# Determine whether $S=\{x,y\in \mathbb{R}^2 : x^2-y^2=1\}$ is disconnected or connected.

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.

• $x^2=1+y^2\ge 1\implies |x|\ge 1$ so $S_1$ and $S_2$ are completely separated by the closed band $|x|\le \frac 12$ for instance. – zwim Nov 28 '18 at 23:53

Hint: consider $$U=\{(x,y):x>0\}$$ and $$V=\{(x,y):x<0\}$$; prove that $$S=(U\cap S)\cup(V\cap S)$$.
• @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. – egreg Nov 29 '18 at 9:58
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 $$-y^2 = 1$$
• That would show that $S$ is not path-connected; but that doesn't answer the original question. – Daniel Schepler Nov 28 '18 at 23:35