The second one is easiest, so I'll do it first. M is the lub of A, N is the lub of B. What does this mean? Take A and M: it means that every x in A is less than M (upper bound) and that there isnt any smaller number that is an upper bound: so if you have any K, say, less than M, then some x in A will not be less than K (thats why it is the LEAST upper bound. If your head is spinning, draw a picture with A as a kind of cloud or blob, and M as a horizontal line just touching the top of the cloud. If you were to lower the line (K) then it would clip off some of the cloud (x) which would be above it, right?) Similarly for B and N.

OK, now, think about A union B. This is like taking both clouds together and treating them as a single cloud. So, is N or M an upper bound of that? Well, one of them has to be (the highest line, right?) That is, max{N,M} is either N or M, depending which is bigger: if N<M then max{N,M}=M, otherwise =N. We will argue by cases. Assume N<M first. If N<M then everything in B is <N which is <M, and everything in A is <M, so everything in (A union B) is <M, so M is an ub (thats all just saying the highest of the two lines is above both clouds); and if there were a smaller one, say K, then K would have to be a ub of (A union B) and hence an ub of A: but it can't be, because M was a lub of A (remember? In other words, if you lower the top line then you lose part of the highest cloud). So M is the lub in this case. If M<N, the argument is exactly the same with B and N instead of A and M, and N is the lub in that case. So: if N<M then the lub is M and max{M,N}=M, and if M<N then the lub is N and max{M,N}=N. Either way, the lub is max{M,N}. (What if M=N? I leave that to you.)

The first problem use that 'leastness' critically. Take some X1 in S - we know there is one because S isn't empty (that condition in the statement of the problem was a clue, in fact) - and ask yourself: is this X1 the lub L of S already? It might be. If it is, we can just define the sequence to be L,L,L,.. on for ever. But if not, then there must be another X2 in S which is closer to L. (Why? Because if there weren't, then X1 would be the upper bound, right? That's just another way of saying the same thing: if there's nothing larger then X1 in S, then X1 is an upper bound of S.) So, if X1 isnt the upper bound, we can find an X2 which is closer to L. In fact, we can find an X2 which is less than half the distance from L that X1 was, i.e. an X2 such that X2>(X1+L)/2. Why? Because if we couldn't, then (X1+L)/2, which is <L, would be an upper bound, so L wouldnt be the LEAST upper bound. So, go on in that way: you have X1 to Xn chosen, and if Xn is L then stick with L from then on, and if not then choose an Xn+1 > (Xn+L)/2, ie which halves the distance between your best hit so far and the lub L. You keep on creeping closer. Then L is the limit of that sequence Xn as n goes to infinity. I'll let you prove that last part. :-) (The point is that it doesnt matter exactly which Xn+1 you pick: they can't be larger than L, and if they are smaller you can always get half-way closer.)

These aren't the shortest or most elegant proofs, and I havnt written them in a nice mathematical style. But if you can see how the necessary thinking goes, you can learn to do that other stuff yourself.