Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I stumbled upon this excerpt as I was reading Graph Theory by Reinhard Diestel:

A polygon is a subset of $\mathbb{R}^2$ which is the union of finitely many straight line segments and is homeomorphic to the unit circle $S^1$, the set of points in $\mathbb{R}^2$ at distance 1 from the origin.

So based on this, how could any polygon be homeomorphic to $S^1$ even though both sets are of different cardinality?

Pardon me if the question is too basic; I'm totally new to topology and I probably am overlooking a detail.

share|improve this question
    
Why do the two sets have different cardinality ? –  Amr Dec 5 '12 at 21:20
    
Why do you think the sets have different cardinality? A polygon is the set of points defined by the union of the sets that are defined by line segments, not the set of line segments. –  Jason Polak Dec 5 '12 at 21:20
2  
OK, why are you assuming it? –  Chris Eagle Dec 5 '12 at 21:24
5  
In that case, the first thing you should do is find a bijection between a line segment like $[0,1)$ and $S^1$. –  Jason Polak Dec 5 '12 at 21:26
1  
I just realized what my mistake was.. Thanks a lot :) –  Khaled Nassar Dec 5 '12 at 21:28

1 Answer 1

up vote 2 down vote accepted

I'll give a conceptual answer to your question. Two curves are homeomorphic if the first can be continuously deformed into the second. Intuitively, if you had a circular loop of wire, you could hammer and bend it into the shape of any polygon (without breaking the wire or adding any new connections).

The cardinality of $S^1$ is the same as the cardinality of any line segment, which in turn has the same cardinality as any finite union of line segments. Cardinality of a set is different from "length" or "measure". For example, though $[1,2]$ is a proper subset of $[0,3]$, the two sets have the same cardinality.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.