# Proving the the x-axis is closed and not open

We have the x-axis $$X=\{ (x,y): y =0 \}$$ . I want to prove it is closed. In other words, We can show the complement is open. That is, we want to find a ball that is completely contained in $$\mathbb{R}^2 \setminus X$$. Let $$(a,b) \in \mathbb{R}^2 \setminus X$$ be arbitrary. A ball of radius $$|b|$$ can do ? Since it is never zero and $$|b| >0$$ then $$B_{\epsilon}((a,b))$$ must be in $$\mathbb{R}^2 \setminus X$$. Indeed. If $$(t,z)$$ is in the ball, then

$$(t-a)^2 + (z-b)^2 < |b|^2$$

We want to prove that $$(t,z)$$ must be in $$\mathbb{R}^2 \setminus X$$ as well. In other words, we need to prove that $$z \neq 0$$.

Notice that from above after distributing we obtain

$$(t-a)^2 + z^2 - 2zb + b^2 < b^2 \implies (t-a)^2<2zb-z^2=z(2b-z)$$

Here is where I get stuck. How can we show that the above is less than $$z$$? that way we get $$z > 0$$ and so proving the result. Is my approach correct?

To show it is not open, isnt the previous part showing this result as well?

• You have a mistake in 5th line.$(a, b)$ should not be in $X$ . – dmtri Sep 26 '18 at 19:00
• Do a proof by contradiction. If $z =0$ what happens? You get... $(t-a)^2 < 0$. Is that possible? – fleablood Sep 26 '18 at 19:04

Your approach is correct, but quite complicated. When you have $$(t-a)^2 + (z-b)^2 < |b|^2$$ you can immediately tell that $$z \neq 0$$. Indeed, if $$z=0$$, you would have $$(t-a)^2 + b^2 < b^2$$ i.e. $$(t-a)^2 < 0$$ which is absurd.

You may consider two different cases for your point $$(a,b)$$

One case where $$b>0$$ and the other case when $$b<0$$

Then it is more straight forward to show that $$z\ne 0$$

Notice $$p=(a,b) \not \in X \iff b \ne 0$$. So take $$\epsilon = |b|$$. Let $$(x,y) \in B(p,\epsilon)$$

Prove that if $$d((x,y),(a,b)) = \sqrt {(x-a)^2 + (y-b)^2} < |b|$$, then $$y \ne 0$$.

And that's fairly easy to do as $$y = 0\implies \sqrt{(x-a)^2 + (y-b)^2} = \sqrt{(x -a)^2 + (-b)^2} \ge \sqrt{(-b)^2} = |b|$$.

Do you see that that proves $$X^c$$ is open?