Question 1210368
.
Google translation to English:


If A, B are two non-empty sets of R and bounded, prove that
inf (A+B) = inf A + inf B
~~~~~~~~~~~~~~~~~~~~~~~~~~~~



            The proof consists of two parts.



<pre>
            <U>First part of the proof</U>


Let a = inf(A),  b = inf(B).


Since the subsets A and B are bounded in R, the values 'a' and 'b' do exist and are defined properly.


The fact that a = inf(A) means that there is an infinite sequence {{{a[i]}}} of elements {{{a[i]}}} in A
which converges to 'a'.


The fact that b = inf(B) means that there is an infinite sequence {{{b[i]}}} of elements {{{b[i]}}} in B
which converges to 'b'.


Then the sequence  {{{a[i]+b[i]}}}  converges to value  a+b.

This simple elementary statement is easy to prove.


It implies that 

     inf(A+B) <= a+b.    (1)



            <U>Second part of the proof</U>


Again, let  a = inf(A),  b = inf(B).


Since the subsets A and B are bounded in R, the values 'a' and 'b' do exist and are defined properly.


Since the subsets A and B are bounded in R, the set of all real numbers of the form {x+y), 
where x is from A and y is from B, is bounded,  too.


Hence, the set of all sums (x+y) has the infinum.  Let z = inf(A+B).


The fact that z = inf(A+B) means that there is an infinite sequence {{{a[i]+b[i]}}} 
with elements {{{a[i]}}} in A  and  {{{b[i]}}}  in B, which converges to z.


Notice that all  {{{a[i]}}}  are not less than 'a',  and  all  {{{b[i]}}}  are not less than 'b', 
due to the definitions of 'a' and 'b'.


It means that

     z = inf(A+B) = ( lim {{{(a[i]+b[i])}}} as i --> {{{infinity}}} ) >= a + b = inf(A) + inf(B).    (2).


Inequalities (1) and (2), taken together,  prove that

    inf(A+B) = inf(A) + inf(B).


QED.
</pre>

Solved, proved and completed.