Assume that $$X=\sum_{i=1}^n X_i$$ with $X_i\subset\Bbb{R}.$ I'd like to show that $\sup X=\sum_{i=1}^n \sup X_i$. I came to this question and its answer from a russian book called Курс математического анализа В 3 томах_Том 1_Кудрявцев" it translates as "Mathematical analysis course in 3 tomes_tom 1_kudriavtsev". I know i can show it by inducction using the case for two sets. However, i'd like to understand the proof given below, hoping to find a russian supporter here. This is what i've got translating with google translator:
We need to verify that: $$1. (\forall x\in X),x\le \sum_{i=1}^n \sup X_i $$ $$2. \left(\forall y\in \Bbb R \text{ such that }y\lt \sum_{i=1}^n \sup X_i \right)\text{ }\exists x\in X\ \text{such that }y\lt x $$
The 1. comes here:
The part 2. is the one i'm struggling with: 
I've translated it as: if $y\lt \sum_{i=1}^k \sup X_i$, then for every k=1,2,...,n there exists a number $y_k$ such that $$y_k\lt \sup X_k,\quad y_1+...+y_n=y.$$ Indeed, if every upper bound $\sup X_1,...,\sup X_n$ is finite, then you can take $$y_k=\sup X_k-\left\{\left\{\sum_{i\neq k} \sup X_i \right\}-y\right\}$$. Then, $y_1+...+y_n=y.$ Obviously, the last equality is not true. Can someone explain me what the author is trying to say? This is not really about solving the problem. It is more about understand it in this way. Probably there is a Russian supporter here, but anyone who can help me with this proof is welcome.
The statement is true and the proof is straight forward.
WLOG we can assume $n=2$.
Then we have $$x_1 + x_2 \le \sup X_1 + \sup X_2$$ so $\sup X_1 + \sup X_2$ is an upper bound for $X_1 + X_2$.
This is (3.9) from the author.
The following statement (3.11) is wrong in general without further assumptions. It's true if all of the $X_i$ are intervals. But if you exchange $=$ by $>$ the statement becomes true again what's enough to complete the proof.
Because: We've already shown that $\sup X_1 + \sup X_2$ is an upper bound of $X_1 + X_2$, we still have to show that it's the least upper bound to let $$\sup(X_1 + X_2) = \sup X_1 + \sup X_2$$ hold
So we have to show that for each $\varepsilon > 0$ there exists an $y \in X_1 + X_2$ s.t. $y > \sup X_1 + \sup X_2 - \varepsilon$ and this is what the author does.
Because $\sup X_i$ is the supremum of $X_i$ we found an $y_i$ s.t. $$y_i > \sup X_i - \frac{\varepsilon}{2}$$
And so we get: $$y := y_1 + y_2 > \sup X_1 - \frac{\varepsilon}{2} + \sup X_2 - \frac{\varepsilon}{2} = \sup X_1 + \sup X_2 - \varepsilon$$
which completes the proof.