Difference between revisions of "Connectedness"
Line 37: | Line 37: | ||
</ul> | </ul> | ||
<span class="equation-number">(2)</span> | <span class="equation-number">(2)</span> | ||
− | <div style="text-align: center;"><math> | + | <div style="text-align: center;"><math>\begin{align} \quad \{ x \in M : d(x, a) = r_0 \} = \emptyset \end{align}</math></div> |
<ul> | <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> | <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> |
Revision as of 11:19, 8 November 2021
Connected and Disconnected Metric Spaces
Definition: A metric space is said to be Disconnected if there exists nonempty open sets and such that and . If is not disconnected then we say that Connected. Furthermore, if then is said to be disconnected/connected if the metric subspace is disconnected/connected.
Intuitively, a set is disconnected if it can be separated into two pieces while a set is connected if it’s an entire piece.
For example, consider the metric space where is the Euclidean metric on . Let , i.e., is an open interval in . We claim that is connected.
Suppose not. Then there exists nonempty open subsets and such that and . Furthermore, and must be open intervals themselves, say and . We must have that . So or and furthermore, or .
If then this implies that (since if then which implies that ). So if . If then and so so . If then and so . If then . Either way we see that .
We can use the same logic for the other cases which will completely show that is connected.
Basic Theorems Regarding Connected and Disconnected Metric Spaces
A metric space is said to be disconnected if there exists , where and:
(1)
We say that is connected if it is not disconnected.
Furthermore, we say that is connected/disconnected if the metric subspace is connected/disconnected.
We will now look at some important theorems regarding connected and disconnected metric spaces.
Theorem 1: A metric space is disconnected if and only if there exists a proper nonempty subset such that is both open and closed.
- Suppose that is disconnected. Then there exists open , , where and .
- Since is open in we have that is closed in . But is also open. Similarly, since is open in , is closed in . So in fact and are both nonempty proper subsets of that are both open and closed.
- Suppose that there exists a proper nonempty subset such that is both open and closed. Let . Then is also both open and closed. Furthermore, since and . Additionally, , so is disconnected.
Theorem 2: If is a connected unbounded metric space, then for every and for all , is nonempty.
- Proof: Let be a connected unbounded metric space and suppose that there exists an and there exists an such that:
(2)
- We will show that a contradiction arises. Let and let . Then is open since it is simply an open ball centered at . Furthermore, is open since is a closed ball centered . is nonempty since and is nonempty since is unbounded (if it were empty then this would imply is bounded). Clearly and . So is a disconnected metric space. But this is a contradiction.
- Therefore the assumption that there exists an and an such that was false.
- So for all and for all the set is nonempty.
Licensing
Content obtained and/or adapted from:
- Connected And Disconnected Metric Spaces, mathonline.wikidot.com under a CC BY-SA license
- Basic Theorems Regarding Connected and Disconnected Metric Spaces, mathonline.wikidot.com under a CC BY-SA license