Given a sequence (b_n)(bn), we say that bb is a cluster point of (b_n)(bn) if every neighborhood of bb contains infinitely many elements of (b_n)(bn).
If we translate this into the real numbers, that means that if L="lim inf"(a_n)L=lim inf(an), then we should be able to create an arbitrarily small interval (L-epsilon, L+epsilon)(L−ε,L+ε) around LL and still have infinitely many a_n in (L-epsilon,L+epsilon)an∈(L−ε,L+ε).
To show this, we need to understand what the limit inferior of a sequence is. In the case of a sequence of real numbers, the limit inferior is an extension of the idea of a greatest lower bound. It is the greatest value such that all numbers in the sequence are eventually greater than that value.
"lim inf"(a_n) =lim inf(an)=
max{L in RR | EEN in NN" s.t. " n>N=>a_n>=L}
It's a lower bound which allows us to discard any number of finite elements of the sequence lower than it.
Now, how can we use this? Well, first of all, notice that (a_n) is bounded. This is important, as it guarantees that we actually have a limit inferior (imagine the sequence (-1)^n n^2, no eventual bounds there!). So, as we are guaranteed of its existence, let's let L be the limit inferior of (a_n).
Now, let's show that L is a cluster point of (a_n) using an argument by contradiction.
Suppose that L is not a cluster point of (a_n). Then there exists some epsilon > 0 such that (L-epsilon, L+epsilon) contains only finitely many elements of (a_n). Let a_(n_1), a_(n_2), ..., a_(n_k) be the ordered list of elements of a_n in (L-epsilon, L+epsilon). Then, for all n > n_k, we have a_n >= L+epsilon. But L+epsilon > L and L is, by definition, the greatest number with that property. Thus we have reached a contradiction, meaning L must be a cluster point of (a_n). ∎
