proving homeomorphism Is there a homeomorphism between $\mathbb{R}^{2} $
  and the set  $ A=\left\{ \left(x,y\right)\in\mathbb{R}^{2}:x^{2}+y^{2}<1\right\} $
  ?
And what if $A=\left\{ \left(x,y\right)\in\mathbb{R}^{2}:x^{2}+y^{2}\leq1\right\} $? 
At the second part I think there is no homeomorphism  since the closed ball is compact space and  $\mathbb{R}^{2} $ isn't. But Im not sure about my arguments. :S 
 A: Have you seen that the interval $(0,1)$ is homeomorphic to the real line? If not, try to prove that. Once we know that, we know we can perform a continuous stretching of that interval into any line.
Imagine the open unit disk (the set A) drawn in the center of a cartesian plane. Fix the center, consider each of the rays out of the center of the circle as rotations of the interval (0,1). We can apply the same continuous stretching as before to each of these rays, and we end up with the entire plane. So that is a homeomorphism from the open unit disk to the plane.
Your reasoning for the second one is correct, but lets remind ourselves why. Homeomorphisms are functions which preserve the topology between two spaces: open sets in one correspond to open sets in the other. Suppose there was a homeomorphism between the plane and the closed unit disk. The plane is not compact, so there is some open cover with no finite subcover. Now take the images of all the sets in that cover. This would produce an open cover on the closed unit disk with no finite subcover, which is impossible. 
A: For the first set, try $(x,y)\mapsto \left( x\cdot \frac{1}{\sqrt{x^2+y^2}+1},y\cdot \frac{1}{\sqrt{x^2+y^2}+1}\right)$.
For the second part, your argument is correct. 
A: For the first try $f:\mathbb{R}^2\longrightarrow A, \ \ f(x,y)=f(re^{i\theta})=\left(\dfrac{2}{e^r+1}-1\right)e^{i\theta}, \ r \in \mathbb{R}, \ \theta \in [0,\pi)$.
(The function $g:\mathbb{R}\longrightarrow (-1,1), r\mapsto \left(\dfrac{2}{e^r+1}-1\right)$ is homeomorphism).

For the second you are right.
