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?


(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. $\displaystyle{\sup (\cup_{k=1}^{n} A_k) = \max_{1\le k \le n}\{\sup A_k\}}$.

  • 2
    $\begingroup$ Supremum of any union is the supremum of the suprema of those sets. Even for uncountbale unions! $\endgroup$ Oct 11, 2020 at 8:45
  • 3
    $\begingroup$ Kavi Rama Murthy is correct under one condition: Namely, that you allow $\infty$ as a valid value for the supremum. Since you are restricting yourself to bounded sets, I am assuming this is not true in your case. The problem with infinite collections of sets (countable or uncountable) is that each of the sets can be bounded, but the union of all of them can still be unbounded. $\endgroup$ Oct 11, 2020 at 16:22
  • $\begingroup$ Understanding analysis book defines the existence of suprema by using the Axiom of Completeness. The second argument, therefore, does not hold (given the book's assumptions and all), since, e.g., if we take $A_k = [1, k]$ we have $\bigcup_k A_k = [1, \infty)$ which is not bounded above, hence not complete. Therefore we cannot find a supremum to $[1, \infty)$, neither does your answer in part (a) hold: e.g., $\max \{ \sup A_1, \sup A_2, \ldots \} = \max \{ 1, 2, \ldots \} = \max \mathbb{N}$ (which does not exist, as even $\sup \mathbb{N}$ does not exist). $\endgroup$ Nov 27, 2021 at 0:14

1 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 $Sup(\cup_{j=1}^{n} A_{n}$)=$Sup(A_{1}\cup(\cup_{j=2}^{n} A_{n})$)= $max\{Sup(A_{1}),max\{Sup(A_{2}),Sup(A_{3}),...Sup(A_{n})\}\}$ $=$ $max\{Sup(A_{1}),...,Sup(A_{n})\}$


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