Exercise 17.20

(a) It is easy to show that $A$ is closed so that $\overline{A}=A$, and that also $\overline{ℝ-A}=A$ so that $\mathrm{Bd}A=A$. It is also easy to see that no basis element and therefore no neighborhood of any point in $A$ is contained entirely within $A$. From this it follows that $\mathrm{Int}A=\varnothing$.

(b) It is easy to show that $B$ is open so that $\mathrm{Int}B=B$. It is likewise not difficult to prove that $\overline{B}=\left\{x\times y\mid x\ge 0\right\}$. We then have from Exercise 17.19 part (c) that $\mathrm{Bd}B=\overline{B}-B=\left\{x\times y\mid x=0\right\}\cup \left\{x\times y\mid x>0\text{ and }y=0\right\}$.

(c) Here we have that $C=A\cup B=\left\{x\times y\mid y=0\right\}\cup \left\{x\times y\mid x>0\right\}$. It is then easy to show that the closure is $\overline{C}=\left\{x\times y\mid y=0\right\}\cup \left\{x\times y\mid x\ge 0\right\}$. We also have that $ℝ-C=\left\{x\times y\mid x\le 0\text{ and }y\ne 0\right\}$ so that $\overline{ℝ-C}=\left\{x\times y\mid x\le 0\right\}$. From these we clearly then have

It is also not difficult to show that $\mathrm{Int}C=\left\{x\times y\mid x>0\right\}$.

(d) Clearly we have that $\overline{D}$ is all of ${ℝ}^2$ as a consequence of the fact that the rationals are order-dense in the reals. Also, since any neighborhood of any point in $D$ will intersect a point $x\times y$ with irrational $x$, it follows that no point of $D$ is in its interior. Thus, $\mathrm{Int}D=\varnothing$ so that $\overline{D}=\mathrm{Int}D\cup \mathrm{Bd}D=\varnothing \cup \mathrm{Bd}D=\mathrm{Bd}D$ by Exercise 17.19 part (a), and hence $\mathrm{Bd}D=\overline{D}={ℝ}^2$.

(e) It should be fairly obvious by this point that

and $\mathrm{Int}E=\left\{x\times y\mid 0<x^2-y^2<1\right\}$. This would be easy but tedious to prove rigorously.

(f) First we clearly have that $\mathrm{Int}F=\left\{x\times y\mid x\ne 0\text{ and }y<1∕x\right\}$. We also have that $\overline{F}=\left\{x\times y\mid x=0\right\}\cup \left\{x\times y\mid x\ne 0\text{ and }y\le 1∕x\right\}$. By Exercise 17.19 part (a) we have that $\overline{F}=\mathrm{Int}F\cup \mathrm{Bd}F$ and that $\mathrm{Int}F\cap \mathrm{Bd}F=\varnothing$ so that $\mathrm{Bd}F=\overline{F}-\mathrm{Int}F$. Thus we have that $\mathrm{Bd}F=\left\{x\times y\mid x=0\right\}\cup \left\{x\times y\mid x\ne 0\text{ and }y=1∕x\right\}$. Again these facts are not difficult to show rigorously but would be tedious.