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**Laurent** Let $\displaystyle L=\inf_n \sup S_n$. You should note first that the sequence $\displaystyle (\sup S_n)_n$ is decreasing hence $\displaystyle L=\lim_n \sup S_n$.

Since $\displaystyle \limsup_n a_n = \sup T$, you must prove:

a) that $\displaystyle L\geq\lim_n a_{\varphi(n)}$ for any convergent subsequence $\displaystyle (a_{\varphi(n)})_n$

b) that, for any $\displaystyle \varepsilon>0$, there is a convergent subsequence $\displaystyle (a_{\varphi(n)})_n$ such that $\displaystyle \lim_n a_{\varphi(n)}>L-\varepsilon$.

I let you try a) (compare $\displaystyle a_{\varphi(n)}$ with terms from the sequence $\displaystyle (\sup S_n)_n$)

As for b), you have to "construct" a convergent subsequence. In fact, instead of using $\displaystyle \varepsilon$ like above, you can directly construct a subsequence that converges toward $\displaystyle \sup T$ (which means that in fact $\displaystyle \sup T=\max T$). To do so, intuitively, choose $\displaystyle a_{\varphi(1)}$ close to $\displaystyle \sup S_1$ (it is a sup, hence it is a limit of values $\displaystyle a_k$ in $\displaystyle S_1$), then $\displaystyle a_{\varphi(2)}$ close(r) to $\displaystyle \sup S_{\varphi(n)+1}$, and so on: the subsequence is constituted by values that almost match the supremums. I let you formalize this idea. (If you don't succeed, ask for additional help)

NB: you'll probably have to distinguish cases depending whether $\displaystyle \sup T$ is finite or not.

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