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?

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


Your approach is correct.

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?


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