Baire's Theorem and Applications - Revision history

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Licensing

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The Baire Category Theorem

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Licensing

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The Baire Category Theorem

Lemma 1: Let
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Created page with "<h1 id="toc0">The Baire Category Theorem</h1> <blockquote style="background: white; border: 1px solid black; padding: 1em;"> <td>Lemma 1: Let <sp..."

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<h1 id="toc0">The Baire Category Theorem</h1>
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Lemma 1: Let <math>(X, \tau)</math> be a topological space and let <math>A \subseteq X</math>. If <math>A</math> is a nowhere dense set then for every <math>U \in \tau</math> there exists a <math>B \subseteq U</math> such that <math>A \cap \bar{B} = \emptyset</math>.</td>
</blockquote>
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
<td>Theorem 1 (The Baire Category Theorem): Every complete metric space is of the second category.</td>
</blockquote>
<ul>
<li>Proof: Let <math>X</math> be a complete metric space. Then every Cauchy sequence <math>(x_n)_{n=1}^{\infty}</math> of elements from <math>X</math> converges in <math>X</math>. Suppose that <math>X</math> is of the first category. Then there exists a countable collection of nowhere dense sets <math>A_1, A_2, ... \subset X</math> such that:</li>
</ul>
<div style="text-align: center;"><math>\begin{align} \quad X = \bigcup_{i=1}^{\infty} A_i \end{align}</math></div>
<ul>
<li>Let <math>U \subset X</math>. For each nowhere dense set <math>A_i</math>, <math>i \in \{ 1, 2, ... \}</math> there exists a set <math>B_i \subset U</math> such that <math>A_i \cap \bar{B_i} = \emptyset</math>.</li>
</ul>
<ul>
<li>Let <math>B_1(x_1, r) \subset U</math> be a ball contained in <math>U</math> such that <math>A_1 \cap \bar{B_1} = \emptyset</math>. Let <math>B_2 \left ( x_2, \frac{r}{2} \right ) \subseteq B_1</math> be a ball contained in <math>B_1</math> whose radius is <math>\frac{r}{2}</math> and such that <math>A_2 \cap \bar{B_2} = \emptyset</math>. Repeat this process. For each <math>n \in \{ 2, 3, ... \}</math> let <math>B_n \left (x_n, \frac{r}{n} \right )</math> be a ball contained in <math>B_{n-1}</math> whose radius is <math>\frac{r}{n}</math> and such that <math>A_n \cap \bar{B_n} = \emptyset</math> and such that <math>A_n \cap \bar{B_n} = \emptyset</math>.</li>
</ul>
<ul>
<li>The sequence <math>(x_n)_{n=1}^{\infty}</math> is Cauchy since as <math>n \in \mathbb{N}</math> gets large, the elements <math>x_n</math> are very close. Since <math>X</math> is a complete metric space, we must have that this Cauchy sequence therefore converges to some <math>p \in X</math>, i.e., <math>\lim_{n \to \infty} x_n = p</math>.</li>
</ul>
<ul>
<li>Now notice that <math>x \in \bar{B_n}</math> for all <math>n \in \mathbb{N}</math> because if not, then there exists an <math>m \in \mathbb{N}</math> such that <math>x \not \in \bar{B_n}</math> for all <math>n \geq m</math>. Hence <math>(\bar{B_n})^c</math> is open and so there exists an open ball <math>X</math> such that <math>x \in B \subset (\bar{B_n})^c</math> but then <math>\lim_{x \to \infty} x_n \neq x</math> because <math>x_m \not \in B</math> for all <math>m \geq n</math>.</li>
</ul>
<ul>
<li>Since <math>x \in \bar{B_n}</math> for all <math>n \in \mathbb{N}</math> then since <math>A_n \cap \bar{B_n} = \emptyset</math> we must have that then <math> x \not\in A_n </math></li>
</ul>

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 ==Licensing==
 Content obtained and/or adapted from:
 * [http://mathonline.wikidot.com/the-baire-category-theorem The Baire Category Theorem, mathonline.wikidot.com] under a CC BY-SA license

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-===The Baire Category Theorem===
+==Sets of the First and Second Categories in a Topological Space==
 <blockquote style="background: white; border: 1px solid black; padding: 1em;">
 <td><strong>Lemma 1:</strong> Let <span class="math-inline"><math>(X, \tau)</math></span> be a topological space and let <span class="math-inline"><math>A \subseteq X</math></span>. If <span class="math-inline"><math>A</math></span> is a nowhere dense set then for every <span class="math-inline"><math>U \in \tau</math></span> there exists a <span class="math-inline"><math>B \subseteq U</math></span> such that <span class="math-inline"><math>A \cap \bar{B} = \emptyset</math></span>.</td>

← Older revision		Revision as of 21:09, 8 November 2021
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	Content obtained and/or adapted from:		Content obtained and/or adapted from:
	* [http://mathonline.wikidot.com/the-baire-category-theorem The Baire Category Theorem, mathonline.wikidot.com] under a CC BY-SA license		* [http://mathonline.wikidot.com/the-baire-category-theorem The Baire Category Theorem, mathonline.wikidot.com] under a CC BY-SA license
		+	* [http://mathonline.wikidot.com/dense-and-nowhere-dense-sets-in-a-topological-space Dense and Nowhere Dense Sets in a Topological Space, mathonline.wikidot.com] under a CC BY-SA license

← Older revision		Revision as of 20:48, 8 November 2021
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		+	==Dense and Nowhere Dense Sets==
		+	===Dense Sets in a Topological Space===
		+	<blockquote style="background: white; border: 1px solid black; padding: 1em;">
		+	<td><strong>Definition:</strong> Let <span class="math-inline"><math>(X, \tau)</math></span> be a topological space. The set <span class="math-inline"><math>A \subseteq X</math></span> is said to be <strong>Dense</strong> in <span class="math-inline"><math>X</math></span> if the intersection of every nonempty open set with <span class="math-inline"><math>A</math></span> is nonempty, that is, <span class="math-inline"><math>A \cap U \neq \emptyset</math></span> for all <span class="math-inline"><math>U \in \tau \setminus \{ \emptyset \}</math></span>.</td>
		+	</blockquote>
		+	<p>Given any topological space <span class="math-inline"><math>(X, \tau)</math></span> it is important to note that <span class="math-inline"><math>X</math></span> is dense in <span class="math-inline"><math>X</math></span> because every <span class="math-inline"><math>U \in \tau</math></span> is such that <span class="math-inline"><math>U \subseteq X</math></span>, and so <span class="math-inline"><math>X \cap U = U \neq \emptyset</math></span> for all <span class="math-inline"><math>U \in \tau \setminus \{ \emptyset \}</math></span>.</p>
		+	<p>For another example, consider the topological space <span class="math-inline"><math>(\mathbb{R}, \tau)</math></span> where <span class="math-inline"><math>\tau</math></span> is the usual topology of open intervals. Then the set of rational numbers <span class="math-inline"><math>\mathbb{Q} \subset \mathbb{R}</math></span> is dense in <span class="math-inline"><math>\mathbb{R}</math></span>. If not, then there exists an <span class="math-inline"><math>U \in \tau \setminus \{ \emptyset \}</math></span> such that <span class="math-inline"><math>\mathbb{Q} \cap U = \emptyset</math></span>.</p>
		+	<p>Since <span class="math-inline"><math>U \in \tau</math></span> we have that <span class="math-inline"><math>(a, b) \subseteq U</math></span> for some open interval <span class="math-inline"><math>(a, b)</math></span> with <span class="math-inline"><math>a, b \in \mathbb{R}</math></span> and <span class="math-inline"><math>a < b</math></span>. Suppose that <span class="math-inline"><math>\mathbb{Q} \setminus U = \emptyset</math></span>. Then we must also have that:</p>
		+	<div style="text-align: center;"><math> \begin{align} \quad \mathbb{Q} \cap U = \mathbb{Q} \cap (a, b) = \emptyset \end{align}</math></div>
		+	<p>The intersection above implies that there exists no rational numbers in the interval <span class="math-inline"><math>(a, b)</math></span>, i.e., there exists no <span class="math-inline"><math>q \in \mathbb{Q}</math></span> such that <span class="math-inline"><math>a < q < b</math></span>. But this is a contradiction since for all <span class="math-inline"><math>a, b \in \mathbb{R}</math></span> with <span class="math-inline"><math>a < b</math></span> there ALWAYS exists a rational number <span class="math-inline"><math>q \in \mathbb{Q}</math></span> such that <span class="math-inline"><math>a < q < b</math></span>, i.e., <span class="math-inline"><math>q \in (a, b)</math></span>. So <span class="math-inline"><math>\mathbb{Q} \cap (a, b) \neq \emptyset</math></span> for all <span class="math-inline"><math>U \in \tau \setminus \{ \emptyset \}</math></span>. Thus, <span class="math-inline"><math>\mathbb{Q}</math></span> is dense in <span class="math-inline"><math>\mathbb{R}</math></span>.</p>
		+	<p>We will now look at a very important theorem which will give us a way to determine whether a set <span class="math-inline"><math>A \subseteq X</math></span> is dense in <span class="math-inline"><math>X</math></span> or not.</p>
		+	<blockquote style="background: white; border: 1px solid black; padding: 1em;">
		+	<td><strong>Theorem 1:</strong> Let <span class="math-inline"><math>(X, \tau)</math></span> be a topological space and let <span class="math-inline"><math>A \subseteq X</math></span>. Then <span class="math-inline"><math>A</math></span> is dense in <span class="math-inline"><math>X</math></span> if and only if <span class="math-inline"><math>\bar{A} = X</math></span>.</td>
		+	</blockquote>
		+	<ul>
		+	<li><strong>Proof:</strong> <span class="math-inline"><math>\Rightarrow</math></span> Suppose that <span class="math-inline"><math>A</math></span> is dense in <span class="math-inline"><math>X</math></span>. Then for all <span class="math-inline"><math>U \in \tau \setminus \{ \emptyset \}</math></span> we have that <span class="math-inline"><math>A \cap U = \emptyset</math></span>. Clearly <span class="math-inline"><math>\bar{A} \subseteq X</math></span> so we only need to show that <span class="math-inline"><math>X \subseteq \bar{A}</math></span>.</li>
		+	</ul>
		+	===Nowhere Dense Sets in a Topological Space===
		+	<blockquote style="background: white; border: 1px solid black; padding: 1em;">
		+	<td><strong>Definition:</strong> Let <span class="math-inline"><math>(X, \tau)</math></span> be a topological space. A set <span class="math-inline"><math>A \subseteq X</math></span> is said to be <strong>Nowhere Dense</strong> in <span class="math-inline"><math>X</math></span> if the interior of the closure of <span class="math-inline"><math>A</math></span> is empty, that is, <span class="math-inline"><math>\mathrm{int} (\bar{A}) = \emptyset</math></span>.</td>
		+	</blockquote>
		+	<p>For example, consider the topological space <span class="math-inline"><math>(\mathbb{R}, \tau)</math></span> where <span class="math-inline"><math>\tau</math></span> is the usually topology of open intervals on <span class="math-inline"><math>\mathbb{R}</math></span>, and consider the set of integers <span class="math-inline"><math>\mathbb{Z}</math></span>. The closure of <span class="math-inline"><math>\mathbb{Z}</math></span>, <span class="math-inline"><math>\bar{\mathbb{Z}}</math></span> is the smallest closed set containing <span class="math-inline"><math>\mathbb{Z}</math></span>. The smallest closed set containing <span class="math-inline"><math>\mathbb{Z}</math></span> is <span class="math-inline"><math>\mathbb{Z}</math></span> since <span class="math-inline"><math>\mathbb{Z}^c</math></span> is open as <span class="math-inline"><math>\mathbb{Z}^c</math></span> is an arbitrary union of open sets:</p>
		+	<div style="text-align: center;"><math> \begin{align} \quad \mathbb{Z}^c = ... (-2, -1) \cup (-1, 0) \cup (0, 1) \cup (1, 2) \cup ... \end{align}</math></div>
		+	<p>So what is the interior of <span class="math-inline"><math>\bar{\mathbb{Z}} = \mathbb{Z}</math></span>? It is the largest open set contained in <span class="math-inline"><math>\bar{\mathbb{Z}} = \mathbb{Z}</math></span>. All open sets of <span class="math-inline"><math>\mathbb{R}</math></span> with respect to this topology <span class="math-inline"><math>\tau</math></span> are either the empty set, an open interval, a union of open intervals, or the whole set (the union of all open intervals). But no open intervals are contained in <span class="math-inline"><math>\mathbb{Z}</math></span> and so:</p>
		+	<div style="text-align: center;"><math> \begin{align} \quad \mathrm{int} (\bar{\mathbb{Z}}) = \emptyset \end{align}</math></div>
		+	<p>Therefore <span class="math-inline"><math>\mathbb{Z}</math></span> is a nowhere dense set in <span class="math-inline"><math>\mathbb{R}</math></span> with respect to the usual topology <span class="math-inline"><math>\tau</math></span> on <span class="math-inline"><math>\mathbb{R}</math></span>.</p>
		+
		+
	===The Baire Category Theorem===		===The Baire Category Theorem===
	<blockquote style="background: white; border: 1px solid black; padding: 1em;">		<blockquote style="background: white; border: 1px solid black; padding: 1em;">

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-<h1 id="toc0"><span>The Baire Category Theorem</span></h1>
+===The Baire Category Theorem===
 <blockquote style="background: white; border: 1px solid black; padding: 1em;">
 <td><strong>Lemma 1:</strong> Let <span class="math-inline"><math>(X, \tau)</math></span> be a topological space and let <span class="math-inline"><math>A \subseteq X</math></span>. If <span class="math-inline"><math>A</math></span> is a nowhere dense set then for every <span class="math-inline"><math>U \in \tau</math></span> there exists a <span class="math-inline"><math>B \subseteq U</math></span> such that <span class="math-inline"><math>A \cap \bar{B} = \emptyset</math></span>.</td>
@@ Line 25: / Line 25: @@
 <li>Since <span class="math-inline"><math>x \in \bar{B_n}</math></span> for all <span class="math-inline"><math>n \in \mathbb{N}</math></span> then since <span class="math-inline"><math>A_n \cap \bar{B_n} = \emptyset</math></span> we must have that then <span class="math-inline"><math> x \not\in A_n </math></span></li>
 </ul>