A homeomorphism $h$ of the plane $\mathbb{R}^2$ onto itself is given by the formula $h(x, y) = (2x, \frac{1}{2}y).$ I am studying for a qualifying exam.  I have been able to do a large percentage of them, but this one in particular has me completely stumped.  Could someone give me a start maybe?
A homeomorphism $h$ of the plane $\mathbb{R}^2$ onto itself is given by the formula $h(x, y) = (2x, \frac{1}{2}y).$ This generates an action of the infinite cyclic group on $\mathbb{R}^2$.  Is the qoutient space Hausdorff?  Is the quotient space compact?  Justify your answers in each case.
 A: The quotient space is Hausdorff if the group is discrete, meaning that if you follow the orbit of every point, it has no limit point. Look at what happens to points not on an axis and points on an axis.
A different way of thinking about it is that an open set in the quotient corresponds to an open set around each equivalence class in the original set. So think of the equivalence class of one point (it will usually be a countable set) and think of an open set around it. Now take another equivalence class of points and do the same thing. Can you make the two open sets disjoint? What about if one point is on the x-axis? What if one is the origin?
For compactness, a group action gives a compact quotient if there is a compact set in the original space which gets 'spread out' by the group action to cover the entire space, i.e. the translates under the group cover the whole space. Try putting a compact set somewhere and seeing what happens. In fact, every compact set is contained in a ball centered at the origin, so see what happens to balls centered at the origin, and say whether its translates cover the entire space. 
