# $\inf\limits_{\{x_n\} \in c_0} \sup \{a_n + x_n\} = \limsup\limits_{n\to\infty} a_n$

Suppose $$a_n$$ is bounded and let $$c_0$$ be the set of all sequences $$x_n$$ such that $$\lim\limits_{n\to\infty} x_n =0$$. I need to show $$\inf\limits_{\{x_n\} \in c_0} \sup \{a_n + x_n\} = \limsup\limits_{n\to\infty} a_n.$$

Since $$x_n$$ converges to $$0$$, I know that $$\limsup\limits_{n\to\infty} a_n = \limsup\limits_{n\to\infty}(a_n + x_n)$$. Also, I thought I had one inequality $$\inf\limits_{\{x_n\} \in c_0} \sup \{a_n + x_n\} \ge \limsup\limits_{n\to\infty} a_n$$ since it first holds for any arbitrary $$x_n \in c_0$$, but now I'm doubting it because it's the $$\inf$$. Any help with the whole shebang would be great.

Take $$x_i=-a_i$$ for $$1 \leq i \leq N$$ and $$x_i=0$$ for $$i >N$$ to see that the left side does not exceed $$\sup_{n>N} a_n$$. Since $$N$$ is arbitrary this gives LHS $$\leq$$RHS. Now let $$x_n \to 0$$. Choose $$N$$ such that $$|x_i|<\epsilon$$ for $$i >N$$. Then $$sup (x_n+a_n) \geq \sup_{n>N} (a_n-\epsilon) =\sup_{n>N} a_n-\epsilon$$. Can you take it from here?
• I do not understand how LHS $\le$ RHS since you picked a specific $x_n$. – user439126 Oct 14 '19 at 3:27
• @user439126 Infimum of any set is less than or equal to any element of the set. Since there is an infimum on LHS we can pick any $\{x_n\}$ with limit $0$ and conclude that LHS $\leq \sup \{a_n+x_n\}$. – Kavi Rama Murthy Oct 14 '19 at 4:46
• Great, that helps a ton. For the last part, are you suggesting I show that $\sup\{a_n+x_n\}$ is the greatest lowerbound of $\sup\limits_{n>N}a_n$? – user439126 Oct 14 '19 at 18:48