Difference between revisions of "Convergent Sequences in Metric Spaces"

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<p>Furthermore, it's not hard to see that this sequence converges to <span class="math-inline"><math>x</math></span>, i.e., <span class="math-inline"><math>\lim_{n \to \infty} x_n = x</math></span>, i.e., <span class="math-inline"><math>\lim_{n \to \infty} d(x_n, x) = 0</math></span> since for all <span class="math-inline"><math>x_n</math></span> we have that <span class="math-inline"><math>d(x_n, x) = 0</math></span>, so <span class="math-inline"><math>\lim_{n \to \infty} d(x_n, x) = \lim_{n \to \infty} 0 = 0</math></span>.</p>
 
<p>Furthermore, it's not hard to see that this sequence converges to <span class="math-inline"><math>x</math></span>, i.e., <span class="math-inline"><math>\lim_{n \to \infty} x_n = x</math></span>, i.e., <span class="math-inline"><math>\lim_{n \to \infty} d(x_n, x) = 0</math></span> since for all <span class="math-inline"><math>x_n</math></span> we have that <span class="math-inline"><math>d(x_n, x) = 0</math></span>, so <span class="math-inline"><math>\lim_{n \to \infty} d(x_n, x) = \lim_{n \to \infty} 0 = 0</math></span>.</p>
 
<p>We will soon see that many of theorems regarding limits of sequences of real numbers are analogous to limits of sequences of elements from metric spaces.</p>
 
<p>We will soon see that many of theorems regarding limits of sequences of real numbers are analogous to limits of sequences of elements from metric spaces.</p>
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===The Boundedness of Convergent Sequences in Metric Spaces===
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<p>If <span class="math-inline"><math>(M, d)</math></span> is a metric space and <span class="math-inline"><math>(x_n)_{n=1}^{\infty}</math></span> is a sequence in <span class="math-inline"><math>M</math></span> that is convergent then the limit of this sequence <span class="math-inline"><math>p \in M</math></span> is unique.</p>
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<p>We will now look at another rather nice theorem which states that if <span class="math-inline"><math>(x_n)_{n=1}^{\infty}</math></span> is convergent then it is also bounded.</p>
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<table class="wiki-content-table">
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<tr>
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<td><strong>Theorem 1:</strong> Let <span class="math-inline"><math>(M, d)</math></span> be a metric space and let <span class="math-inline"><math>(x_n)_{n=1}^{\infty}</math></span> be a sequence in <span class="math-inline"><math>M</math></span>. If <span class="math-inline"><math>(x_n)_{n=1}^{\infty}</math></span> is convergent then the set <span class="math-inline"><math>\{ x_1, x_2, ..., x_n, ... \}</math></span> is bounded.</td>
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</tr>
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</table>
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<ul>
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<li><strong>Proof:</strong> Let <span class="math-inline"><math>(M, d)</math></span> be a metric space and let <span class="math-inline"><math>(x_n)_{n=1}^{\infty}</math></span> be a sequence in <span class="math-inline"><math>M</math></span> that converges to <span class="math-inline"><math>p \in M</math></span>, i.e., <span class="math-inline"><math>\lim_{n \to \infty} x_n = p</math></span>. Then <span class="math-inline"><math>\lim_{n \to \infty} d(x_n, p) = 0</math></span>. So for all <span class="math-inline"><math>\epsilon > 0</math></span> there exists an <span class="math-inline"><math>N \in \mathbb{N}</math></span> such that if <span class="math-inline"><math>n \geq N</math></span> then <span class="math-inline"><math>d(x_n, p) < \epsilon</math></span>. So for <span class="math-inline"><math>\epsilon_0 = 1 > 0</math></span> there exists an <span class="math-inline"><math>N(\epsilon_0) \in \mathbb{N}</math></span> such that if <span class="math-inline"><math>n \geq N(\epsilon_0)</math></span> then:</li>
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</ul>
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<div style="text-align: center;"><math>\begin{align} \quad d(x_n, p) < \epsilon_0 = 1 \end{align}</math></div>
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<ul>
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<li>Now consider the elements <span class="math-inline"><math>x_1, x_2, ..., x_{N(\epsilon_0)}</math></span>. This is a finite set of elements and furthermore the set of distances from these elements to <span class="math-inline"><math>p</math></span> is finite:</li>
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</ul>
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<div style="text-align: center;"><math>\begin{align} \quad \{ d(x_1, p), d(x_2, p), ..., d(x_{N(\epsilon_0) - 1}, p) \} \end{align}</math></div>
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<ul>
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<li>Define <span class="math-inline"><math>M ^*</math></span> to be the maximum of these distances:</li>
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</ul>
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<div style="text-align: center;"><math>\begin{align} \quad M^* = \max \{ d(x_1, p), d(x_2, p), ..., d(x_{N(\epsilon_0) - 1}, p) \} \end{align}</math></div>
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<ul>
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<li>So if <span class="math-inline"><math>1 \leq n < N(\epsilon_0)</math></span> we have that <span class="math-inline"><math>d(x_n, p) \leq M^*</math></span> and if <span class="math-inline"><math>n \geq N(\epsilon_0)</math></span> then <span class="math-inline"><math>d(x_n, p) < 1</math></span>. Let <span class="math-inline"><math>M = \max \{ M^*, 1 \}</math></span>. Then for all <span class="math-inline"><math>n \in \mathbb{N}</math></span>, <span class="math-inline"><math>d(x_n, p) \leq M</math></span>. So consider the open ball <span class="math-inline"><math>B(p, M + 1)</math></span>. Then <span class="math-inline"><math>x_n \in B(p, M + 1)</math></span> for all <span class="math-inline"><math>n \in \{1, 2, ... \}</math></span> so:</li>
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</ul>
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<div style="text-align: center;"><math>\begin{align} \quad \{ x_1, x_2, ..., x_n, ... \} \subseteq B(p, M+1) \end{align}</math></div>
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<ul>
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<li>Therefore <span class="math-inline"><math>\{ x_1, x_2, ..., x_n, ... \}</math></span> is a bounded set in <span class="math-inline"><math>M</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
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</ul>
  
 
== Licensing ==  
 
== Licensing ==  
 
Content obtained and/or adapted from:
 
Content obtained and/or adapted from:
 
* [http://mathonline.wikidot.com/limits-of-sequences-in-metric-spaces Limits of Sequences in Metric Spaces, mathonline.wikidot.com] under a CC BY-SA license
 
* [http://mathonline.wikidot.com/limits-of-sequences-in-metric-spaces Limits of Sequences in Metric Spaces, mathonline.wikidot.com] under a CC BY-SA license

Revision as of 10:50, 8 November 2021

Limits of Sequences in Metric Spaces

Recall that if a sequence of real numbers is an infinite ordered list where for every . We will now generalize the concept of a sequence to contain elements from a metric space .

Definition: Let be a metric space. An (infinite) Sequence in denoted is an infinite ordered list of elements for all .

Finite sequences in a metric space can be defined as a finite ordered list of elements in but their study is not that interesting to us.

We can also define whether a sequence of elements from a metric space converges or diverges.

Definition: Let be a metric space. A sequence in is said to be Convergent to the element written if and the element is said to be the Limit of the sequence . If no such exists, then is said to be Divergent.

There is a subtle but important point to make. In the definition above, represents the limit of a sequence of elements from the metric space to an element while represents the limit of a sequence of positive real numbers to - such limits we already have experience with.

Convergent sequence in metric space

For example, if is any nonempty set, is the discrete metric, and , then the sequence defined by for all , then the sequence:

Furthermore, it's not hard to see that this sequence converges to , i.e., , i.e., since for all we have that , so .

We will soon see that many of theorems regarding limits of sequences of real numbers are analogous to limits of sequences of elements from metric spaces.

The Boundedness of Convergent Sequences in Metric Spaces

If is a metric space and is a sequence in that is convergent then the limit of this sequence is unique.

We will now look at another rather nice theorem which states that if is convergent then it is also bounded.

Theorem 1: Let be a metric space and let be a sequence in . If is convergent then the set is bounded.
  • Proof: Let be a metric space and let be a sequence in that converges to , i.e., . Then . So for all there exists an such that if then . So for there exists an such that if then:
  • Now consider the elements . This is a finite set of elements and furthermore the set of distances from these elements to is finite:
  • Define to be the maximum of these distances:
  • So if we have that and if then . Let . Then for all , . So consider the open ball . Then for all so:
  • Therefore is a bounded set in .

Licensing

Content obtained and/or adapted from: