# 1-Variable Real Analysis True/False questions on Supremum, Infimum, and Inequalities with them

(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!

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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.

1. Start with any $a_1\in A$, then #$B(a_1)\leq n-1$.

2. Take $a_2\in B(a_1)$, then #$B(a_2)\leq n-2$.

3. 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 $a\in A$).

4. 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 $B = (1,2)$.

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a) Correct. You can formally prove this by induction. For the case of one point, observe that $x=\sup\{x\}\in\{x\}$. Suppose that it is true for sets containing $n$ real numbers, where $n$ is a positive integer. If $\{x_1,x_2,\ldots,x_{n+1}\}$ is a set of $n+1$ real numbers, then by hypothesis $\sup\{x_1,\ldots,x_n\}=x_k$ for some $k\in\{1,\ldots,n\}$, and $\sup\{x_1,\ldots,x_{n+1}\}=\max\{x_k,x_{n+1}\}\in\{x_1,\ldots,x_{n+1}\}$.

b) Correct. Can you give an example where the equality holds? (Update: Gortaur has now given an example.)

c) See b)

d) Correct, it is true. See this other question.

e) Correct, for the correct reason. Can you give an explicit example? (Hint: See Gortaur's example for b).)

Note that for the ones that are false (b, c, e), all you need to do to make your answers more formal is to give an example. For the ones that are true (a,d), a proof would be needed.

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