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I know it's quite obvious that $\limsup(a\cdot a_n)=a\cdot \limsup(a_n)$ for $a$ a real number >0, but I don't know how to prove it.

My second question is whether the following proof works for: $\limsup(a + b) \leq \limsup(a) + \limsup(b)$ for a and b as sequences.

http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2001;task=show_msg;msg=0119.0001.0001 Thanks!

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    For the inequality, you can look at these questions: [How to prove these inequalities in real analysis?](http://math.stackexchange.com/questions/205346/how-to-prove-these-inequalities-in-real-analysis), [Properties of $\liminf$ and $\limsup$ of sum of sequences](http://math.stackexchange.com/questions/70478/properties-of-liminf-and-limsup-of-sum-of-sequences) or [Subadditivity of the limit superior](http://math.stackexchange.com/questions/69391/subadditivity-of-the-limit-superior).2012-12-04

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Assuming $ a>0$.

\begin{equation} \limsup x_n = \lim_n (\sup \{x_m : m\ge n\}) \end{equation} Thus, assuming $a>0$ and $\sup a_n \ge 0$ we have. \begin{eqnarray} \limsup( a \cdot a_n \_n) &=& \lim_n (\sup \{a \cdot a_m : m\ge n\}) \\ &=& \lim_n [a \cdot(\sup \{ a_m : m\ge n\})]\\ &=& a \cdot \lim_n (\sup \{a_m : m\ge n\})\\ &=& a \cdot \limsup a_n. \end{eqnarray} Also, \begin{eqnarray} \limsup (a_n + b_n ) &=& \lim_n (\sup \{a_m + b_m : m\ge n\}) \\ &\le& \lim_n [(\sup \{a_m : m\ge n\}) + (\sup \{a_m : m\ge n\})]\\ &=& \lim_n (\sup \{a_m : m\ge n\} + \lim_n (\sup \{b_m : m\ge n\} \\ &=& \limsup a_n + \limsup b_n . \end{eqnarray}

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    Ok, I must assume that $ \sup a_n \ge 0$, In fact, if this is done and $S= \sup a_n$, we have $ a\cdot a_m \le a \cdot S $ and for any \varepsilon > 0 given, we can choose $ a_m $ such that | s - a_m | < a / \varepsilon . So, | a S - a a_m | = a |a - \_m | < \varepsilon. 2012-12-02
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$\{{a_m}+{b_m}:m\geqslant n\}\subseteq \{{a_m}+{b_k}:m,k\geqslant n\}$

since we are pairing elements from two sets together in the first set while drawing each elements at random from two sets in the second set. By taking the supremum we have:

$\sup\{{a_m}+{b_m}:m\geqslant n\}\leqslant\sup\{{a_m}+{b_k}:m,k\geqslant n\}\\=\sup\{\{{a_m}:m\geqslant n\}+\sup\{\{{b_m}:m\geqslant n\}$

Using $\textbf{lemma}$ : $\sup (A+B)=\sup A+ \sup B$ , where $(A+B)=\{a+b:a\in A,b\in B\}$ Taking limit of above iequality gets:

$\lim_{n\to\infty}\sup\{{a_m}+{b_m}:m\geqslant n\}\leqslant \lim_{n\to\infty}\sup\{{a_m}+{b_k}:m,k\geqslant n\}\\=\lim_{n\to\infty}\sup\{\{{a_m}:m\geqslant n\}+\lim_{n\to\infty} \sup\{\{{b_m}:m\geqslant n\}$ $Q.E.D$

Proof of $\textbf{lemma}$:

$\forall c\in A+B,\exists a\in A,b\in B,s.t.c=a+b\leqslant \sup A +\sup B$

So $A+B$ is bounded by $\sup A +\sup B$

$\forall \varepsilon \gt 0,\exists a \in A,b \in B ,s.t.a \gt \sup A-\varepsilon ,b \gt \sup B -\varepsilon ,a+b\gt \sup A +\sup B -2\varepsilon$

So any number less than $\sup A +\sup B $ is not an upper bound. Thus $\sup A +\sup B $ is the least upper bound.

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    In fact, you don't need the lemma: since $a_m \le \sup A$ and $b_k \le \sup B$, we must have $\sup (A+B) \le \sup A + \sup B$, whence the desired result follows.2015-11-17