left ( 0, \frac{1}{2}, \right ), \left ( 0, \frac{2}{3} \right ) , \left ( 0, \frac{3}{4} \right ) , ... \right \} \end{align}</math></div>
<p>Clearly <span class="math-inline"><math>\mathcal F</math></span> is an infinite subcovering of <span class="math-inline"><math>(0, 1)</math></span> and furthermore:</p>
<div style="text-align: center;"><math>\begin{align} \quad (0, 1) \subseteq \bigcup_{k=2}^{\infty} \left ( 0, 1 - \frac{1}{k} \right ) \end{align}</math></div>
<p>Let <span class="math-inline"><math>\mathcal F^*</math></span> be a finite subset of <span class="math-inline"><math>\mathcal F</math></span> containing <span class="math-inline"><math>p</math></span> elements. Then:</p>
<div style="text-align: center;"><math>\begin{align} \quad \mathcal F^* = \left \{ \left (0, 1 - \frac{1}{n_1}\right ), \left (0, 1 - \frac{1}{n_2}\right ), ..., \left (0, 1 - \frac{1}{n_p}\right ) \right \} \end{align}</math></div>
<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^* > 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>

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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