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As part of a proof, I want to show that

$$\sup_{k \geq n}\{x_k\} + \sup_{k \geq n}\{y_k\} = \sup_{j, k \geq n}\{x_j + y_k\}$$

where we can assume that the sups of $x_k$ and $y_k$ are finite.

However, I'm not sure how to argue this equality holds. I'm thinking along the lines of: we know that the biggest element in $\{x_j + y_k\}$ would have to be the sums of the sups of $x_j$ and $y_k$ - otherwise, the sum wouldn't be the largest - and as a result, we can conclude the above equality for some $j, k$. it seems a little hand-wavy to me, but is that a correct argument?

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    I think your idea is correct. You can make it rigorous by plugging in some $\epsilon$ (i.e. for any $\epsilon > 0$, there exists $j\geq n$ such that $x_j + \epsilon \geq \operatorname{sup}x$, etc.). – WhatsUp Sep 23 '21 at 19:45
  • To make it rigorous, take split the problem up: can you show that $\sup_{k \geq n}{x_k} + \sup_{k \geq n}{y_k}$ is an upper bound on $x_j + y_k$ for $j, k \ge n$? Sure, because $x_j \le \sup_{k \geq n}{x_k}$ and $y_k \le \sup_{k \geq n}{y_k}$ (excuse the abuse of notation here, using $k$ both as a fixed and dummy variable). You then need to show it's the least upper bound, without tacitly assuming a maximum exists, which you can do using WhatsUp's idea. – Theo Bendit Sep 23 '21 at 20:12
  • See for example https://math.stackexchange.com/q/4551/42969. – Martin R Sep 23 '21 at 20:13

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