Difference between revisions of "Bounded Sets and Bounded Functions in a Metric Space"
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<td><strong>Definition:</strong> Let <span class="math-inline"><math>(M, d)</math></span> be a metric space. A subset <span class="math-inline"><math>S \subseteq M</math></span> is said to be <strong>Bounded</strong> if there exists a positive real number <span class="math-inline"><math>r > 0</math></span> such that <span class="math-inline"><math>S \subseteq B(x, r)</math></span> for some <span class="math-inline"><math>x \in M</math></span>. The set <span class="math-inline"><math>S</math></span> is said to be <strong>Unbounded</strong> if it is not bounded.</td> | <td><strong>Definition:</strong> Let <span class="math-inline"><math>(M, d)</math></span> be a metric space. A subset <span class="math-inline"><math>S \subseteq M</math></span> is said to be <strong>Bounded</strong> if there exists a positive real number <span class="math-inline"><math>r > 0</math></span> such that <span class="math-inline"><math>S \subseteq B(x, r)</math></span> for some <span class="math-inline"><math>x \in M</math></span>. The set <span class="math-inline"><math>S</math></span> is said to be <strong>Unbounded</strong> if it is not bounded.</td> |
Revision as of 14:06, 9 November 2021
Bounded Sets in a Metric Space
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.
Licensing
Content obtained and/or adapted from:
- Bounded Sets in a Metric Space, mathonline.wikidot.com under a CC BY-SA license