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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 $\Bbb R^p$]

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    $\begingroup$ Your reasoning is, in principle, correct, while some formulations are not. It would be helpful in case of such a long text if you'd add some marker in the text which makes it easier to refer to them in comments. Example: 'Is it alright to prove a set is empty by assuming it has one elemenent..' - no, it is not. You may assume, though, it has 'at least one element' -- which is what you actually did earlier. (Do you see the difference?) (Some people are suspicious in general regarding proofs by contradiction, is it up to you how you deal with this ;-) $\endgroup$ – Thomas Feb 16 '14 at 7:53
  • $\begingroup$ @Thomas What other formulations do you think are incorrect? $\endgroup$ – Ishfaaq Feb 16 '14 at 8:01
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    $\begingroup$ @Thomas OP's shown the second part, and it's fairly common to cover the cardinality of the rationals (and finite products thereof) before delving into topology. In all, OP's work seems good to me. $\endgroup$ – Jonathan Y. Feb 16 '14 at 8:19
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    $\begingroup$ @Ishfaaq certainly: yes to all (modulu Thomas's remark regarding the distinction between 'an element' and 'one element'). $\endgroup$ – Jonathan Y. Feb 16 '14 at 8:34
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    $\begingroup$ @Jonathan Y.: Thanks loads. Greatly appreciated.. $\endgroup$ – Ishfaaq Feb 16 '14 at 8:36
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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.

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  • $\begingroup$ Thanks for the suggestion. But I'm actually okay with dealing with the multiple dimensions. My issue is with the empty set its complement and using multiple levels of assumptions before arriving at a contradiction. That was what I was unsure about. I can do this as an exercise though. Thanks for suggesting it.. $\endgroup$ – Ishfaaq Feb 16 '14 at 8:14
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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 $C^c = \emptyset \implies C = \Bbb R^p$

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