Showing that open balls in $\mathbb{R}^2$ are homeomorphic. 
Show that open balls in $\mathbb{R}^2$ are homeomorphic.

I'm trying to see if this works with examples. Suppose I have the open balls $ B_s = \{(x,y) \in \mathbb{R}^2 \mid (x-1)^2+(y-1)^2 < \frac12 \}$ and $B_l = \{(x,y) \in \mathbb{R}^2 \mid (x-5)^2+(y-5)^2 < 1 \}. $
If I define $t: \mathbb{R^2} \to \mathbb{R}^2, x \mapsto x+a$, where $a$ is a translation vector.
I would need have that all points of the smaller ball would map to the larger one. If I pick $a = ((1,1),(5,5))$ so the vector is the one shifting the center of the smaller ball to be in the center of the larger one. Now take $x = ((0,0),( \frac12,1)) \in B_s$. Applying $t$ to $x$ I have $$t(x) = ((0,0),(\frac12,1)) + ((1,1),(5,5)) = ((1,1),(\frac{11}{2},6)) \in B_l$$ so the point $x$ gets mapped correctly to the larger ball. However picking $x' =((0,0),(\frac32,1))$ I have $$t(x')=((0,0),(\frac{3}{2},1))+((1,1),(5,5)) = ((1,1), (\frac{13}{2},6)) \notin B_l.$$
What might be happening here? Is it the case that $t$ is not a suitable homeomorphism?
 A: $f(x)=rx+p$ maps $B(0,1)$ homeomorphically to the ball $B(p,r)$, where $p \in \Bbb R^2$ and $r>0$. So all open balls are mutually homeomeorphic. Works in any $\Bbb R^n$ BTW.
A: If you have two open balls $B(p,r)$ and $B(q,s)$, the idea is to first move $B(p,r)$ to be centered at the origin with translation $x\mapsto x - p$. Now, you can scale the resulting ball to have the radius $s$ with scaling $x\mapsto \frac sr x$. Finally, move the ball to be centered at $q$ with translation $x\mapsto x + q$. If we compose all these maps we get $$f(x)= \frac sr(x-p) + q.$$
This map is bijective since you can just change the symbols to conclude the inverse is $$g(x)= \frac rs(x-q)+p.$$
It's a rather quick check that these are really inverses of each other.
Let us show that $f(B(p,r)) = B(q,s)$ and $g(B(q,s))=B(p,r)$.
So, let $x\in B(p,r)$, i.e. $\lVert x - p \rVert < r$. We want to show that $f(x) \in B(q,s)$, i.e. $\lVert f(x) - q \rVert < s$:
\begin{align}\lVert f(x) - q \rVert < s &\iff \lVert (\frac sr(x-p) + q) - q \rVert < s \\ &\iff \frac sr \lVert x - p \rVert < s \\ &\iff \lVert x-p \rVert < r.\end{align}
This shows $f(B(p,r)) \subseteq B(q,s)$. Analougously $g(B(q,s)) \subseteq B(p,r).$ Combine the two to get $$ f(B(p,r)) \subseteq B(q,s) \implies g(f(B(p,r))) \subseteq g(B(q,s)) \implies B(p,r) \subseteq g(B(q,s))$$ and analougusly $B(q,s) \subseteq f(B(p,r))$. This finishes the proof that $f,g$ are mapping $B(p,r)$ and $B(q,s)$ to each other.
Continuity of $f$ and $g$ follows since they are compositions of continuous functions. I leave that as an exercise.
Notice that the argument is general and works in any $\mathbb R^n$.
