Find a nonempty proper subset, $A$ of $\mathbb{R} \times \mathbb{R}$ with the following characteristics

Question

Suppose $(X,d)=(\mathbb{R} \times \mathbb{R}, L^\infty)$

Find a nonempty proper subset, $A$ of $\mathbb{R} \times \mathbb{R}$ with the following characteristics:

• The set $A$ is infinite.
• The complement of the set $\mathbb{R^2} \setminus A$ is infinite.
• The set $A$ is open.

Definitions I'm using: A set $B$ is infinite if there exists an injection $f:\mathbb{N} \to B$.

The subset $A \subseteq X$ is open if $\forall a \in A, \exists \varepsilon \in (0,\infty) \ni B_\varepsilon(a) \subseteq A$.

In a metric space $(X,d)$, a subset $U \subseteq X$ is open iff its complement $U^c$ is closed, where $U^c:=X \setminus U$.

I know that a set $S$ is closed if $(a_n)$ is convergent $\in S$, then $(a_n) \to x$, where $x \in S$.

But I'm not sure which set would work here. Would appreciate any suggestions!

Edit

Let $C=\mathbb{R} \times \mathbb{Z}$, and $A=C^c=\mathbb{R} \times (\mathbb{R} \setminus \mathbb{Z})$.

To show that $A$ and $C$ are infinite, define $f_A: \mathbb{N} \to A$ by $n \mapsto (n,\frac{1}{2})$, and $f_C: \mathbb{N} \to C$ by $n \mapsto (n,1)$.

Suppose $f_A(a)=f_A(b)$. Then $(a,\frac{1}{2})=(b,\frac{1}{2}) \iff a=b$. Thus $f_A$ is an injection.

Suppose $f_C(a)=f_C(b)$. Then $(a,1)=(b,1) \iff a=b$. Thus $f_C$ is an injection.

Therefore, $A$ and $C$ are infinite.

Note that $B_{\varepsilon}(a)=\{y \in \mathbb{R} \times \mathbb{R} | d((y_1,y_2),(a_1,a_2))<\varepsilon\}$

To prove that $A$ is open, let $a \in A$. Then $a=(a_1,a_2)$, where $a_1 \in \mathbb{R}$ and $a_2 \in \mathbb{R} \setminus \mathbb{Z}$. Choose $\varepsilon:=\min(a_2-\lfloor a_2\rfloor, 1+\lfloor a_2\rfloor-a_2)$, so that $y \in B_{\varepsilon}(a) \iff d((y_1,y_2),(a_1,a_2))=max(|y_1-a_1|,|y_2-a_2|)<min(a_2-\lfloor a_2\rfloor, 1+\lfloor a_2\rfloor-a_2)$.

As $y \in \mathbb{R} \times \mathbb{R}, y=(y_1,y_2)$ with $y_1,y_2 \in \mathbb{R}$. Thus $B_{\varepsilon}(a) \subseteq A \iff y_2 \in \mathbb{R} \setminus \mathbb{Z}$.

Since $\varepsilon$ is either $a_2-\lfloor a_2\rfloor$ or $1+\lfloor a_2\rfloor-a_2$, consider the following two cases:

Case 1: If $\varepsilon=a_2-\lfloor a_2\rfloor$, then $|y_2-a_2|<a_2-\lfloor a_2\rfloor$. Now, $a_2 - \lfloor a_2 \rfloor \in [0,1)$, so $0 \le a_2- \lfloor a_2 \rfloor <1$. But $a_2 \in \mathbb{R} \setminus \mathbb{Z}$, so $a_2-\lfloor a_2 \rfloor \in (0,1)$. Thus $|y_2-a_2|=0 \iff y_2=a_2 \iff y_2 \in \mathbb{R} \setminus \mathbb{Z}$.

Case 2: If $\varepsilon=1+\lfloor a_2\rfloor-a_2$, then $|y_2-a_2|<1+\lfloor a_2\rfloor-a_2<a_2-\lfloor a_2 \rfloor \Rightarrow |y_2-a_2|<a_2-\lfloor a_2 \rfloor$. Thus $|y_2-a_2|=0 \iff y_2=a_2 \iff y_2 \in \mathbb{R} \setminus \mathbb{Z}$.

In either case, $y_2 \in \mathbb{R} \setminus \mathbb{Z}$.

Thus $y \in A$ for $d(a,y)<\varepsilon$, and hence $\forall a \in A, \exists \varepsilon>0 \ni B_{\varepsilon}(a) \subseteq A$. Therefore, $A$ is open.

Answer 1

Let $C=\mathbb R\times \mathbb Z$ and take $A:=C^c$. Then $A$ is the union of long stripes.

We have to prove $A$ and $C$ are infinite. This is true, just take $f_A\colon \mathbb N\to A, n\mapsto (n,\frac 12)$ and $f_C\colon \mathbb N\to C, n\mapsto (n,1)$.

It is left to show, that $A$ is open. Let $a\in A$ be arbitrary. We have $a_2\notin \mathbb Z.$ When we choose $\varepsilon :=\min(a_2-\lfloor a_2\rfloor, 1+\lfloor a_2\rfloor-a_2)$, can you prove $B_{\varepsilon}(a)\subset A$?