Which of the following are Dense in $\mathbb{R}^2$? 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. 
 A: 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.
A: 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.
A: *

*No. It's a bunch of parallel lines.  These are vertical and the go through the integer points on the $x$-axis.

*No. It's a circle.

*Yes. It's the plane with the $x$ and $y$ axes excised.

*Interesting. It is a union of parallel lines with slope -1 and $y$-intercept at the various rationals. It's dense in the plane.
A: *

*is not dense. The set of verticle lines with natural number $x$ coordinate is not dense. 

*This is dense. Given $(x,y)$, let $r = x + y$. If $r$ is irrational , let $q$ be any rational number close to $r$. Then $(x - (r - q), y)$ has rational sum and gets close to $(x,y)$. 

*This is a circle of radius of radius $\sqrt{5}$ which is not dense. 

*This is dense. You can get arbitrary close to any $(x,y)$ without intersecting the $x$ or the $y$ axis. 
