# How can I prove that $\sup(\bigcup_{i \in I} A_{i}) =\sup\{\sup \,A_{i} : i \in I \}$? [closed]

Let $I$ be non-empty set (of "indexes") and for all $i \in I$ let $A_{i} \subset \mathbb{R}$ be non-empty and upper bounded set. How can I prove that $$\sup \left( \bigcup_{i \in I} A_{i} \right) = \sup\{\sup A_{i} \mid i \in I \}$$

## closed as off-topic by Najib Idrissi, Lost1, Micah, Carl Mummert, anomalyOct 27 '14 at 2:10

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## 1 Answer

Let $x\in\bigcup_{i\in I}A_i$. Then, there exists some $j\in I$ such that $x\in A_j$. Therefore, $$x\leq\sup A_j\leq\sup\{\sup A_i\,|\,i\in I\}.$$ Since this is true for any $x\in\bigcup_{i\in I}A_i$, it follows that $$\sup\left\{\bigcup_{i\in I}A_i\right\}\leq \sup\{\sup A_i\,|\,i\in I\}.$$

For the converse, pick any $j\in I$ and $x\in A_j$ (this is possible by the assumption that $A_j$ is not empty). Then, since $x\in\bigcup_{i\in I}A_i$, it follows that $$x\leq\sup\left\{\bigcup_{i\in I}A_i\right\}.$$ This is true for all $x\in A_j$, so $$\sup A_j\leq\sup\left\{\bigcup_{i\in I}A_i\right\}.$$ Since $j\in I$ was arbitrary as well, one can conclude that $$\sup\{\sup A_i\,|\,i\in I\}\leq\sup\left\{\bigcup_{i\in I}A_i\right\}.$$

Note: Even though $A_i$ is assumed to be bounded from above for each $i\in I$ (so $\sup A_i$ exists), there is no guarantee that $\bigcup_{i\in I}A_i$, too, is bounded from above. Therefore, it may occur that the two suprema whose equality has just been established are actually infinite.

• Thanks I really appreciate it. That is what I was looking for. – VirrageS Oct 26 '14 at 19:54