Limits and continuity. I need help with these exercises of Analysis about limits and continuity.


*

*Construct a set $ A \subset  [0,1] \times [0,1]$ such that $A$ has at most one point en each horizontal line and one in each vertical line and $ \partial A = [0,1] \times [0,1]. $

*Consider a transformation $ f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n} $ which satisfies the following properties: (i) $f(K) $is closed and bounded for each $K$ closed bounded subset of $\mathbb{R}^{n}. $  (ii) If $ (K_{s})_{s=0}^{ \infty} $ is a decreasing sequence (i.e. $K_{0} \supseteq K_{1}  \supseteq K_{2} \ldots$) of bounded closed subsets of $\mathbb{R}^{n},$ then 
$f(\bigcap_{s=0}^{ \infty} K_{s}) = \bigcap_{s=0}^{ \infty} f(K_{s}).$
Prove that $f$ is continuous.
 A: *

*Let $h: \mathbb R^2\to\mathbb R^2$ be the rotation given by $$(x,y)\mapsto(x\cos1-y\sin1,x\sin1+y\cos1).$$ The set $\mathbb Q^2$ is dense in $\mathbb R^2$. Since $h$ is a homeomorphism, $h(\mathbb Q^2)$ must also be dense in $\mathbb R^2$. Therefore $A=h(\mathbb Q^2)\cap[0,1]^2$ is dense in $[0,1]^2$. We only have to show that $A$ has the required property. First, we show that $h(\mathbb Q^2)\cap(\{x\}\times\mathbb R)$ has at most one element for each $x\in\mathbb R$. So, let $(p_1,p_2),(q_1,q_2)\in\mathbb Q^2$ be such that $$p_1\cos1-p_2\sin1=q_1\cos1-q_2\sin1.$$ Then $(p_1-q_1)\cos1=(p_2-q_2)\sin1$. This implies that $p_1-q_1=p_2-q_2=0$, since $\{\cos1,\sin1\}$ is linearly independent over $\mathbb Q$. (This follows from the fact that $1$ is an irrational multiple of $2\pi$.) The proof that $h(\mathbb Q^2)\cap(\mathbb R\times\{x\})$ has at most one element, follows the same idea.

*Suppose $(x_j)_{j\in\mathbb N}$ is a convergent sequence in $\mathbb R^n$ and $x$ its limit. Define compact sets $K_m =\{x_j|\;j\geq m\}\cup\{x\}$. The sequence $(K_m)_{m\in\mathbb N}$ is decreasing and $\bigcap_{m\in\mathbb N}K_n=\{x\}$. By the assumptions, this implies that $f(\{x\})=\bigcap_{m\in\mathbb N}f(K_m)$. This implies that $$\lim_{m\to\infty}f(x_m) = f(x).$$ To see this, note that $(f(K_m))_{m\in\mathbb N}$ is a decreasing sequence of compact sets. There are now two possible cases for each $m\in\mathbb N$: either $f(K_m)\setminus f(K_{m+1})$ is empty or it contains precisely one element: $f(x_m)$. In the first case, $f(x_m)$ is either equal to $f(x)$ or is an isolated point of $f(K_m)$: this is because $f(x_m)=f(x_k)$ is true at most for finitely many $k\in\mathbb N$ if $f(x_m)\neq f(x)$ and therefore for the maximal $k$ with this property, we have that $f(K_k)\setminus f(K_{k+1})$ contains precisely one element: $f(x_k)$. This element must be an isolated point of $f(K_k)$, since otherwise $f(K_{k+1})$ wouldn't be closed. In the second case, $f(x_m)$ must again be an isolated point of $f(K_m)$, for the same reason. This means that $f(K_1)$ is a compact set which consists of isolated points and $f(x)$. This implies that the sequence $(f(x_m))_{m\in\mathbb N}$ converges to $f(x)$: since $f(K_1)$ is compact, all accumulation points of this sequence must be elements of $f(K_1)$. But since $f(x)$ is the only non-isolated point of the set, it must be the only accumulation point of the sequence. Since the sequence is bounded, this accumulation point must be the limit of the sequence. We have shown that for each convergent sequence $x_j\to x$ in $\mathbb R^n$, $$\lim_{j\to\infty}f(x_j)=f(x)$$ holds. Therefore $f$ is continuous.

