# is the set A={ (x,y)∈R² : x²+ y² =1} connected?

• May 21st 2011, 11:21 AM
sorv1986
is the set A={ (x,y)∈R² : x²+ y² =1} connected?
is the set A={ (x,y)∈R² : x²+ y² =1} connected? how do we prove it?
is there any general proof? or some quick method by which we can tell any set is connected or not
• May 21st 2011, 11:26 AM
Tinyboss
It's a subset of the plane, so you can visualize it or draw a picture. What does it look like?
• May 21st 2011, 11:27 AM
sorv1986
Quote:

Originally Posted by Tinyboss
It's a subset of the plane, so you can visualize it or draw a picture. What does it look like?

looks like a unit circle centred at origin
• May 21st 2011, 11:50 AM
Tinyboss
And is that connected, under the subspace topology inherited from R^2? Here's a hint: a continuous image of a connected set is connected. Can you express the unit circle as the continuous image of some set you know is connected?
• May 21st 2011, 12:11 PM
sorv1986
Quote:

Originally Posted by Tinyboss
And is that connected, under the subspace topology inherited from R^2? Here's a hint: a continuous image of a connected set is connected. Can you express the unit circle as the continuous image of some set you know is connected?

If f : [0,1] -> R2 defined by f(t) = (cost, sint) then the map is continuous and the closed interval [0,1] is connected. So the image of f should be connected.and i think the image of f is exactly the unit circle centerd at origin. is it right?

but is there any general method for proving or disproving any subset of R2 connected?
• May 21st 2011, 12:13 PM
Tinyboss
I don't know of a general way.
• May 21st 2011, 02:28 PM
tonio
Quote:

Originally Posted by sorv1986
If f : [0,1] -> R2 defined by f(t) = (cost, sint) then the map is continuous and the closed interval [0,1] is connected. So the image of f should be connected.and i think the image of f is exactly the unit circle centerd at origin. is it right?

but is there any general method for proving or disproving any subset of R2 connected?

Well, yes, but it isn't that easy to implement: a space X is NOT connected iff there is a

continuous surjection $X\rightarrow \{0,1\}$ , with the second set with the inherited

topology from the reals (is thus a discrete space).

Tonio