Why is $\mathbb{R^2} \setminus (0, 0)$ connected but not simply connected? Why is $\mathbb{R^2} \setminus (0, 0)$ connected but not simply connected?
Simply connected means path connected, intuitively it seems that for any point in $\mathbb{R^2} \setminus (0, 0)$ we can have an unbroken path to any other point in this set? So what am I misunderstanding?
 A: Simply connected does not mean path-connected.
Simply connected means that if a path starts at a point $p$ and returns in to $p$, then that loop can be contracted to $p$.
If a path starts at $1$ and winds once clockwise around $0$ and returns to $1$, then that path cannot be contracted to $1$ within the space $\mathbb R^2\setminus \{(0,0)\}$.  It gets "caught" on the point $(0,0)$.
A: Your definition is incorrect: simply connected means that any loop in the space can be continuously shrunk to a point. But a loop around the missing point of $\mathbb R^2-\{(0,0)\}$ (for instance, a parameterization of the unit circle centered at the origin) cannot be shrunk to a point in a continuous manner without going through the missing point $(0,0)$; there are formal proofs of this, but I think it's obvious enough to "see".
A: Path connectedness is a more basic concept. Simple connectedness means any two paths between two points  can be 'deformed continuously' to each other.
Take the unit circle centred at origin. The upper and lower semi-cricular arcs can be thought of as two paths from $(-1,0)$ to $(1,0)$. When we try to push one of these arcs  towards the other you have a hurdle at (0,0). SOme intermediate path has to use the origin which unfortunately has been removed. 
