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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?

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Lemon
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1 Answers1

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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 $C_2$

5xum
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  • What about my comments about $A$? Were they correct? – Lemon Apr 24 '14 at 05:32
  • 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 05:42
  • Do I have to prove that they are closed? – Lemon Apr 24 '14 at 05:44
  • You have to prove that their union is the whole space and that their intersection is empty and that they are both open or that they are both closed. Equivalently, you can take only one of the sets and prove that it is both open and closed. You should know this as it is the definition of conectedness. – 5xum Apr 24 '14 at 05:49
  • How about $A_1 = A \cap \mathbb{R}^2_+$ (first and fourth quadrant) – Lemon Apr 24 '14 at 05:50
  • OK, what does that prove? – 5xum Apr 24 '14 at 05:50
  • This proves $A_1$ is open – Lemon Apr 24 '14 at 05:52
  • Yes. And a similar set can be found to prove $A_2$ is open. – 5xum Apr 24 '14 at 05:55