Suppose S ? T ? R . Show, if T is bounded above and S is non-empty, then both S

Question

Suppose S ? T ? R. Show, if T is bounded above and S is non-empty, then both S and T have least upper bounds and moreover, sup(S) ? sup(T).
I know that there is a theorem that states that a non empty subset of R that is bounded above has a least upper bound but is that sufficient enough in reasoning? Any help would be appreciated. And I'm looking more for guidence towards the proof rather than just the proof itself. Thanks! Suppose S ? T ? R. Show, if T is bounded above and S is non-empty, then both S and T have least upper bounds and moreover, sup(S) ? sup(T).
I know that there is a theorem that states that a non empty subset of R that is bounded above has a least upper bound but is that sufficient enough in reasoning? Any help would be appreciated. And I'm looking more for guidence towards the proof rather than just the proof itself. Thanks!

Explanation / Answer

S subset T, T subset R.

S is non-empty and T is bounded above.

Define B = { M : M >= x for all x in T} = { M : M is upper bound of T}

Now since B subset of R so each pair of numbers are comparable.

Let L in B such that L<= M for all M in B. Then this L is the least upper bound of T.

Since S subset of T and is non-empty. And, clearly L is a real number such that L>=x for all x in S.

And by applying same argument we get that S has a least upper bound.

Next, sup(S) <= sup(T)

Since S and T both has least upper bound say K and L, respectively.

Clearly, sup(S)<=K, since K >=x for all x in S

and sup(T)<=L, since L>=x for all x in T.

Since, L is an upper bound of S and T, therefore K <= L, since K is least upper bound of S.

Since, K <=sup(T)

Thus, sup(S)<=K <= sup(T) <= L

We get sup(S)<= sup(T)

Note that K and L may not lie inside S and T, respectively. But, sup(S) and sup(T) lie in S and T, respectively.

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