Is $A = \{ (x,y): x^2 - y^2 = -1 \}$ path connected in $\mathbb R^2$ ?
From its graph, I would conclude that it's not path connected.
Take $(0,-1)$ and $(0,1)$ in A. Take any path $f: [0,1] \to A$ with $f(0) = (0,-1)$, $f(1) = (0,1)$. How can we show that f can't be continuous?
Take two points $(0, 1)$ and $(0, -1)$. If there is a path connecting them, then (by Intermediate Value Theorem) there is some point at curve with coordinates $(x, 0)$. In order for this point to lie on $x^2-y^2 = -1$ its $x$-coordinate should satisfy $x^2 = -1$, which is impossible for real $x$.
Hint Since path-connected implies connected, it suffices to show that $A$ is not connected: Show that the sets $A \cap \{y > 0\}$ and $A \cap \{y < 0\}$ comprise a separation of $A$.
$x^2-y^2=-1$ is equivalent to saying that $f(x)=y=\sqrt{x^2+1}$ or $g(x)=y=-\sqrt{x^2+1}$ clearly, the graphs of $f$ and $g$ don't intersect since $f(x)>0, g(x)<0$.
