$\mathbb{R}^2 - \{x\mbox{-axis}\}$ is open

problem: $\mathbb{R}^2 - \{x\mbox{-axis}\}$ is open

My attempt: We want to show that there exists a neighborhood $D(x,r)$ around $x$ such that $D(x,r) \subseteq \mathbb{R}^2 - \{x\mbox{-axis}\}$. In other words, we need to show that $D(x,r) \cap \mathbb{R}^2 - \{x\mbox{-axis}\}$ is the empty set. Can we argue by contradiction, and say $D(x,r) \cap \mathbb{R}^2 - \{x\mbox{-axis}\}$ is non-empty? How can we choose $r$ so that we can get a contradiction?

thanks,

• If the point $x=(x_1,x_2)$, try $r=|x_2|/2$. Draw a picture! – Jyrki Lahtonen Aug 28 '13 at 20:32
• An argument by contradiction makes the problem needlessly complicated. I like the hint above. – Stefan Smith Aug 29 '13 at 0:34

If $y$ is the y-coördinate of the point $x$ ($y\neq 0$), take $r=\frac12 |y|>0$. We now know that $$d(x,\{x\mbox{-axis}\})=|y|$$ (with $d$ the distance), and $$d(x,z)\leq r=\frac12|y|<|y| \;\;\;\;\;\;\;\;\;\;\;\forall z\in D(x,r)$$ because $|y|>0$.

We now see that $D(x,r)\cap\{x\mbox{-axis}\}=\emptyset$, so $\mathbb{R}-\{x\mbox{-axis}\}$ is open.

Hint: Prove that the $x$-axis is closed.

• Solid hint. fistbump – rschwieb Aug 28 '13 at 20:26
• @rschwieb Thanks. – Git Gud Aug 28 '13 at 20:26
• @Citizen One could continue your proof, but the natural way to finding a contradiction in your proof is actually mimicing the proof suggested by my hint. – Git Gud Aug 28 '13 at 20:27

You can try taking the function $f(x,y)=y$ which is continuous. Since $\{0\}$ is a closed set of $\mathbb{R}$, the set $A=\{(x,0)\mid x\in\mathbb{R}\}$ is closed (because it is the preimage of the closed set $\{0\}$), then your set ($A^c$) is open.

The topology on $\mathbb{R}^2$ is the initial topology generated by the projections. Hence $f(x,y) = y$ is continuous and furthermore $$f^{-1}( \mathbb{R}\setminus \{0\})= \mathbb{R}^2 \setminus\{ x\text{-axis}\}$$ As $\mathbb{R}\setminus \{0\}$ is open, your set is the preimage of an open set under a continuous function and hence open.

Let $S$ be the set in question, namely $\{(x,y)\mid y\ne 0\}$. For each point $p=(x,y)$ of $S$, let $G_p$ be the open ball around $p$ of radius $\frac y2$. Note that since $G_p$ was carefully chosen to go only halfway from $p$ to the $x$-axis, it does not intersect the axis and is therefore a subset of $S$.

Let $$S' = \bigcup_{p\in S} G_p.$$

Since $S'$ is a union of subsets of $S$, it is contained in $S$. But $S'$ includes each point $p$ of $S$, so it contains $S$. So $S'=S$.

$S'$ is certainly open, being a union of open sets, and therefore $S$ is open.