Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

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## #1 2008-07-05 16:13:17

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

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## #2 2008-07-06 03:59:24

Ricky
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Registered: 2005-12-04
Posts: 3,791

### Re: Path-connected subspaces

I don't have your book, so in order to be a full proof, you'd also need to show:

1. ~ is an equivalence relation (May be proved in your book, if so you should reference the proof)
2. An equivalence class is path connected (1 line, maybe less)
3. State that equivalence classes are disjoint.  (Literally, "Equivalence classes are disjoint").

In a proper proof, you should be explicitly using each hypothesis and explicitly concluding each conclusion.  Certainly in your proof, the logical steps for 2 and 3 are so obvious to any one who has studied path connected spaces that their brain immediately makes the leap.  But that does not mean that you explicitly saying so should not be there, it needs to be.

As an algebra professor of mine once said, "The proof is correct, modulo details."

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #3 2008-07-06 05:09:43

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: Path-connected subspaces

I didnt take the result from any book  I discovered it myself!

Youre right in that the part on equivalence relation should not have been taken for granted. But I already knew that symmetry follows from the existence of inverse paths and transitivity from the definition of product paths. In addition, continuity of product paths follows from a result in general topology sometimes called the pasting lemma. So, yes. I shouldnt have glossed over the equivalence-relation bit, but ought to have spent a few more lines explaining why its an equivalence relation.

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