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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.

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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
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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
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4 Answers

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.

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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.

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  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.

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You seem to have the wrong order. –  Andres 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
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  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.

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