Definition of connected set A connected space is a topological space that cannot be represented as the union of two or more disjoint non-empty open subsets.
I do not understand the definition precisely since the set $x^2-y^2=1$ is not connected but it can only be divided into two closed subset. Obviously closed subset is not the same as open subset right?
 A: Let $C$ be your set and $U=C\cap \{(x,y): x>0\}$, $V=C\cap \{(x,y): x<0\}$. Then $U$ and $V$ are disjoint open subsets of $C$ whose union is $C$. 
A: I think your confusion is that you are thinking of the two branches of the hyperbola as subsets of $\mathbb R^2$. As such, they are of course closed but not open subsets.
But then you are looking at the space $\mathbb R^2$, not at the space that consists only of those parabola branches. But $\mathbb R^2$ is not the space that's disconnected.
Now if you want to look at the hyperbola as topological space, you have to remove everything that is not part of the hyperbola. For the topology you essentially do the same: A set is open in the subspace topology if it is the intersection of an open set of the full space with the subset.
So let's look at how that works out for the hyperbola, $H=\{(x,y)\in\mathbb R^2|x^2-y^2=1\}$
Consider the set $P=\{(x,y)\in\mathbb R^2|x>0\}$. Quite obviously this is an open set in $\mathbb R^2$. This means that $P\cap H$ is an open set in the subspace topology of $H$. But that is exactly the right branch of the hyperbola.
Similarly $M=\{(x,y)\in\mathbb R^2|x<0\}$ is open in $\mathbb R^2$, therefore its intersection with $H$ is open in $H$. And this is the left branch of $H$.
And of course, the complete hyperbola is the union of its two branches. Since both branches are open sets in the subspace topology, the space $H$ is indeed not connected.
A: Your topological space is $H:=\{(x,y) \in \mathbb R^2: x^2-y^2=1\}$ with the topologie induced by the absolute value.
Let $H_1:=\{(x,y) \in \mathbb R^2: x>0, x^2-y^2=1\}$ and $H_2:=\{(x,y) \in \mathbb R^2: x<0, x^2-y^2=1\}$.
Then $H= H_1 \cup H_2$ and $H_1$ and$ H_2 $ are open in the topology of $H$
