1) Let Produce an open set in the product topology, which here would be like
, such that
2) Closed sets are the complement of open sets. Excluding the empty set and all of the only
open sets of are intervals of the form , so, excluding the empty set and all of
all closed sets are of the form .
Then use that the closure of a set is the intersection of all closed sets containing that set.
3) If , then for some a > |x|, or .
Thus if , you can say something about U and V.
That tells you what the basic open "boxes" in look like.
So an open set will be an arbitrary union of those, which will put serious restrictions on the kinds of "shapes and fills" they can have.
Take a guess at the symmetries & conditions required to be an open set in ,
then prove that a set is open if and only if it has those symmetries & conditions.
(Think about the boundary (a curve), in the usual sense, of an open set of in the 1st quadrant.)
With the topology of so characterized, you'll be able to say when the inverse image
of an open interval of (i.e. a basis set) is open in ,
and so set the condition required for continuity of any .