Define $\limsup_{k \rightarrow \infty}a_k$ and $\liminf_{k \rightarrow \infty}a_k$ by
$$\limsup_{k \rightarrow \infty}a_k = \lim_{j\rightarrow \infty}b_j = \inf_{j}\{\sup_{k\geq j}a_k \} $$
$$\liminf_{k \rightarrow \infty}a_k = \lim_{j\rightarrow \infty}c_j = \sup_{j}\{\inf_{k\geq j}a_k \}. $$
Prove that $L = \limsup_{k \rightarrow \infty}a_k$ if and only if $(i)$ there is a subsequence $\{a_{k_j} \}$ of $\{a_k\}$ which converges to $L$, and $(ii)$ if $L'>L$, there is an integer $K$ such that $a_k < L'$ for $k\geq K$.
My attempt: Let $b_j = \sup_{k\geq j}a_k$. $b_j$ is monotone decreasing and converges to his infimum $L$. Then $$\forall \varepsilon >0, \exists j_0: j\geq j_0 \Rightarrow |b_j - L| <\varepsilon/2. $$
Since $b_j=\sup_{k\geq j}a_k$, there exists $\{a_{k_j}\}$ such that $|a_{k_j} - b_j|<\varepsilon/2$ and the subsequence $\{a_{k_j}\}$ converges to $L$ by the triangle inequality.
Now let $L'>L$. Then, there exists $\varepsilon>0$ such that $L' = L+\varepsilon>b_{j_0} = \sup_{k \geq j_0} a_k$ for some $j_0$ since $L$ is the infimum of $b_j$. Then $L' > a_k$ takign $K=j_0$.
Is this correct? How can i finish the whole proof?
Answer
This is correct for one direction- i.e. you have shown that if $L=\lim \sup_{k\to\infty} a_k$ the (i) and (ii) hold. You now only need to show the opposite. It turns out that (i) implies $L\leq\lim \sup_{k\to\infty} a_k$ and (ii) implies $L\geq \lim \sup_{k\to\infty} a_k$, so you can do these separately. Aim to prove these statements indirectly e.g. the first inequality is the same as $\forall \epsilon, \lim\sup_{k\to\infty} a_k\geq L-\epsilon$.
Suppose first that (i) is true. This implies $\forall \epsilon, \exists j_0: j\geq j_0 \implies a_{k_j}\geq L-\epsilon$. Consider what this implies for the $b_{k_j}$. The other part is similar. Based on what you have done so far, you should be able to finish off from here.
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