Hello, there may be a simpler way, but this proved it for me:
a) Let . Then, for we have that , hence is open.
For b) and c), adapt the same argument accordingly.
Consider the metric space ([0, 1], | · |).In this space, no number smaller than 0 or more than 1 exists.
a) Show that any interval in the form (r,s), where 0 < r < s < 1 is open in this space.
b)Show that any interval in the form [0, s), where s < 1, is open in this space.
c) Show that any interval in the form (r, 1], where r > 0, is open in this space.