Without worrying about formal proofs for the moment, decide if the following statements about suprema and infima are true or false. For those that are false, supply an example where the claim in question does not appear to hold.
a) A finite, nonempty set always contains its supremum.
b) If a < L for every element a in the set A, then sup A < L
c) If A and B are sets with the property that a < b for every a in A and every b in B, then it follows that sup A < inf B.
d) If sup A = s and sup B = t, then sup A + B = s + t. The set A + B is defined as A+B = {a+b : a in A and b in B}.
e) If sup A <= sup B, then there exists an element b in B that is an upperbound for A.

A) TRUE

B) FALSE

C)  FALSE

D) TRUE

E)  FALSE

Step-by-step explanation:

A)    A finite, nonempty set always contains its supremum. This is TRUE. Do remember that a "finite" set is one with finitely elements,not a "bounded" set. There is always a maximum element, and the server as the supremum.

B) If a∠L for every element a in the set A, then sup. A∠ L FALSE Let A=(0,1), the open interval Let L=1. then sup. A=L

C)  If A and B are sets with the property that a∠b for every a ∈ A and every b ∈ B, then it follows that sup A ∠ inf B. FALSE. We use open intervals again. Let A= (0,1) and B =(1,2). Then sup A=inf B =1

D) If sup A= s and B=t, then sup(A+B)=s+t. The set A+B is defined as A+B= (a+b : a ∈ A and b ∈ B) TRUE

E) If sup A ≤ sup B, then there exists an elements b ∈ B that is an upper bound for A. FALSE. We can take both A and B to be the open interval (0,1). The superma are of course the same (1), and there is no element of B that is an upper bound of A.

step-by-step explanation: