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Let $\{x_n\}_{n=1}^\infty$ and $\{y_n\}_{n=1}^\infty$ be sequences of real numbers. Verify that each of the following holds, provided the right-hand side makes sense. $$\limsup(x_n+y_n)\geq\limsup(x_n)+\liminf(y_n)$$

For this problem we are allowed to assume the following holds: $$\limsup(x_n+y_n)\leq\limsup(x_n)+\limsup(y_n)$$ $$\limsup(-x_n)=-\liminf(x_n)\text{ and } \liminf(-x_n)=-\limsup(x_n)$$

What I've done so far:

See that \begin{equation} \begin{split} \limsup(x_n)=\limsup(x_n+y_n-y_n) & \leq\limsup(x_n+y_n)+\limsup(-y_n) \\ & = \limsup(x_n+y_n) -\liminf(y_n) \end{split} \end{equation}

So we obtain $$\limsup(x_n+y_n)\geq\limsup(x_n)+\liminf(y_n)\text{ as desired.}$$

I feel like this should hold, but I'm curious if this approach covers all potential exceptions. What if we are talking about the extended real numbers? What if $\{y_n\}$ or $\{x_n\}$ diverge?

Martin Sleziak
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Aidan
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    When you're working in $\bar{\mathbb{R}} = \mathbb{R} \cup {\pm \infty }$, the $\limsup$ and $\liminf$ of a sequence always exist, whether the sequence converges or not. That's part of what makes them so useful. – David Jan 24 '16 at 22:40
  • So, that is to say on the extended reals, it's not even necessarily a different consideration? – Aidan Jan 24 '16 at 22:42
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    The only possible problem here in the extended reals is that you can't add $+\infty$ and $-\infty$. A $\limsup$ of a sequence of real numbers can never be $-\infty$. So when you write $\limsup(x_n + y_n) + \limsup(-y_n)$, this problem can't occur. You have an inequality $A \leq B + C$, where none of the three numbers is $-\infty$, and you want to conclude that $A - C \leq B$ provided $A - C$ makes sense. The inequality is vacuously true if $B$ or $C$ is $+\infty$. The only case left is that $B$ and $C$ are both finite. In that case, $A$ too must be finite. – David Jan 24 '16 at 22:55
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    See also this post and other questions linked there – Martin Sleziak Jul 19 '16 at 08:01

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