Connectedness - Revision history

Lila at 17:45, 8 November 2021

2021-11-08T17:45:35Z

Lila: /* Licensing */

2021-11-08T17:44:46Z

Licensing

Lila: /* Licensing */

2021-11-08T17:43:48Z

Licensing

Lila at 17:20, 8 November 2021

2021-11-08T17:20:15Z

Lila at 17:19, 8 November 2021

2021-11-08T17:19:33Z

Lila at 17:18, 8 November 2021

2021-11-08T17:18:52Z

Lila: /* Licensing */

2021-11-08T17:14:48Z

Licensing

Lila at 17:11, 8 November 2021

2021-11-08T17:11:55Z

Lila at 17:11, 8 November 2021

2021-11-08T17:11:25Z

Lila at 17:10, 8 November 2021

2021-11-08T17:10:43Z

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 <p>If <span class="math-inline"><math>c = a</math></span> then this implies that <span class="math-inline"><math>f = b</math></span> (since if <span class="math-inline"><math>d = b</math></span> then <span class="math-inline"><math>A = (a, b)</math></span> which implies that <span class="math-inline"><math>B = \emptyset</math></span>). So if <span class="math-inline"><math>A \cup B = (c, d) \cup (e, f) = (a, d) \cup (e, b) \text{ we must have that } a < d, e < b</math></span>. If <span class="math-inline"><math>d = e</math></span> then <span class="math-inline"><math>A \cup B = (a, d) \cup (d, b)</math></span> and so <span class="math-inline"><math>d \not \in (a, b)</math></span> so <span class="math-inline"><math>A \cup B \neq (a, b)</math></span>. If <span class="math-inline"><math>d < e</math></span> then <span class="math-inline"><math>A \cup B = (a, d) \cup (e, b)</math></span> and <span class="math-inline"><math>(d, e) \not \in (a, b)</math></span> so <span class="math-inline"><math>A \cup B \neq (a, b)</math></span>. If <span class="math-inline"><math>d > e</math></span> then <span class="math-inline"><math>A \cap B = (e, d) \neq \emptyset</math></span>. Either way we see that <span class="math-inline"><math>(a, b) \neq A \cup B</math></span>.</p>
 <p>We can use the same logic for the other cases which will completely show that <span class="math-inline"><math>(a, b)</math></span> is connected.</p>
 === Basic Theorems Regarding Connected and Disconnected Metric Spaces ===
@@ Line 47: / Line 49: @@
 <li>So for all <span class="math-inline"><math>a \in M</math></span> and for all <span class="math-inline"><math>r > 0</math></span> the set <span class="math-inline"><math>\{ x \in M : d(x, a) = r \}</math></span> is nonempty. <span class="math-inline"><math>\blacksquare</math></span></li>
 </ul>
 ===Continuous Functions on Connected Sets of Metric Spaces===
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 <li>Hence the assumption that <span class="math-inline"><math>f(C)</math></span> was disconnected is false. Therefore, if <span class="math-inline"><math>C</math></span> is a connected set in <span class="math-inline"><math>S</math></span> and <span class="math-inline"><math>f : C \to T</math></span> is continuous then <span class="math-inline"><math>f(C)</math></span> is connected in <span class="math-inline"><math>T</math></span></li>
 </ul>

← Older revision		Revision as of 17:44, 8 November 2021
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	* [http://mathonline.wikidot.com/connected-and-disconnected-metric-spaces Connected And Disconnected Metric Spaces, mathonline.wikidot.com] under a CC BY-SA license		* [http://mathonline.wikidot.com/connected-and-disconnected-metric-spaces Connected And Disconnected Metric Spaces, mathonline.wikidot.com] under a CC BY-SA license
	* [http://mathonline.wikidot.com/basic-theorems-regarding-connected-and-disconnected-metric-s Basic Theorems Regarding Connected and Disconnected Metric Spaces, mathonline.wikidot.com] under a CC BY-SA license		* [http://mathonline.wikidot.com/basic-theorems-regarding-connected-and-disconnected-metric-s Basic Theorems Regarding Connected and Disconnected Metric Spaces, mathonline.wikidot.com] under a CC BY-SA license
		+	* [http://mathonline.wikidot.com/continuous-functions-on-connected-sets-of-metric-spaces Continuous Functions on Connected Sets of Metric Spaces, mathonline.wikidot.com] under a CC BY-SA license

@@ Line 47: / Line 47: @@
 <li>So for all <span class="math-inline"><math>a \in M</math></span> and for all <span class="math-inline"><math>r > 0</math></span> the set <span class="math-inline"><math>\{ x \in M : d(x, a) = r \}</math></span> is nonempty. <span class="math-inline"><math>\blacksquare</math></span></li>
 </ul>
 == Licensing ==

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 === Basic Theorems Regarding Connected and Disconnected Metric Spaces ===
 <p>A metric space <span class="math-inline"><math>(M, d)</math></span> is said to be disconnected if there exists <span class="math-inline"><math>A, B \subseteq M</math></span>, <span class="math-inline"><math>A, B \neq \emptyset</math></span> where <span class="math-inline"><math>A \cap B = \emptyset</math></span> and:</p>
-<span class="equation-number">(1)</span>
 <div style="text-align: center;"><math>\begin{align} \quad M = A \cup B \end{align}</math></div>
 <p>We say that <span class="math-inline"><math>(M, d)</math></span> is connected if it is not disconnected.</p>
@@ Line 36: / Line 36: @@
 <li><strong>Proof:</strong> Let <span class="math-inline"><math>(M, d)</math></span> be a connected unbounded metric space and suppose that there exists an <span class="math-inline"><math>a \in M</math></span> and there exists an <span class="math-inline"><math>r_0 > 0</math></span> such that:</li>
 </ul>
-<span class="equation-number">(2)</span>
 <div style="text-align: center;"><math>\begin{align} \quad \{ x \in M : d(x, a) = r_0 \} = \emptyset \end{align}</math></div>
 <ul>

@@ Line 37: / Line 37: @@
 </ul>
 <span class="equation-number">(2)</span>
-<div style="text-align: center;"><math>\\begin{align} \quad \{ x \in M : d(x, a) = r_0 \} = \emptyset \end{align}</math></div>
+<div style="text-align: center;"><math>\begin{align} \quad \{ x \in M : d(x, a) = r_0 \} = \emptyset \end{align}</math></div>
 <ul>
 <li>We will show that a contradiction arises. Let <span class="math-inline"><math>A = \{ x \in M : d(x, a) < r_0 \}</math></span> and let <span class="math-inline"><math>B = \{ x \in M : d(x, a) > r_0 \}</math></span>. Then <span class="math-inline"><math>A</math></span> is open since it is simply an open ball centered at <span class="math-inline"><math>a</math></span>. Furthermore, <span class="math-inline"><math>B</math></span> is open since <span class="math-inline"><math>B^c</math></span> is a closed ball centered <span class="math-inline"><math>a</math></span>. <span class="math-inline"><math>A</math></span> is nonempty since <span class="math-inline"><math>a \in A</math></span> and <span class="math-inline"><math>B</math></span> is nonempty since <span class="math-inline"><math>(M, d)</math></span> is unbounded (if it were empty then this would imply <span class="math-inline"><math>(M, d)</math></span> is bounded). Clearly <span class="math-inline"><math>A \cap B = \emptyset</math></span> and <span class="math-inline"><math>M = A \cup B</math></span>. So <span class="math-inline"><math>(M, d)</math></span> is a disconnected metric space. But this is a contradiction.</li>

← Older revision		Revision as of 17:14, 8 November 2021
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	Content obtained and/or adapted from:		Content obtained and/or adapted from:
	* [http://mathonline.wikidot.com/connected-and-disconnected-metric-spaces Connected And Disconnected Metric Spaces, mathonline.wikidot.com] under a CC BY-SA license		* [http://mathonline.wikidot.com/connected-and-disconnected-metric-spaces Connected And Disconnected Metric Spaces, mathonline.wikidot.com] under a CC BY-SA license
		+	* [http://mathonline.wikidot.com/basic-theorems-regarding-connected-and-disconnected-metric-s Basic Theorems Regarding Connected and Disconnected Metric Spaces, mathonline.wikidot.com] under a CC BY-SA license

← Older revision		Revision as of 17:11, 8 November 2021
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	<p>If <span class="math-inline"><math>c = a</math></span> then this implies that <span class="math-inline"><math>f = b</math></span> (since if <span class="math-inline"><math>d = b</math></span> then <span class="math-inline"><math>A = (a, b)</math></span> which implies that <span class="math-inline"><math>B = \emptyset</math></span>). So if <span class="math-inline"><math>A \cup B = (c, d) \cup (e, f) = (a, d) \cup (e, b) \text{ we must have that } a < d, e < b</math></span>. If <span class="math-inline"><math>d = e</math></span> then <span class="math-inline"><math>A \cup B = (a, d) \cup (d, b)</math></span> and so <span class="math-inline"><math>d \not \in (a, b)</math></span> so <span class="math-inline"><math>A \cup B \neq (a, b)</math></span>. If <span class="math-inline"><math>d < e</math></span> then <span class="math-inline"><math>A \cup B = (a, d) \cup (e, b)</math></span> and <span class="math-inline"><math>(d, e) \not \in (a, b)</math></span> so <span class="math-inline"><math>A \cup B \neq (a, b)</math></span>. If <span class="math-inline"><math>d > e</math></span> then <span class="math-inline"><math>A \cap B = (e, d) \neq \emptyset</math></span>. Either way we see that <span class="math-inline"><math>(a, b) \neq A \cup B</math></span>.</p>		<p>If <span class="math-inline"><math>c = a</math></span> then this implies that <span class="math-inline"><math>f = b</math></span> (since if <span class="math-inline"><math>d = b</math></span> then <span class="math-inline"><math>A = (a, b)</math></span> which implies that <span class="math-inline"><math>B = \emptyset</math></span>). So if <span class="math-inline"><math>A \cup B = (c, d) \cup (e, f) = (a, d) \cup (e, b) \text{ we must have that } a < d, e < b</math></span>. If <span class="math-inline"><math>d = e</math></span> then <span class="math-inline"><math>A \cup B = (a, d) \cup (d, b)</math></span> and so <span class="math-inline"><math>d \not \in (a, b)</math></span> so <span class="math-inline"><math>A \cup B \neq (a, b)</math></span>. If <span class="math-inline"><math>d < e</math></span> then <span class="math-inline"><math>A \cup B = (a, d) \cup (e, b)</math></span> and <span class="math-inline"><math>(d, e) \not \in (a, b)</math></span> so <span class="math-inline"><math>A \cup B \neq (a, b)</math></span>. If <span class="math-inline"><math>d > e</math></span> then <span class="math-inline"><math>A \cap B = (e, d) \neq \emptyset</math></span>. Either way we see that <span class="math-inline"><math>(a, b) \neq A \cup B</math></span>.</p>
	<p>We can use the same logic for the other cases which will completely show that <span class="math-inline"><math>(a, b)</math></span> is connected.</p>		<p>We can use the same logic for the other cases which will completely show that <span class="math-inline"><math>(a, b)</math></span> is connected.</p>
		+
		+	== Licensing ==
		+	Content obtained and/or adapted from:
		+	* [http://mathonline.wikidot.com/connected-and-disconnected-metric-spaces Connected And Disconnected Metric Spaces, mathonline.wikidot.com] under a CC BY-SA license

@@ Line 8: / Line 8: @@
 <p>For example, 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 on <span class="math-inline"><math>\mathbb{R}</math></span>. Let <span class="math-inline"><math>S = (a, b) \subset \mathbb{R}</math></span>, i.e., <span class="math-inline"><math>S</math></span> is an open interval in <span class="math-inline"><math>\mathbb{R}</math></span>. We claim that <span class="math-inline"><math>S</math></span> is connected.</p>
 <p>Suppose not. Then there exists nonempty open subsets <span class="math-inline"><math>A</math></span> and <span class="math-inline"><math>B</math></span> such that <span class="math-inline"><math>A \cap B = \emptyset</math></span> and <span class="math-inline"><math>(a, b) = A \cup B</math></span>. Furthermore, <span class="math-inline"><math>A</math></span> and <span class="math-inline"><math>B</math></span> must be open intervals themselves, say <span class="math-inline"><math>A = (c, d)</math></span> and <span class="math-inline"><math>B = (e, f)</math></span>. We must have that <span class="math-inline"><math>A \cup B = (c, d) \cup (e, f)</math></span>. So <span class="math-inline"><math>c = a</math></span> or <span class="math-inline"><math>e = a</math></span> and furthermore, <span class="math-inline"><math>d = b</math></span> or <span class="math-inline"><math>f = b</math></span>.</p>
-<p>If <span class="math-inline"><math>c = a</math></span> then this implies that <span class="math-inline"><math>f = b</math></span> (since if <span class="math-inline"><math>d = b</math></span> then <span class="math-inline"><math>A = (a, b)</math></span> which implies that <span class="math-inline"><math>B = \emptyset</math></span>). So if <span class="math-inline"><math>A \cup B = (c, d) \cup (e, f) = (a, d) \cup (e, b) \text{ we must have that } a < d, e < b</math></span>. If <span class="math-inline"><math>d = e</math></span> then <span class="math-inline"><math>A \cup B = (a, d) \cup (d, b)</math></span> and so <span class="math-inline"><math>d \not \in (a, b)</math></span> so <span class="math-inline"><math>A \cup B \neq = (a, b)</math></span>. If <span class="math-inline"><math>d < e</math></span> then <span class="math-inline"><math>A \cup B = (a, d) \cup (e, b)</math></span> and <span class="math-inline"><math>(d, e) \not \in (a, b)</math></span> so <span class="math-inline"><math>A \cup B \neq (a, b)</math></span>. If <span class="math-inline"><math>d > e</math></span> then <span class="math-inline"><math>A \cap B = (e, d) \neq \emptyset</math></span>. Either way we see that <span class="math-inline"><math>(a, b) \neq A \cup B</math></span>.</p>
+<p>If <span class="math-inline"><math>c = a</math></span> then this implies that <span class="math-inline"><math>f = b</math></span> (since if <span class="math-inline"><math>d = b</math></span> then <span class="math-inline"><math>A = (a, b)</math></span> which implies that <span class="math-inline"><math>B = \emptyset</math></span>). So if <span class="math-inline"><math>A \cup B = (c, d) \cup (e, f) = (a, d) \cup (e, b) \text{ we must have that } a < d, e < b</math></span>. If <span class="math-inline"><math>d = e</math></span> then <span class="math-inline"><math>A \cup B = (a, d) \cup (d, b)</math></span> and so <span class="math-inline"><math>d \not \in (a, b)</math></span> so <span class="math-inline"><math>A \cup B \neq (a, b)</math></span>. If <span class="math-inline"><math>d < e</math></span> then <span class="math-inline"><math>A \cup B = (a, d) \cup (e, b)</math></span> and <span class="math-inline"><math>(d, e) \not \in (a, b)</math></span> so <span class="math-inline"><math>A \cup B \neq (a, b)</math></span>. If <span class="math-inline"><math>d > e</math></span> then <span class="math-inline"><math>A \cap B = (e, d) \neq \emptyset</math></span>. Either way we see that <span class="math-inline"><math>(a, b) \neq A \cup B</math></span>.</p>
 <p>We can use the same logic for the other cases which will completely show that <span class="math-inline"><math>(a, b)</math></span> is connected.</p>