Let L = sup{S1} and M = sup{S2}.

So L is the least upper bound of nonempty bounded set S1, and M is the least upper bound of nonempty bounded set S2. (Note that any nonempty bounded set of real numbers has a least upper bound.)

We need to prove that max(L, M) is the least upper bound of S1 ∪ S2.

i) We need to prove that max(L, M) is an upper bound of S1 ∪ S2.

Note that L <= max(L, M) and M <= max(L, M).

Since L is an upper bound of S1, no elements of S1 exceed max(L, M).

Since M is an upper bound of S2, no elements of S2 exceed max(L, M).

If x is in S1 ∪ S2, then x is in S1 or x is in S2.

In either of these two cases, x does not exceed max(L, M).

Therefore, max(L, M) is an upper bound of S1 ∪ S2.

ii) We need to show that any y that is less than max(L, M) is not an upper bound of of S1 ∪ S2.

If y < max(L, M) then either y < L or y < M.

In the first case, some x in S1 exceeds y, because L is the *least* upper bound of S1. Since S1 is a subset of S1 ∪ S2, some x in S1 ∪ S2 exceeds y.

In the second case, some x in S2 exceeds y, because M is the *least* upper bound of S2. Since S2 is a subset of S1 ∪ S2, some x in S1 ∪ S2 exceeds y.

So in either case, some x in S1 ∪ S2 exceeds y. Thus, y is not an upper bound of of S1 ∪ S2.

From i) and ii), we conclude that max(L, M) is the least upper bound of S1 ∪ S2.

Thus, sup {S1 ∪ S2} = max{sup{S1}, sup{S2}}.

Lord bless you today!

If A is any set and p is any number, to prove sup A = p, you do two things:

(1) prove that p is an upper bound for A.

(2) prove that if q is a number and q < p, then q is not an upper bound for A.

[The proof of (1) also has a standard form: you fix an arbitrary element a of A and prove that p >= a.]

We will follow this with A = S1 union S2 and p = max(sup(S1), sup(S2)).

I claim that max(sup(S1), sup(S2)) is an upper bound for S1 union S2. To see this, fix x in S1 union S2. From the definition of S1 union S2, either x is in S1 or x is in S2. We consider the cases separately:

If x is in S1, then by the definition of sup(S1) we have x <= sup(S1). By the definition of "max", we have sup(S1) <= max(sup(S1), sup(S2)). Whenever a, b, and c are numbers with a <= b and b <= c we can deduce that a <= c. So we conclude from the two inequalities just mentioned that x <= max(sup(S1), sup(S2)).

If x is in S2, then by definition of sup(S2) we have x <= sup(S2). By the definition of "max", we have sup(S2) <= max(sup(S1), sup(S2)). So we conclude as before that x <= max(sup(S1),sup(S2)).

[If you look closely at this argument you could condense it into a single paragraph as follows. "If x is in S1 union S2 then from the definition of S1 union S2 there is some i, either 1 or 2, with the property that x is in Si. Then x <= sup(Si) by definition, and sup(Si) <= max(sup(S1),sup(S2)), so x <= max(sup(S1),sup(S2)).]

In either case we have shown that x <= max(sup(S1),sup(S2)). Since x was an arbitrary element of S1 union S2, we have proved that for *all* x in S1 union S2, x <= max(sup(S1),sup(S2)). This proves that max(sup(S1),sup(S2)) is an upper bound for S1 union S2.

Suppose that q is a number and q < max(sup(S1), sup(S2)). Now, max(sup(S1),sup(S2)) is one of the numbers sup(S1), sup(S2). So there is a number i (either 1 or 2) with the property that x = max(sup(S1),sup(S2)) = Si. From q < max(sup(S1), sup(S2)) we deduce that q < sup(Si). Since sup(Si) is by definition the least upper bound of Si and q is less than it, we deduce that q is not an upper bound for Si. This means that there is some x in Si with the property that q < x. Since x is in Si (and i is 1 or 2), we conclude that x is in S1 union S2. So there is an element x of S1 union S2 with the property that q < x. This proves that q is not an upper bound for S1 union S2.

Since q was an arbitrary number less than max(sup(S1),sup(S2)) this shows that no smaller number than max(sup(S1),sup(S2)) can be an upper bound for S1 union S2. So together with what we proved previously this shows that max(sup(S1),sup(S2)) is the least upper bound of S1 union S2, or in symbols, max(sup(S1),sup(S2)) = sup(S1 union S2).

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