Finding a topology on $ \mathbb{R}^2 $ such that the $x$-axis is dense 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
 A: The simplest answer is to give $\Bbb R^2$ the indiscrete topology, whose only open sets are $\varnothing$ and $\Bbb R^2$. The cofinite topology, whose members are $\varnothing$ and all subsets of $\Bbb R^2$ whose complements in $\Bbb R^2$ are finite, also works and is $T_1$.
To get an even nicer topology, let $\varphi:\Bbb R^2\to\Bbb R$ be a bijection; this answer to an earlier question discusses in detail how to find such a $\varphi$. Let $X=\Bbb R^2\setminus(\Bbb Q\times\{0\})$. For each $x\in X$ let $\langle q_x(k):k\in\Bbb N\rangle$ be a sequence of rational numbers converging monotonically to $\varphi(x)$. It’s easy to see that if $x,y\in X$ and $x\ne y$, then the sequences $\langle q_x(k):k\in\Bbb N\rangle$ and $\langle q_y(k):k\in\Bbb N\rangle$ can have only finitely many terms in common. For $x\in X$ and $n\in\Bbb N$ let $B_n(p)=\{p\}\cup\big\{\langle q_x(k):k\ge n\big\}$. For $x\in X$ let $\mathscr{B}(x)=\{B_n(x):n\in\Bbb N\}$, and topologize $\Bbb R^2$ by making each point of $\Bbb Q\times\{0\}$ isolated and taking $\mathscr{B}(x)$ as a local base at $x\in X$. The resulting space is Tikhonov and zero-dimensional and has $\Bbb Q\times\{0\}$ (and hence of course $\Bbb R\times\{0\}$) as a dense subset.
But we can do better yet. Let $C$ be the middle-thirds Cantor set, and let $D=[0,1]\setminus C$; $D$ is dense in $[0,1]$, and $|D|=|C|=|[0,1]|=|\Bbb R^2|$. Let $X=\Bbb R\times\{0\}$, and let $Y=\Bbb R^2\setminus X$; $|X|=|Y|=|\Bbb R^2|$ as well, so there are bijections $\varphi_X:X\to D$ and $\varphi_Y:Y\to C$. Define
$$\varphi:\Bbb R^2\to[0,1]:x\mapsto\begin{cases}
\varphi_X(x),&\text{if }x\in X\\
\varphi_Y(x),&\text{if }x\in Y\;.
\end{cases}$$
Finally, topologize $\Bbb R^2$ by making $\varphi$ a homeomorphism from $\Bbb R^2$ to $[0,1]$ with the usual topology: $U\subseteq\Bbb R^2$ is open iff $\varphi[U]$ is open in $[0,1]$ with the usual topology. $D$ is dense in $[0,1]$, so $X=\varphi^{-1}[D]$ is dense in $\Bbb R^2$, as desired, and this time we have a compact metric topology on $\Bbb R^2$ in which $X$ is dense!
A: 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.
A: I think that you already get the solution. You also can put for open set the vertical strip  $| |$(assuming $\mathbb R^2$ a vertical strip)
