In the derivation of a cycloid - problem in Stewart

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The discussion centers on the derivation of a cycloid, specifically the relationship between the arc PT and the line segment OT. Participants debate whether these two are homeomorphisms, with some asserting that they are not, as homeomorphisms are functions rather than mere sets of points. The concept of isomorphism is also discussed, with clarification that it applies to algebraic structures and not to the curves in question. There is a consensus that while arc PT and OT can be considered homeomorphic, the terminology used must be precise to avoid confusion. Overall, the conversation emphasizes the importance of careful language in mathematical discussions.
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In the derivation of a cycloid, it says (pg. 655 Stewart, 5th ed) |OT| = \text{arc} \ PT = r \theta.

I was just wondering if if this implies that \text{arc}\ PT and the line segment OT are homeomorphisms?

When I first learned about cycloids, I didn't know that homeomorphisms were. But now, having learned about them, it seems that they are in fact homeomorphisms.
 
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i would go even further and say they are isomorphic, they're both just curves
 
Careful - you've not said what you mean by isomorphic. Sure, as sets they are, as topological spaces, but not all curves are isomorphic in all categories (eg, rational curves with rational maps).
 
I would say you need to be more careful with your use of words! No, arc PT and line segment OT are not homeomorphisms because hemeomorphisms are functions, not sets of points. It is true, of course, that they are homeomorphic.

My understanding of "isomorphism" is that it is a relationship between algebraic structures. In you do not have operations on the set, then "isomorphic" does not apply.
 
I would say you need to be more careful with your use of words! No, arc PT and line segment OT are not homeomorphisms because hemeomorphisms are functions, not sets of points. It is true, of course, that they are homeomorphic.

My understanding of "isomorphism" is that it is a relationship between algebraic structures. In you do not have operations on the set, then "isomorphic" does not apply.
 
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