Hi, I've been stuck on this problem for a while now and I'm scared it's just me failing with the inequalities, but it might be something more fundamentally wrong with my understanding of open sets.
From what I've learnt, a set S is open if for all z in S, there exists an r>0 such that D(z;r) is a subset of S.
My problem is here: I need to prove that
My attempt at a proof so far gets here:
Let . Then if,
So my aim now is to show that , equivalently, that
I figure I start with , but from there I don't know where to go.
I've been on this problem a while, and I think if I can figure this out, I should be okay with problems in the future.
Thankyou for reading, and I appreciate all help.
METHOD 1: (The best) In general, if are continuous, then is continuous.
Thus in , is open, hence is open, hence is open.
This is the general way to handle this. Now just note that by and are continuous, and you're done.
METHOD 2: Let , which is open in .
Note that , so that , where is given by .
(There, is the open interval in .) Since is a composition of continuous functions, is open in , and hence is open in .
(That's because open in the relative topology implies for some open in . But since is open in , conclude that is open in ).
METHOD 3: "Read" what S is. S is the set of z that are closer to 1 than they are to -i. That's clearly the upper part of the plane that's split by the perpendicular bisector from 1 to -i., which is graphed in the plane as y > -x. Thus . An open ball around any such point is would be .
METHOD 4: The algebraic way to do method 3: Let . Have:
Simplifying that gives:
Now let , let (note ), and let .
Then . Then do the algebra to prove that .
I didn't work it out, but you get the idea. From what's already been said, it *must* be true that that suffices to prove .