That course sounds like the book Elements of Modern Algebra by Gilbert. Maybe check it out and compare it with a standard 4th year algebra text in Dummit/Foote where you will go into more details of groups/rings/fields and then more advanced topics in algebra.
This is not true.
This is not true either
To show it is a topology, set up your usual arbitrary union of open sets and take the complement in X. There is a set theory identity you use here to show that is it countable. Do the same for the finite intersection of open sets. This is essentially...
Two possibilities have been eliminated. It's not hard to find the answer, and to deduct the reason for it (although the definitions are made precise so you can solve this kind of problem by going back to your notes)
Really? Z, Q?
Each of your open balls are bounded, so wouldn't the intersection be bounded? I admit I know nothing of nsa, but I don't precisely see the question.
You are being asked to show that Q = Cl (IntQ)
You know how to define Int and Bd, so write out the definitions explicitly (in set theory notation) to show that they coincide.
You are told that F9 = {a+bi | a,b in F9}. The first part is asking you to try out all the elements. So for example, 1+2i is an element.
The second part is asking you to show that for all a+bi in F9, there is some c+di such that (a+bi)(c+di) = 1. Try writing the inverse out using a and b in some...
What are you having trouble understanding? What is the natural metric topology of R? This is essentially the same that is put on RxR, pointwise. Can you see how this coincides with open disks? It is not a hard problem.