Let
. You should note first that the sequence
is decreasing hence
.
Since
, you must prove:
a) that
for any convergent subsequence
b) that, for any
, there is a convergent subsequence
such that
.
I let you try a) (compare
with terms from the sequence
)
As for b), you have to "construct" a convergent subsequence. In fact, instead of using
like above, you can directly construct a subsequence that converges toward
(which means that in fact
). To do so, intuitively, choose
close to
(it is a sup, hence it is a limit of values
in
), then
close(r) to
, 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
is finite or not.
--
About brackets in LaTeX: they're obtained through the commands \{ and \}. There's the same pattern for other restricted characters, like \# for #.