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Which of the following sets are dense in $\Bbb R^2$ with respect to the usual topology.

  1. $\{(x,y) \in\mathbb{R}^2:x\in \mathbb{N}\}$
  2. $\{(x,y) \in\mathbb{R}^2:x+y \text{ is a rational number}\}$
  3. $\{(x,y) \in\mathbb{R}^2:x^2+y^2=5\}$
  4. $\{(x,y) \in\mathbb{R}^2:xy\neq 0\}$

Clearly 1 is false.
3 is false as it is bounded and closed
4 is true as it is the set of all points that are not on the axes x and y.
Am I correct.
But I am not sure about 2 but my guess is true as rationals/ irrationals are dense and 2 holds iff either both are rational or conjugate irrational .

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Looks good to me. –  Clayton Dec 17 '12 at 5:51
I assume that the $\mathbb{R}$ in the title is intended to be $\mathbb{R}^2$. –  André Nicolas Dec 17 '12 at 5:59
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2 Answers

Let $(a,b)$ be a point in $\mathbb{R}^2$, and let $\epsilon\gt 0$. We show that there exist rational numbers $s,t$ such that $\sqrt{(a-s)^2+(b-t)^2}\lt \epsilon$. In particular, $s+t$ is rational.

Since the rationals are dense in $\mathbb{R}$, there is a rational $s$ such that $|a-s|\lt \dfrac{\epsilon}{\sqrt{2}}$. Similarly, there is a rational $t$ such that $|b-t|\lt \dfrac{\epsilon}{\sqrt{2}}$.

It follows that $\sqrt{(a-s)^2+(b-t)^2}\lt \epsilon$.

A similar argument but somewhat simpler argument takes care of (4).

The fact that the set is bounded is enough for (3). Or the fact that there is no point on the circle that is anywhere near $(0,0)$.

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so only 2 and 4 are correct.am i right?? –  priti Dec 17 '12 at 6:07
Yes, you had the right answers in all cases. I wrote out some details only because you expressed some doubt. –  André Nicolas Dec 17 '12 at 6:12
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The set in 2 contains $\mathbb Q \times \mathbb Q$ and so is dense in $\mathbb R \times \mathbb R$.

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