The following exercise was in the Elements of Real Analysis by Bartle.
Give an example of a set $A$ in $\Bbb R^p$ such that $A^{\circ} = \emptyset$ and $A ^ - = \Bbb R^p$. Can such a set be countable?
$A^{\circ}$ and $A ^ -$ respectively stand for the interior and closure of $A$.
The question "is it countable" prompted me to consider $\Bbb Q^p$ - the set of all rational points in $\Bbb R^p$ with rational coordinates. I came up with the following justification but am a little iffy about it. I need some help in verifying whether it is correct please.
Argument 1: $A^{\circ} = \emptyset$:
Suppose $B \subseteq \Bbb Q^p$ is an open set. Suppose it is non-empty. Then $\exists x = (x_1, x_2,...., x_p) \in B$. Since B is open there must exist an open ball $G_x$ centred at $x$ entirely contained in $B$. Suppose such a ball exists with radius $r_x \gt 0$. Then $y = (x_1 + \frac {r_x}{\sqrt 2}, x_2, ..., x_p) \in G_x $ but $ y \not \in B$. There can be no such open ball centred at $x$. Therefore there can be no element $x \in B \implies B = \emptyset$.
Argument 2: $A^{-} = \Bbb R^p$:
Now suppose there is a closed set $C \subseteq \Bbb R^p$ which contains $\Bbb Q^p$. Then $C^c$ is open and does not contain $\Bbb Q^p$. Suppose $\exists z = (z_1, z_2,.., z_p) \in C^c$. As before there must be an open ball $G_z$ entirely contained in C with radius $r_z \gt 0$. Then if $y = (y_1, y_2, ..., y_p)$ where $y_i$ is a rational number in the interval $(z_i - \frac {r_z}{\sqrt p}, z_i + \frac {r_z}{\sqrt p})$ then $y \in G_z$. But this is absurd since $y \in \Bbb Q^p$. Therefore $C^c = \emptyset \implies C = \Bbb R^p$
Q1: Is it alright to say that since a complement of a subset of $\Bbb R^p$ is empty then the original set is $\Bbb R^p$?
Q2: Is it alright to prove that a set is empty by assuming it has one element and then forming a contradiction?
Q3: Even so are the above arguments good enough?
Please comment. Would be grateful for any help provided. Thanks in advance.
[I am using the standard Euclidean norm in defining balls in $\Bbb R^p$]
Yes; $\mathbb Q^p$ does the job. Maybe it will also be helpful to show that Cl$A \times B$=Cl($A)$$\times$Cl$(B)$, after showing Cl$\mathbb Q=\mathbb R$. To show $\mathbb Q$ has empty interior, you can also use decimal representations to show that between any two rationals, there is an irrational: given $a,b$ Rationals; $a=a_0.a_1a_2.... ; b=b_0.b_1b_2....$, with $a<b$ , there is a finite $k$ so that $a_k<b_k$. Then create the expansion of a term $c; a<c<b , $from $a_{k+1}$ on, so that there is no periodic repetition of the terms, i.e., $c:=a_0.a_1a_2....a_ka_{k+1'}a_{k+2'}....$ , so that the terms $a_{k+j}$ do not have any period. You can, e.g., attach the decimal expandion of $\pi$ beyond the $(k+1)-st $ term.