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

1. ## 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

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

3. 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

4. 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?

5. 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?

6. I don't know of a general way.

7. 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 $\displaystyle X\rightarrow \{0,1\}$ , with the second set with the inherited

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

Tonio