Difference between revisions of "Connectedness"

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<div style="text-align: center;"><math>\\begin{align} \quad \{ x \in M : d(x, a) = r_0 \} = \emptyset \end{align}</math></div>
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<div style="text-align: center;"><math>\begin{align} \quad \{ x \in M : d(x, a) = r_0 \} = \emptyset \end{align}</math></div>
 
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<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.

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