I am just reading about limit supremum concept in Rudin.
Rudin defines the limit supremum as follows:
Let $\{a_n\}$ be a sequence in $[\infty,\infty]$.
$b_k = \sup \{a_k, a_{k + 1}, \ldots, \}, \ldots (k = 1,2,3, \ldots)$
$\beta = \lim \sup = \inf \{b_1, b_2, \ldots, \}$
There is something that I am confused about. It states that limit supremum is the largest number with the property that there is a subsequence
$\{a_{n_i} \}$ of $\{a_n\}$ such that $a_{n_i}$ such that $a_{n_i} \rightarrow \beta$ as $i \rightarrow \infty$, and $\beta$ is the largest number with this property. I don’t understand how to prove that this property.
1) Extract a subsequence $b_k$ from the definition of $\beta$ that converges to the infimum $\beta$.
2) Extract subsequences $a_{k_n}$ from $a_n$ which converge to $b_k$ for each $k$.
3) Pick any positive sequence $\epsilon_k\rightarrow 0$, and for each $k$, find an $i$ such that $|a_{k_i}-b_k|<\epsilon_k$.
4) With the above subsequence of $a_n$, conclude the result by invoking the triangle inequality on the appropriate terms.
To show $\beta$ is the largest such number you can assume by contradiction that there's a $\gamma>\beta$ with the same property. This would give you a subsequence $a_{m_k}$ that converges to $\gamma$. This means that at some point $a_{m_k}>\beta+\epsilon$ for all $k\geq N$ (where say, $\epsilon=(\gamma-\beta)/2$). This implies that $b_{m_k}>\beta+\epsilon$ for all $k>N$, and hence the infimum over all such $\beta_k$ would be strictly greater than $\beta$, a contradiction.