Difference between revisions of "Bounded Sets and Bounded Functions in a Metric Space"
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<p>By the definition above, we see that <span class="math-inline"><math>S</math></span> is bounded if there exists some open ball with a finite radius that contains <span class="math-inline"><math>S</math></span>.</p> | <p>By the definition above, we see that <span class="math-inline"><math>S</math></span> is bounded if there exists some open ball with a finite radius that contains <span class="math-inline"><math>S</math></span>.</p> | ||
<p>For example, consider the metric space <span class="math-inline"><math>(M, d)</math></span> where <span class="math-inline"><math>d</math></span> is the discrete metric defined for all <span class="math-inline"><math>x, y \in M</math></span> by:</p> | <p>For example, consider the metric space <span class="math-inline"><math>(M, d)</math></span> where <span class="math-inline"><math>d</math></span> is the discrete metric defined for all <span class="math-inline"><math>x, y \in M</math></span> by:</p> | ||
− | <div style="text-align: center;"><math>\begin{align} \quad d(x, y) = \left\{\begin{matrix} 0 & | + | <div style="text-align: center;"><math>\begin{align} \quad d(x, y) = \left\{\begin{matrix} 0 & \mathrm{if} \; x = y\\ 1 & \mathrm{if} \; x \neq y \end{matrix}\right. \end{align}</math></div> |
<p>Let <span class="math-inline"><math>S \subseteq M</math></span>. Then by the definition of the discrete metric, for all <span class="math-inline"><math>x, y \in S</math></span> we have that <span class="math-inline"><math>d(x, y) \leq 1</math></span>. Therefore, if we consider any point <span class="math-inline"><math>x \in S</math></span> and take <span class="math-inline"><math>r = 2</math></span> then:</p> | <p>Let <span class="math-inline"><math>S \subseteq M</math></span>. Then by the definition of the discrete metric, for all <span class="math-inline"><math>x, y \in S</math></span> we have that <span class="math-inline"><math>d(x, y) \leq 1</math></span>. Therefore, if we consider any point <span class="math-inline"><math>x \in S</math></span> and take <span class="math-inline"><math>r = 2</math></span> then:</p> | ||
<div style="text-align: center;"><math>\begin{align} \quad S \subseteq B(x, 2) \end{align}</math></div> | <div style="text-align: center;"><math>\begin{align} \quad S \subseteq B(x, 2) \end{align}</math></div> |
Revision as of 14:03, 9 November 2021
Definition: Let be a metric space. A subset is said to be Bounded if there exists a positive real number such that for some . The set is said to be Unbounded if it is not bounded.
By the definition above, we see that is bounded if there exists some open ball with a finite radius that contains .
For example, consider the metric space where is the discrete metric defined for all by:
Let . Then by the definition of the discrete metric, for all we have that . Therefore, if we consider any point and take then:
Therefore, is bounded. This shows that every subset of is bounded with respect to the discrete metric. In fact, the wholeset is also bounded and for any .
For another example, consider the metric space where is the Euclidean metric. Consider the following set:
The set above is the first octant of , and is actually unbounded. To prove this, suppose that instead is bounded. Then there exists a maximal distance between some pair of points , say:
Then . For consider the point . Then:
Since for each we see that:
But then implies which is a contradiction. Therefore our assumption that was bounded is false.