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==Every path-connected space is connected== |
==Every path-connected space is connected== |
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Let ''S'' be path-connected and suppose, for contradiction, that ''S'' is not connected. Then <math>S = A \cup B</math> for nonempty disjoint [[Open (topology)|open]] sets ''A'' and ''B''. Let <math>x \in A, y \in B</math>. Since ''S'' is path-connected, there exists a [[continuous]] path <math>\gamma: [0, 1] \to S</math> with <math>\gamma(0) = x</math> and <math>\gamma(1) = y</math>. |
Let ''S'' be path-connected and suppose, for contradiction, that ''S'' is not connected. Then <math>S = A \cup B</math> for nonempty disjoint [[Open (topology)|open]] sets ''A'' and ''B''. Let <math>x \in A, y \in B</math>. Since ''S'' is path-connected, there exists a [[continuous]] path <math>\gamma: [0, 1] \to S</math> with <math>\gamma(0) = x</math> and <math>\gamma(1) = y</math>. |
Revision as of 00:47, 8 September 2008
Every path-connected space is connected
Let S be path-connected and suppose, for contradiction, that S is not connected. Then for nonempty disjoint open sets A and B. Let . Since S is path-connected, there exists a continuous path with and .
Since f is continuous, and are open subsets of [0, 1]. Moreover and are nonempty since they contain 0 and 1, respectively. They are disjoint since A and B are disjoint, and , so [0, 1] is not connected, a contradiction.
Therefore S is connected.
A locally path-connected space is path-connected if and only if it is connected
A path-connected space is always connected. We therefore focus on the converse.
Let S be connected and locally path-connected. We for points a and b in S, we denote by the relation a ~ b that there exists a path between a and b.
Lemma 1: ~ is an equivalence relation
We check:
- Reflexivity: a ~ a, as evidenced by the trivial path .
- Symmetry: Suppose a ~ b, and let be the associated path between a and b. Then defines a continuous path from b to a, and thus b ~ a.
- Transitivity: Suppose a ~ b, with associated path , and b ~ c, with associated path . Then let
is then a continuous path from a to c, so a ~ c.
Lemma 2: For a point a in S, the equivalence class [a] is open
Let . Then since S is locally path-connected, there is a neighborhood U of p so that, for every , there is a path from p to q. But then [q] = [p] = [a], so and p is an interior point of [a]. Hence [a] is open.
Lemma 3: [a] is closed
Let C denote the complement of [a]. Then
The union of open sets is open, and each term of the union is open by Lemma 1, so C is open, and hence [a] is closed.
Proof of theorem
Let x and y be two points of S. Then by Lemmas 1 and 2, [x] is clopen, so since S is connected, [x] = S. Hence , and there exists a path between x and y, and S is path-connected.