Supremum of supremums is equal to the supremum of the union My question is essentially the same as This previous question. That is, Let $C=\cup A_n:n\in \mathbb{N}$, show that $\sup\{\sup(A_n:n\in \mathbb{N})\}=\sup C$. The only difference in my case is $A_n \subset (-\infty,10]$.
It is easy to show that $\sup\{\sup(A_n:n\in \mathbb{N})\}\le\sup C$, but I don't understand how the previous answers show $\sup\{\sup(A_n:n\in \mathbb{N})\}\ge\sup C$.
In one answer it is proposed that $x<a\le \sup A_{i_0}\le\sup\{\sup(A_n:n\in \mathbb{N})\}$. An argument is given for $x<a\le \sup A_{i_0}$, but not for $\sup A_{i_0}\le\sup\{\sup(A_n:n\in \mathbb{N})\}$ which is the point. The other answers make similar arguments so what am I not understanding about this argument?
 A: 1) Because $C=\cup A_n\supset A_n\quad\forall n\in\mathbb N$ it follows that $\sup C\ge \sup A_n\quad\forall n\in\mathbb N$. Now the LHS is a fixed number and we can take again $\sup$ over all $n\in\mathbb N$, so we get $\sup C\ge \sup \left\{\sup A_n,\,\, n\in\mathbb N\right\}$.
2) Now we need to show that equality holds. To show this, assume that it is strict inequality, i.e $\sup C> \sup \left\{\sup A_n,\,\, n\in\mathbb N\right\}$ and let denote the number $M=\sup \left\{\sup A_n,\,\, n\in\mathbb N\right\}$. 
Then $M$ is not a $\sup$ for the set $C=\cup A_n$, which by definition means that $\exists a\in\cup A_n$ such that $M<a\leq \sup \cup A_n$. Let $a\in A_{n_0}$ for some index $n_0$. Then $a>\sup \left\{\sup A_n,\,\, n\in\mathbb N\right\}\ge\sup A_n,\quad\forall n\in\mathbb N$. In particular $a>\sup A_{n_0}$ which is impossible since $a\in A_{n_0}$. This contradistion shows that indeed we have an equality.
Note that since $A_n\subset (-\infty,10],\quad\forall n\in\mathbb N$ it follows that the set $C=\cup A_n\subset (-\infty, 10]$ which means that $\sup$ exists.
