TITLE: 1-Variable Real Analysis True/False questions on Supremum, Infimum, and Inequalities with them
QUESTION [7 upvotes]: (S. Abbott. Understanding Analysis 1 ed. pp 18 question 1.3.9) is asking me to answer the following questions without any formal proofs. I have some intuition for them, but I was hoping to get some external input as well. It would be great if you could maybe give some rigorous explanations to the intuition behind.
a) A finite, nonempty set always contains its  supremum. - I think this is True.
b) If $a < L$ for every element $a$ in the set $A$, then sup$A < L$. - I think this is False, because it could be $\leq$.
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$B$. - I think that again, it's $\leq$ and not strictly less.
d) If sup$A$ = $s$, and sup$B$ = $t$, then sup($A+B$)=$s+t$. $A+B={a+b | a \in A, b \in B}$. - I think it's True, after thinking of several examples with sets that do and do not contain their suprema, but I am not sure.
e) If sup$A \leq$ sup$B$, then there exists an element $b \in B$ that is an upper bound for $A$. - I thought this was False, because we can set $B=A$, and the condition on superma would hold, yet pick such an $A$ that doesn't contain its own supremum.
Thank you!

REPLY [4 votes]: Since you asked us to provide a formal proof, I hope that my answer is relevant.
a) I believe that this is about a subset of $\mathbb{R}$. Let us call this finite set $A$ and put #$A = n$. For each two elements $a,b\in A$ there are three possibilities: $a<b$ or $b<a$ or $a=b$. For each $a\in A$ there is $B(a) = \{b\in A: a>b\}$ which is also finite. 

Start with any $a_1\in A$, then #$B(a_1)\leq n-1$. 
Take $a_2\in B(a_1)$, then #$B(a_2)\leq n-2$.
Note that there will be $k\leq n$ such that #$B(a_k) = 0$. (Suppose contrary - then #$B(a_{n+1})\leq -1$ - contradiction with non-negativity of #$B(a)$ for any $a\in A$).
For each $a\in A$ we have $a\leq a_k$. (Suppose contrary - then #$B(a_k)\geq 1$, contradiciton). Moreover $a_k\in A$, so $a_k=\max\limits_{a\in A} \,\,\,a$.

b) Consider $a\in [0,1)$ and $L=1$.
c) Consider $A = [0,1)$ and $B = (1,2)$.
For d) Jonas gave already a link, e) you proved formally.