Let $f:ℝ²→ℝ²$ be an affine, one-to-one and continuous function. Assuming that there exist a compact set $D$ such that $f(D)∩D$ contain an infinite, closed and bounded set $C$.
My question is: Can we deduce that there exist a non empty compact set $A$ such that $f(A)⊂A$. If not can one add some assumptions to get this result.
I have no idea to start.
How about the following?
Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be given by $f(x,y) = (x,y)+(1,0)$. This function is certainly continuous and one-to-one. Also let $D$ be the closed unit ball centred at the origin with radius $2$. Then $f(D)$ is the closed unit ball centred at $(1,0)$ of radius $2$ and therefore $f(D)\cap D$ is an infinite closed and bounded set. However $f(A)\subset A$ is impossible for a closed and bounded set.
Suppose that $(\xi,\gamma)\in f(A)$ implies that $(\xi,\gamma)\in A$ then $(\xi+1,\gamma)\in A$ for every $(\xi,\gamma)\in A$ but then $A$ must be infinite.
If you want something linear consider $f(x,y) = 2(x,y)$ then for $D$ equal to the closed unit ball we find that $f(D)\cap D = D$. Also if $(\xi,\gamma)\in f(A)$ implies that $(\xi,\gamma)\in A$ then $(2x,2y)\in A$ for every $(x,y)\in A$ which implies that $A$ is infinite since $(2^nx,2^ny)\in A$ for every $n$.