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So I want to prove its disconnected. That is $S_1\cup S_2= S$ for $S_1,S_2$ being non empty, disjoint relatively open sets.

My proposed sets are $S_1=\{x,y\in \mathbb{R}^2 : x>0, x^2-y^2=1\}$ and $S_2=\{x,y\in \mathbb{R}^2 :x<0, x^2-y^2=1\}$

$S_1, S_2$ are nonempty since $(-1,0)$ and $(1,0)$ are in $S_1$ and $S_2$ respectively.

$S_1\cup S_2=S$ since $(0,y)\not\in S, \forall y\in \mathbb{R}$

$S_1\cap S_2=\emptyset$ since $\forall x_1\in S_1$ and $\forall x_2 \in S_2$, $x_1>x_2$

My issue is showing they are relatively open.

I want to show that taking some arbitrary $p=(p_1,p_2)\in S_1$, that for some $\epsilon >0$, $B_\epsilon(p)\cap S \subseteq S_1$.

My idea would be to fix $\epsilon <1$ since the closest point to $S_2$ should be the point $(-1,0)$ so this epsilon should work.

So then fix some $q=(q_1,q_2)\in B_\epsilon(p)$, then I believe I should have that $q\in S$ so that $q_1^2-q_2^2=1$ and I want to show that since $d(p,q)<1$ that this means $q_1<0$ and thus is in $S_1$. But I can't figure out how to get there.

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  • $\begingroup$ $x^2=1+y^2\ge 1\implies |x|\ge 1$ so $S_1$ and $S_2$ are completely separated by the closed band $|x|\le \frac 12$ for instance. $\endgroup$ – zwim Nov 28 '18 at 23:53
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Hint: consider $U=\{(x,y):x>0\}$ and $V=\{(x,y):x<0\}$; prove that $S=(U\cap S)\cup(V\cap S)$.

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  • $\begingroup$ What is different about this method and what I have done with my sets? $\endgroup$ – AColoredReptile Nov 28 '18 at 23:45
  • $\begingroup$ @AColoredReptile Not much, but less complicated. A relatively open set in $S$ is the intersection of an open set with $S$. You don't need proofs with balls. $\endgroup$ – egreg Nov 29 '18 at 9:58
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If they are connected, then there would be some continuous path in $S$ from $(1,0)$ to $(-1,0).$ In which case, it must pass $y$ axis, and have a $y$ intercept.

But there is no $y$ such that $-y^2 = 1$

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  • $\begingroup$ That would show that $S$ is not path-connected; but that doesn't answer the original question. $\endgroup$ – Daniel Schepler Nov 28 '18 at 23:35

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