Difference between revisions of "Bounded Sets and Bounded Functions in a Metric Space"
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<div style="text-align: center;"><math>\begin{align} \quad d(\mathbf{x}, k\mathbf{y}) \geq d(\mathbf{x}, \mathbf{y}) = d_{\mathrm{max}} \end{align}</math></div> | <div style="text-align: center;"><math>\begin{align} \quad d(\mathbf{x}, k\mathbf{y}) \geq d(\mathbf{x}, \mathbf{y}) = d_{\mathrm{max}} \end{align}</math></div> | ||
<p>But then <span class="math-inline"><math>\mathbf{y} \not \in B(\mathbf{x}, d_{\mathrm{max}})</math></span> implies <span class="math-inline"><math>\mathbf{y} \not \in S</math></span> which is a contradiction. Therefore our assumption that <span class="math-inline"><math>S</math></span> was bounded is false.</p> | <p>But then <span class="math-inline"><math>\mathbf{y} \not \in B(\mathbf{x}, d_{\mathrm{max}})</math></span> implies <span class="math-inline"><math>\mathbf{y} \not \in S</math></span> which is a contradiction. Therefore our assumption that <span class="math-inline"><math>S</math></span> was bounded is false.</p> | ||
+ | |||
+ | == Licensing == | ||
+ | Content obtained and/or adapted from: | ||
+ | * [http://mathonline.wikidot.com/bounded-sets-in-a-metric-space Bounded Sets in a Metric Space, mathonline.wikidot.com] under a CC BY-SA license |
Revision as of 14:05, 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.
Licensing
Content obtained and/or adapted from:
- Bounded Sets in a Metric Space, mathonline.wikidot.com under a CC BY-SA license