Limsup of a sequence is greater then the limsup of a subsequence?

Question

Consider a sequence of real numbers $\{a_n\}_n$. Let $\{a_{n_j}\}_j$ be a subsequence of $\{a_n\}_n$. Suppose I have shown that $\limsup_{j \rightarrow \infty} a_{n_j}=L$ with $|L|<\infty$. Is it true that $\limsup_{n \rightarrow \infty} a_{n}\geq L$?

Answer 1

By definition $\limsup_{n\to\infty}\, a_n$ is the greatest number that you can achieve with a convergent subsequence of $\{a_n\}$. Since $\{a_{n_j}\}$ is a subsequence of $\{a_n\}$ we have that every subsequence of $\{a_{n_j}\}$ is a subsequence of $\{a_n\}$. Therefore $$\limsup_{n\to\infty}\, a_n\geq \limsup_{n\to\infty}\, a_{n_j}$$ because of the $\sup$'s properties.