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Consider the following assertion:

Let $\mathbb{R}^*=\mathbb{R}\cup\{+\infty, -\infty\}$. For each $i \in \mathbb{N}$, let $S_i\subset\mathbb{R}^*$ a set of non-negative extended real numbers. Define $\sum_{i=1}^\infty S_i = \{\sum_{i=1}^\infty x_i : x_i\in S_i, i\in\mathbb{N}\}$. Then $\inf\sum_i S_i\leq\sum_i\inf S_i$.

Is this statement true? How can I prove it, since there are any countable collection of sets?

To clarify, I aim to generalize the assertion over a finite sum of sets:

$\inf(A+B) = \inf A +\inf B$,

where $A,B\subset \mathbb{R}$ are nonempty limited subsets and

$A+B:=\{a+b : a\in A~\mbox{and}~b\in B\}.$

In the "general" case above, I need only the inequality "$\leq$" instead of the equality.

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    that notation is really weird, why does $S_i$ has a sum sign? – 2017-01-24
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    over what set is the first inf being taken over? can you state the assertion exactly as it was given to you? – 2017-01-24
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    I am worried when the sum in brackets on the definition of $\sum S_i$, for example, $S_1=\cdots = S_n = \cdots = \{-1,1\}$ and consider $1-1+1-1+\cdots$. – 2017-01-24
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    (Oh, I forgot to write the phrase "... is not well-defined" in my first sentence.) – 2017-01-24
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    @JorgeFernándezHidalgo I change some things to clarify what I mean. – 2017-01-24
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    @DanielXiang Please read again. Maybe there are some mistake, maybe a simple "no" will help. – 2017-01-24
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    @HanulJeon Please Hanul, read again – 2017-01-24
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    Its proof is (surprisingly) easy: for each $i$ take $a_i\in S_i$ such that $\inf S_i \ge a_i - \varepsilon/2^i$ then... – 2017-01-24
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    @HanulJeon Yes, that was simple!! Thank you man =D (a question: is the equality also true?) – 2017-01-24
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    Yes, you can prove the reverse inequality so they are equal. – 2017-01-24
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    the reverse is much easier. Thanks Hanul – 2017-01-24

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