TITLE: Finding a topology on $ \mathbb{R}^2 $ such that the $x$-axis is dense
QUESTION [5 upvotes]: The problem is the following
Put a topology on $ \mathbb{R}^2$ with the property that the line $\{(x,0):x\in  \mathbb{R}\}$ is dense in $\mathbb{R}^2$
My attempt
If (a,b) is in $R^2$, then define an open sets of $(a,b)$ as the strip between $d$ and $c$, inclusive where $c$ has the opposite sign of $b$ and $d>b$ if $b$ is positive otherwise $d<b$ . Clearly this always contains the set $ \mathbb{R}\times\{0\}$. It also obeys the the topological laws, ie, the intersection of two open sets is open. The union of any number of open sets is open.
Thanks for your help

REPLY [2 votes]: The following generalizes all solutions (EDIT: not all solutions, just those which give a topology on $\mathbb{R}^2$ homeomorphic to the standard topology).  It doesn't have much topological content, but it serves to show how basic set theory can often be used to trivialize problems in other fields.  A famous example of this phenomenon is Cantor's proof of the existence of (infinitely many) transcendental numbers.
So, let $X$ be a dense subset of $\mathbb{R}^2$ be such that 
$$\left|X\right| = \left|\mathbb{R}^2\setminus X\right| = \left|\mathbb{R}^2\right|$$
For instance, $X$ could be the set of points with irrational first coordinate.  Now let $f$ and $g$ be bijections:
$$f : \mathbb{R}\times\{0\} \to X$$
$$g : \mathbb{R}\times (\mathbb{R}\setminus\{0\}) \to \mathbb{R}^2\setminus X$$
Then let $F = f \cup g$ is a bijection from the plane to itself which will map the $x$-axis onto $X$.  The topology we want consists of those sets $A$ for which $F[A]$ is open in the standard topology.