How do I show $\sup(\bigcup_{n=0}^{\infty} A_n) \leqslant \sum_{n=0}^\infty \sup(A_n)$?

Question

How do I show that $\sup\left(\displaystyle\bigcup_{n=0}^\infty A_n\right) \leqslant \displaystyle\sum_{n=0}^\infty \sup(A_n)$, where $A_n \subseteq [0,\infty)$? I am looking for a proof of this inequality. Can anyone suggest an idea on how to prove this?

Answer 1

This statement, as written, seems false.

Let $A_1 = \{-5\}$, let $A_2 = \{3\}$ and let $A_i = \{0\}$ for all $i \geq 3$.

Then $sup(\cup_{i=1}^{\infty} A_i) = 3 \geq -2 = \sum_{i=1}^{\infty} sup(A_i)$.

Answer 2

Given a conditionally complete ordered set $(A, R)$ -- i.e. an ordered set having the property that any nonempty upper-bounded subset $M \subseteq A$ admits a supremum -- one actually has the following:

Proposition. Let $(A, R)$ be conditionally complete and $X \in \left(\mathscr{P}(A)\setminus\{\varnothing\}\right)^I$ be a nonempty family of nonempty subsets, which also means that $I \neq \varnothing$. Then the following two statements are equivalent: a) The union $Y\colon=\displaystyle\bigcup_{i \in I}X_i$ is nonempty and upper bounded and b) for each $i \in I$ the subset $X_i$ is nonempty and upper-bounded and furthermore the set of suprema $\left\{\sup X_i\right\}_{i \in I}$ is non-empty and upper-bounded. When these conditions are satisfied one has:

$$\sup\left(\displaystyle\bigcup_{i \in I}X_i\right)=\sup \left\{\sup X_i\right\}_{i \in I}$$

Proof. The argument follows easily from the definition. Assume first that hypothesis a) holds. The general hypothesis tells us that $X_i \neq \varnothing$ for every index $i \in I$ and since $X_i \subseteq Y$, whatever upper bound $a \in A$ there exists for the union $Y$ will also be an upper bound for $X_i$. Thus $X_i$ admits a supremum which by definition satisfies $\sup X_i \leqslant_R a$. The set $T\colon=\left\{\sup X_i \right\}_{i \in I}$ of suprema is clearly nonempty since $I \neq \varnothing$ and the previous relation -- that $\sup X_i \leqslant_R a$ for every $i \in I$ -- means nothing else than that $a$ is also an upper bound for $T$.

Conversely, assume now that b) holds. Let $b \in A$ be an upper bound for the set $T$ of suprema. Given arbitrary $x \in Y$, there exists $i \in I$ such that $x \in X_i$ and we have by definition that $\sup X_i$ is an upper bound of $X_i$, thus in particular $x \leqslant_R \sup X_i$. Since we also have $\sup X_i \leqslant_R b$, it follows that $x \leqslant_R b$ holds for any $x \in Y$, so that $b$ is established as an upper bound of $Y$ as well. That $Y \neq \varnothing$ is indeed nonempty follows easily from the hypothesis, since $I \neq \varnothing$ allows us to fix a certain index $\alpha \in I$ and claim that $\varnothing \subset X_{\alpha} \subseteq Y$ (where I am referring to strict inclusion).

Finally, in order to establish the mentioned relation, let us set $s\colon=\sup \displaystyle\left(\bigcup_{i \in I}X_i\right)$ and $t\colon=\displaystyle\sup_{i \in I} \sup X_i$. It is easy to prove that if $\varnothing \neq S \subseteq T$ and $T$ is upper-bounded, then $\sup S \leqslant_R \sup T$ (the supremum of $T$, being an upper bound for the larger $T$, will also be an upper bound for the subset $S$). This applies in particular to $\varnothing \neq X_i \subseteq Y$, entailing $\sup X_i \leqslant_R \sup Y=s$, whence by definition of the supremum we furthermore have $t\leqslant_R s$. Conversely, consider an arbitrary $x \in Y$. There exists $i \in I$ such that $x \in X_i$ and therefore $x \leqslant_R \sup X_i \leqslant_R t$. This means that $t$ is in particular an upper bound for $Y$, whence $s \leqslant_R t$, which by antisymmetry of $R$ leads to $s=t$. $\Box$

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In your concrete case, you seem to be considering a (nonempty) family $X \in \left(\mathscr{P}([0, \infty))\setminus\{\varnothing\}\right)^I$ of nonempty sets of positive real numbers. Since with the standard order the extended positive semi-axis $[0, \infty]$ does indeed form a complete (and thus conditionally complete) ordered set, the above proposition applies to the effect that $\sup\left(\displaystyle\bigcup_{i \in I}X_i\right)=\displaystyle\sup_{i \in I}\sup X_i$. In order to prove the desired inequality, it therefore suffices to show that for any family $x \in [0, \infty]^I$ of extended positive reals the relation: $$\displaystyle\sup_{i \in I}x_i \leqslant \sum_{i \in I}x_i \tag{*}$$ holds. However, from the elementary characterisation of summability of arbitrary families of extended positives we know that: $$\displaystyle\sum_{i \in I}x_i=\displaystyle\sup_{F \in \mathscr{F}(I)} \sum_{i \in F}x_i,$$ where by $\mathscr{F}(M)\colon=\{X \subseteq M|\ |X| \in \mathbb{N}\}$ I am denoting the set of all finite subsets of a given set $M$. In other words, the sum of an arbitrary family of extended positive reals exists indeed and is given by the supremum of the family of all finite subsums (i.e. sums of finite subfamilies of the given family). Since in particular the singletons $\{i\} \in \mathscr{F}(I)$ are finite subsets of $I$ for each index $i \in I$, it follows that $\{x_i\}_{i \in I} \subseteq \left\{\displaystyle\sum_{i \in F}x_i\right\}_{F \in \mathscr{F}(I)}$, whence by virtue of the increasing character of the $\sup$ operator we may deduce the desired claim $(^*)$.

P.S. To state this last remark it is useful to introduce the notion of support of an arbitrary family $y \in \overline{\mathbb{R}}^I$ of extended real numbers by $\mathrm{Supp}(y)=\{i \in I|\ y_i \neq 0\}$. It can be proved without great difficulty that given a nonempty family $u \in [0, \infty]^I$ of extended positives such that $\displaystyle\sup_{i \in I} u_i<\infty$, we have equality $\displaystyle\sup_{i \in I}u_i=\displaystyle\sum_{i \in I}u_i$ in the above general relation $(^*)$ if and only if $|\mathrm{Supp}(u)| \leqslant 1$, in other words if and only if the support of $u$ is at most a singleton.

Answer 3

First, note that the inequality might be undefined as any of the following might hold

$$\sup(A_i)=\infty\text{ for some }i=1,2,...$$

$$\sup\left(\bigcup_i^\infty A_i\right)=\infty$$

$$\sum_{i=1}^\infty \sup(A_i)=\infty$$

$$A_i=\emptyset\text{ for some }i=1,2,...$$

However, assuming that these do not occur (and you can treat these as separate cases if you are willing to say that $\infty=\infty$ and $x\leq \infty$ for $x\in\mathbb{R}$) then the inequality is true. One definition of the supremum states: for all $\epsilon>0$, there exists $x\in A$ such that $0\leq \sup(A)-x<\epsilon$. We will use this definition.

Let $\epsilon>0$ be arbitrary. Then there exists

$$\alpha\in \bigcup_i^\infty A_i$$

such that

$$0\leq \sup\left(\bigcup_i^\infty A_i\right)-\alpha<\epsilon$$

$$\alpha\leq \sup\left(\bigcup_i^\infty A_i\right)<\alpha+\epsilon$$

Let $k$ be the smallest natural such that $\alpha\in A_k$. This implies

$$\alpha\leq \sup(A_k)$$

$$\sup\left(\bigcup_i^\infty A_i\right)<\alpha+\epsilon\leq \sup(A_k)+\epsilon$$

But then

$$\sup\left(\bigcup_i^\infty A_i\right)< \sup(A_k)+\epsilon\leq \epsilon+\sum_{i=1}^\infty \sup(A_i)$$

Since $\epsilon$ was arbitrary, we conclude

$$\sup\left(\bigcup_i^\infty A_i\right)\leq \sum_{i=1}^\infty \sup(A_i)$$

Answer 4

We assume that $A_n\neq \emptyset$ for all $n$. For each $n\in\mathbb{N}$, denote $M_{n}=\sup A_{n}\in[0,\infty]$. Let $x\in\cup_{n}A_{n}$, then $x\in A_{k}$ for some $k\in\mathbb{N}$. Therefore, $x\leq M_{k}\leq\sum_{n=1}^{\infty}M_{n}$. This shows that $\sum_{n=1}^{\infty}M_{n}$ is an upper bound of the set $\cup_{n}A_{n}$ and hence $\sup\left(\cup_{n}A_{n}\right)\leq\sum_{n=1}^{\infty}\sup(A_{n})$.