is path connected if is countable
Theorem 1.
Let be a countable subset of . Then is path connected.
We use simply as an example; an analogous proof will work for any .
Proof.
Fix a point not in . The strategy of the proof is to construct a path from any to . If we can do this then for any we may construct a path from to by first following and then following in reverse.
Fix , and consider the set of all (straight) lines through . There are uncountably many of these and they meet in the single point , so not all of them contain a point of . Choose one that doesn’t and move along it: your distance from takes on uncountably many values, and hence at some point this distance from is not shared by any point of . The whole of the circle with radius , centre , lies in so we may move around it freely.
Corollary 1.
Let be continuous and onto. Then is uncountable.
Proof.
Suppose that is countable. can be written as the disjoint union
where the last two sets are open (as is continuous), non-empty (as is onto) and disjoint. Since pathwise connected is the same as connected for Hausdorff spaces, we have that is not path connected, contradicting the theorem. ∎
Title | is path connected if is countable |
---|---|
Canonical name | mathbbR2setminusCIsPathConnectedIfCIsCountable |
Date of creation | 2013-03-22 16:09:11 |
Last modified on | 2013-03-22 16:09:11 |
Owner | silverfish (6603) |
Last modified by | silverfish (6603) |
Numerical id | 5 |
Author | silverfish (6603) |
Entry type | Theorem |
Classification | msc 54D05 |