Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

  1. $\{ (x, y)\in\mathbb R^2 : x\in\mathbb N\}$

  2. $\{ (x, y)\in\mathbb R^2 : x+y\in\mathbb Q\}$

  3. $\{ (x, y)\in\mathbb R^2 : x^2 + y^2 = 5\}$

  4. $\{ (x, y)\in\mathbb R^2 : xy\neq 0\}$.

Any hint is welcome.

share|cite|improve this question
Do you mean "Which of the following are Dense in $\mathbb{R}^2$"? – William Jun 11 '12 at 23:17
I edited your question for formatting, and changed "dense in $R$" to "dense in $\mathbb R^2$". Is this correct? – Alex Becker Jun 11 '12 at 23:18
Think about what those sets really are. For example 3. is a circle of radius $5$. Can that ever be dense? Similar considerations apply to 1. whereas 4. consists of all of $\mathbb{R}^2$ minus the coordinate axes, so clearly it's dense. etc, etc. – user12014 Jun 11 '12 at 23:20
What do you think? Have you tried anything yet? It's often better to include any working you've already done when posting on here! – Edward Hughes Jun 11 '12 at 23:21
@AlexBecker: Thanks a lot. – preeti Jun 11 '12 at 23:22

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.

share|cite|improve this answer

A set is dense in

  1. 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.

  2. 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 $\mathbb R$.

  3. This is not dense. The neighborhood surrounding the origin with $r=1$ contains no points in this set.

  4. 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.

share|cite|improve this answer
  1. No. It's a bunch of parallel lines. These are vertical and the go through the integer points on the $x$-axis.

  2. No. It's a circle.

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

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

share|cite|improve this answer
You seem to have the wrong order. – Andrés E. Caicedo Jun 11 '12 at 23:24
I put in the order 1,3,4,2 and the \LaTeX engine seems to have renumbered them. – ncmathsadist Jun 11 '12 at 23:26
I was surprised by that. – ncmathsadist Jun 11 '12 at 23:29
  1. is not dense. The set of verticle lines with natural number $x$ coordinate is not dense.

  2. 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 $(x,y)$.

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.