Whether $\sup(\sum\limits_{i=1}^{\infty}X_i)$ is equal to $\sum\limits_{i=1}^{\infty}(\sup X_i)$

Question

We have $\sup(A+B)=\sup(A)+\sup(B) $.Thus, we have $\sup(\sum\limits_{i=1}^{n}X_i)=\sum\limits_{i=1}^{n}(\sup X_{i})$ for every finite integer $n\in\mathbb{N}$. However, what about the set sequence? More specifically, we have a set sequence $X_i$ and $X_i \subset \mathbb{R}^{+}$ for each $i\in\mathbb{N}$, that is to say, for each $x\in X_i$, $x\geq 0$, this guarantees the existence of $\sum\limits_{i=1}^{\infty}x_i(x_i\in X_i)$, whatever it is finite or infinite, so that we can talk about $\sup(\sum\limits_{i=1}^{\infty}X_i)$. I wonder whether $\sup(\sum\limits_{i=1}^{\infty}X_i)$ is equal to $\sum\limits_{i=1}^{\infty}(\sup X_i)$ just as finite circumstance. I get a wrong proof like this: fix $b\in\sum\limits_{i=2}^{\infty}X_i$,then for every $a\in X_1, a=a+b-b$, like we did in finite circumstance. But i soon realize we can't have $a=a+b-b$, since b maybe is infinite. If $\sup(\sum\limits_{i=1}^{\infty}X_i)=\sum\limits_{i=1}^{\infty}(\sup X_i)$, how to prove it. If not, how to get a countexample. Thanks!

Answer 1

For any $\epsilon > 0$, there are $x_i \in X_i$ with $x_i > \sup(X_i) - 2^{-i} \epsilon$, so $$ \sup\left(\sum_{i=1}^\infty X_i\right) \ge \sum_{i=1}^\infty x_i \ge \sum_{i=1}^\infty \left( \sup(X_i) - 2^{-i} \epsilon \right) = \sum_{i=1}^\infty \sup(X_i) - \epsilon $$ so taking $\epsilon \to 0$, we get $$\sup\left(\sum_{i=1}^\infty X_i\right) \ge \sum_{i=1}^\infty \sup(X_i)$$ On the other hand, for any sequence with $x_i \in X_i$, $x_i \le \sup(X_i)$, so $$ \sum_{i=1}^\infty x_i \le \sum_{i=1}^\infty \sup(X_i)$$ so that $$\sup\left(\sum_{i=1}^\infty X_i \right) \le \sum_{i=1}^\infty \sup(X_i)$$

Answer 2

They seems equal.

To make sense, we assume that all the number sequences converge at $\mathbb{R}\cup\{\pm\infty\}$.

$$a\in\sum_{i=1}^\infty X_i\implies a\le \sum_{i=1}^\infty \sup(X_i). $$Hence $$\sup(\sum_{i=1}^\infty X_i)\le \sum_{i=1}^\infty \sup(X_i). $$

On the other hand, $\forall \varepsilon>0, \forall i\in\mathbb{N}$, there exists $x_i\in X_i$ such that $x_i<\sup(X_i)-\varepsilon^i$. Then $$\sum_{i=1}^\infty x_i=\sum_{i=1}^\infty \sup(X_i)-\varepsilon^i=\sum_{i=1}^\infty \sup(X_i)-\frac{\varepsilon}{1-\varepsilon}. $$Hence we have $$\sup(\sum_{i=1}^\infty X_i)\ge \sum_{i=1}^\infty \sup(X_i)-\frac{\varepsilon}{1-\varepsilon}.$$

As $\varepsilon\to 0$, it is obvious that $$\sup(\sum_{i=1}^\infty X_i)\ge \sum_{i=1}^\infty \sup(X_i).$$