Supremum of the union of countably finite and infinite number of sets

Question

I am a beginner at Real Analyis, and I am loving reading from the book Understanding Analysis by Stephen Abbott. I would like to ask: is my proof for the supremum of the union of two sets is technically correct and rigorous?

Additionally, how should I think about a countably infinite union of sets?

How would I know, the supremum of all the sets in a countably infinite collection, unless the set $A_n$ is given by some general formula?

Let $A_1,A_2,A_3,\ldots$ be a collection of non-empty sets, each of which is bounded above.

(a) Find a formula for $\sup (A_1 \cup A_2)$. Extend this to $\displaystyle{\sup (\cup_{k=1}^{n} A_k)}$.

(b) Consider $\displaystyle{\sup (\cup_{k=1}^{n} A_k)}$. Does this formula extend to the infinite case?

Proof.

(a) Claim: $A_1 \cup A_2$ is bounded above.

We know that, $A_1 \cup A_2 := \{x : (x \in A_1)\lor (x \in A_2)\}$. There exists upper bounds $u_i$, such that $x \le u_i (\forall x \in A_i)$, $i=1,2$. By the trichotomy property, atleast one $u_1 < u_2, u_1 = u_2, u_1 > u_2$ holds. Let us assume, for simplicity, $u_2 > u_1$. Then, $x \le u_1 < u_2$ for all $x \in A_1$ and $x \le u_2$ for all $x \in A_2$. Thus, $x \le \max (u_1,u_2)$ for all elements $x \in A_1 \cup A_2$.

Claim: $m = \max \{\sup A_1, \sup A_2\}$ is an upper bound for $A_1 \cup A_2$.

We can argue as above $m$ is an upper bound for $A_1 \cup A_2$. This verifies part (i) of the definition of the supremum of a set.

Claim: If $v$ is any other upper bound of $A_1 \cup A_2$, then $m \le v$.

Let $v$ be any other upper bound for $A_1 \cup A_2$. Then, $x \le v$ for all elements $x \in (A_1 \cup A_2)$. This implies two things : $x \le v$ for all elements $x$ in $A_1$ and $x \le v$ for all elements $x \in A_2$. So, $v$ an upper bound for $A_1$ and $A_2$. But, we know that, $\sup A_1 \le v$. And $\sup A_2 \le v$. Hence, $m \le v$.

This closes the proof.

In general, we can extend this formula to $n$ terms. $\displaystyle{\sup (\cup_{k=1}^{n} A_k) = \max_{1\le k \le n}\{\sup A_k\}}$.

Answer 1

Countable union of bounded sets need not be bounded above ,Consider the case $A_{n}=[1,n]$ .Each of the sets is bounded above for a given natural number but $Sup(\cup_{j=1}^{n} A_{j}$)=$Sup\{n:n \in \mathbb{N}\}$=$\infty$

So the formula cannot be extended to countable union and also the above sequence $Sup(\bigcup_{j=1}^{n} A_{j}$) is a increasing sequence of real numbers and is convergent iff its bounded above. For finite union of bounded sets sets it is true that $Sup(\cup_{j=1}^{n} A_{n}$)=$max\{Sup(A_{i}):i \in \mathbb{N}\}$ as $Sup(\cup_{j=1}^{n} A_{n}$)=$Sup(A_{1}\cup(\cup_{j=2}^{n} A_{n})$)= $max\{Sup(A_{1}),max\{Sup(A_{2}),Sup(A_{3}),...Sup(A_{n})\}\}$ $=$ $max\{Sup(A_{1}),...,Sup(A_{n})\}$