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Suppose ${{A_1}}$=[1,3] and ${{A_2}}$=[2,4], then ${{A_1} \cup {A_2}}$=[1,4] now $\sup \left( {{A_1} \cup {A_2}} \right)$ is clearly 4. so, $\sup \left( {{A_1} \cup {A_2}} \right) = \max \left( {\sup {A_1},\sup {A_2}} \right)$ is true.

Confusion with definition:
s is least upper bound for a set $A \subseteq R$ if two criterion are met
(1) s is an upper bound for A
(2) if b is any upper bound for A, then $s \le b$
In the proof if I take ${s_1}$ to be ${\sup {A_1}}$ and ${s_2}$ to be ${\sup {A_2}}$, then if I apply definition then least of ${s_1}$ and ${s_2}$ is $\sup \left( {{A_1} \cup {A_2}} \right)$, which is certainly not true. What is exactly am I missing here?

Then once proved how can I extend it to $\sup \left( { \cup _{k = 1}^n{A_k}} \right)$ ? May be if i get clear with the base case then it will not be required.
Edit: ${{A_1}}$ and ${{A_2}}$ are nonempty sets which are bounded above.

Asaf Karagila
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  • "I apply definition then least of ${s_1}$ and ${s_2}$ is $\sup \left( {{A_1} \cup {A_2}} \right)$, which is certainly not true" -- no, certainly not. What makes you think that "applying the definition" compels that wrong conclusion? More details about how your failed attempt to apply the definition, please. – hmakholm left over Monica Jan 20 '17 at 12:36
  • The smaller $s_i$ will probably not even be an upper bound of the union. Just as 3 isn't an upper bound of $[1,4]$ – Blaza Jan 20 '17 at 12:36
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    Why are you taking the least of $s_1$ and $s_2$, your formula involves a $\max$ and not a $\min$. – Michael Burr Jan 20 '17 at 12:38
  • The example contradicts it. I know I am doing something wrong here because I started as ${{A_1}}$ and ${{A_2}}$ as different set, but when union is taken they are merged with "or" condition. So it was intuitive that maximum of supremum of two sets will be correct answer. How do I do it in mathematical language is where I am confused. The criterion (2) is certainly not understood by me correctly, and how should I take this condition from single set to set obtained by operation over multiple sets is also not understood. –  Jan 20 '17 at 12:45
  • See answer below. The "leastness" applies to other upper bounds of the set in question in this case $A_1\cup A_2$ so if $s_1<s_2$ then $s_1$ is NOT and ub of $A_1\cup A_2$, it is just an ub of $A_1$ – Ansel B Jan 20 '17 at 13:09

2 Answers2

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$\newcommand{\eps}{\varepsilon}$Suggestion: To prove that a specific real number $c$ is the supremum of a non-empty set $A$ of real numbers (that is bounded above), it's often helpful to use the second condition of the definition in contrapositive form:

  1. $c$ is an upper bound for $A$, i.e., for every $x$ in $A$, $x \leq c$;

  2. For every $\eps > 0$, there exists an $x$ in $A$ such that $c - \eps < x$.

So, to show that $c := \max(\sup A_{1}, \sup A_{2}) = \sup(A_{1} \cup A_{2})$, it suffices to show:

  1. If $x \in A_{1} \cup A_{2}$, then $x \leq \max(\sup A_{1}, \sup A_{2})$;

  2. For every $\eps > 0$, there exists an $x$ in $A_{1} \cup A_{2}$ such that $\max(\sup A_{1}, \sup A_{2}) - \eps < x$.

Andrew D. Hwang
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  • Can you please explain how contrapositive form is used. Please if possible explain regrading contrapositive form and normal form, how the ingenuity of contrapositive is so useful. Taking above as an example case. –  Jan 21 '17 at 07:44
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    My answer to When to use the contrapositive to prove a statment explains the reasons for considering the contrapositive in this type of situation. Here, you know that for every $\eps > 0$, there exists an $x_{1}$ in $A_{1}$ such that $\sup A_{1} - \eps < x_{1}$, and similarly for $A_{2}$; your goal is to find an $x$ in $A_{1} \cup A_{2}$ such that $\max(\sup A_{1}, \sup A_{2}) - \eps < x$. If you can show $x$ may as well be one of the $x_{i}$, you've shown 2. [...] – Andrew D. Hwang Jan 21 '17 at 12:58
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    Once you have a "strategy", it's a good idea to "test" or "debug" it by seeing how it works with specific sets. (I'm deliberately not saying too much here because the only way to learn these ideas is to wrestle with them yourself. It often helps to think of "picking the supremum" as a game of solitaire: You have a set $A$, you pick an $x$, and you try to prove that $x$ satisfies two conditions. If you succeed, then $x = \sup A$.) – Andrew D. Hwang Jan 21 '17 at 13:01
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It will be the case that $s1\leq s_2$ or $s_2\leq s_1$, but we can take wlog $s_1\leq s_2$ as the argument will be mirrored in the dual case. So then $s_2$ is a uper bound (ub) of both $A_1$ and $A_2$ and the union as well. So then if u is a ub of $A_1\cup A_2$ it is a ub of $A_2$ as well so then it must be that $s_2\leq u$ and so $s_2$ is the least upper bound of the union, that is $\sup(A_1\cup A_2)=\max\{\sup(A_1),\sup(A_2)\}$. And yes you are correct it can be extended to any $n\in\mathbb{N}$, by a trivial induction argument. Just to note the definition of the supremum tells you that if you want to show that $x$ is the supremum of a set $Z$, first show that $x$ is and ub of $Z$ and if $u$ is another ub of $Z$, then $x\leq u$.

Ansel B
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  • Thanks A B, this statement was really helpful, "So the $s_2$ is a uper bound (ub) of both $A_1$ and $A_2$ and the union as well. So then if u is a ub of $A_1\cup A_2$ it is a ub of $A_2$ as well so then it must be that $s_2\leq u$ and so $s_2$ is the least upper bound of the union." –  Jan 21 '17 at 07:23
  • Yes, that is correct. This is vitally important you understand the definition for the supremum and infimum, as it is crucial for any Analysis course, or for any concepts involving a set with a partial order on it (Lattice theory etc). The below characterization given by Mr Hwang is also very useful sets of real numbers. I hope you can show it equivalent to the definition you give above. You should also be able to show that if $s=\sup{A}$, where $A\subset\mathbb{R}$ then there is a sequence in $(a_n)\subset A$ such that $a_n\rightarrow s$ as $n\rightarrow\infty$, and dually for the infimum. – Ansel B Jan 22 '17 at 11:08