# Example of disconnected set on $\mathbb{R}^2$ (hyperbola)

The hyperbola given by

$$A = \{ (x,y): x^2 -y^2 = 1 \}$$

is disconnected.

I believe the reason is because

$$A = A_1 \cup A_2 = \{ (x,y): x = \sqrt{1 + y^2 } \} \cup \{ (x,y): x = -\sqrt{1 + y^2 } \}$$

Both $A_1$ and $A_2$ closed in $A$ (and in $\mathbb{R^2}$) and they are open in $A$ (but not in $\mathbb{R}^2$).

Idea is that the subspace topology on $A$ are intervals on the hyperbola.

Now Ibelieve there is something wrong with my reasoning because the set

$$C = \{(x,y) : x^2 + y^2 = 1 \}$$

is connected, but I can break $C = C_1 \cup C_2$

where $C_1 = \{(x,y) : x = -\sqrt{1 - y^2} \} = -C_2$. By the arguments above, then actually $C_1$ and $C_2$ are clopen, but this contradicts that the unit circle is connected.

Could someone point out my mistake?

A couple of things are wrong. For one, you need two sets with an empty intersection to prove a space is not connected. You did not find such sets, since $$C_1\cap C_2 = \{(0,1),(0,-1)\}\neq \emptyset.$$
The second thing is that $C_1$ and $C_2$ are not open subsets of $C$. Every open set containing $(0,1)$ in $\mathbb R^2$ will also contain $(\epsilon, \sqrt{1-\epsilon})$ and $(\epsilon, -\sqrt{1-\epsilon})$ for some small enough value of $\epsilon$, so $(0,1)$ is not in the interior of either $C_1$ or $C_2$
• What about my comments about $A$? Were they correct? – Hawk Apr 24 '14 at 5:32
• They were correct, but they were not justified. You haven't proven that $A_1$ and $A_2$ are open. You have to prove this, and your example with $C$ shows why: if you would try to prove that $C_1$ and $C_2$ are open, you would see that they are not and you would see that they are not a good example. – 5xum Apr 24 '14 at 5:42
• How about $A_1 = A \cap \mathbb{R}^2_+$ (first and fourth quadrant) – Hawk Apr 24 '14 at 5:50