TITLE: Which of the following are Dense in $\mathbb{R}^2$?
QUESTION [7 upvotes]: Which of the following sets are dense in $\mathbb R^2$ with respect to the usual topology.

$\{ (x, y)\in\mathbb R^2 : x\in\mathbb N\}$
$\{ (x, y)\in\mathbb R^2 : x+y\in\mathbb Q\}$
$\{ (x, y)\in\mathbb R^2 : x^2 + y^2 = 5\}$
$\{ (x, y)\in\mathbb R^2 : xy\neq 0\}$. 

Any hint is welcome.

REPLY [5 votes]: For a set to be dense in $\mathbb{R}^2$ (or in any other metric space, for that matter) it is necessary and sufficient to check that ot intersects every open disc. So, to prove that a set isn't dense, it's enough to find one open disc that includes no points of the set. For example, in (1), take $D((\frac{1}{2},0)\frac{1}{4})$ (a disk with radius $\frac{1}{4}$ around $(\frac{1}{2},0)$). It contains no point of the set in (1). Hence the set is not dense.
To prove (4),  take any open disk $D((x,y),r)$. If $r<min\{|x|,|y|\}$ all points of the disk are in the given set. Else, take $s=min\{|x|,|y|\}$ and take any point in $D((x,y),s)$. This point is both in $D((x,y),r)$ and in the given set. Hence the set is dense.
You can prove all other cases in the same manner.

REPLY [4 votes]: A set is dense in 

This is not dense.  For example, the neighborhood with $r=1/3$ surrounding $(1/2,0)$ contains no points in this set (since $x\in\mathbb N$), so this point cannot be a limit point.
This is dense.  It contains $\{(x,y):x,y\in \mathbb Q\}$ which is dense.  The proof for its density is similar to the proof that $\mathbb Q$ is dense in $\mathbb R$.
This is not dense.  The neighborhood surrounding the origin with $r=1$ contains no points in this set.
This is dense.  Take $x,y \in \mathbb R$ such that $xy=0$.  This is the complement of the set specified in the question with respect to $\mathbb R^2$.  Then, $x=0$ or $y=0$.

Take a neighborhood around this set with radius $r$.  Then, if $x=0$ and $y=0$, take the point $(r/2,r/2)$.  This is a member of the neighborhood, so this point is a limit point.
If $x=0$ and $y\not =0$, then take the point $(r/2,y)$.  This is a member of the neighborhood, so the point is a limit point.
Similarly, if $x\not=0$ and $y=0$, take the point $(x,r/2)$.  The same argument as above shows that this is a limit point.