Show that the unit circle is connected

[Demon117]

1. Show that S:= {(x,y)an element of R^2 : x^2 + y^2 =1} is connected.



2. Relevant theorems

1. Path-connected implies connected.

The Attempt at a Solution

Define f: [0,2pi] --> R^2 by f(x) = (cos(x),sin(x)).
This map is continuous, and its image is S^1. The interval [0,2pi] is connected, so its image is as well; which means S is path connected. Hence, by theorem S is connected.

Does this proof make sense, what else should I include?

 

[lanedance]

that looks reasaonable to me

note the paramterisation you've given can also be used to show there is a path within the space from any point to another

 

[Demon117]

that looks reasaonable to me

note the paramterisation you've given can also be used to show there is a path within the space from any point to another

That is a nice feature.

If f:S-->R was a continuous function, how might you show that f(x)=f(-x) for some x in S?

Here are my thoughts:

Since f is continuous, let epsilon = 1. Then for p, (-p) in S we have |f(p)-f(-p)|<1. Thus -1<f(p)-f(-p)<1 so that -1+f(-p)<f(p)<1+f(-p). It follows that f(p)>-1+f(-p) and f(p)<1+f(-p). This suggests that f(p)<f(-p) and f(p)>f(-P). I must show that these inequalities define open sets. But I am unsure how to proceed. Any suggestions?