Assume that A and B are nonempty sets, that A is bounded above, and that B is contained in A. Prove that B is bounded above and that the least upper bound of B is less than or equal to the least upper bound of A.
Definition: Least Upper bound: Let s be the least upper bound then:
1)s is an upper bound (a set A contained in R is bounded above if there exists an element b in R such that a is less than or equal to b for all elements a in A.)
2)if b is any upper bound of A then s is less than or equal to b.
The Attempt at a Solution
Assume A and B are nonempty sets and that A is bounded above. Meaning, A is contained in R and that there exists an element c in R so that c is greater than or equal to a, for all a in A.
Now assume that B is contained in A, meaning every element in B is also an element of A. Since A has an upper bound (c) and every element of B is also an element of A, that means c is also an upper bound for for B. Therefore B is bounded above
**not sure if this is a good argument or not**
If B is contained in A that means, that there are no element in B that are greater than any of the elements in A. Therefore, the least upper bound of B must be less than or equal to the least upper bound of A.
**this seems rather short, I feel like i need to say more**
Any help/tips would be appreciated!