Problem. Prove that a bounded sequence is converging if and only if it has exactly one partial limit.
Proof. We prove the two implications separately.
First direction:
Let \(\left(a_{n}\right)_{n=1}^{\infty}\) be a bounded sequence and let us assume that it’s converging to \(L\in\mathbb{R}\).
Existence: by the Bolzano-Weierstrass theorem, \(\left(a_{n}\right)\) has a converging sub-sequence.
Uniqueness: Let \(\left(a_{n_{k}}\right)_{k=1}^{\infty}\) be a converging sub-sequence of \(\left(a_{n}\right).\)
Then from the inheritance theorem (every sub-sequence of a convergent sequence converges to the same limit), it also converges to \(L\), as needed.
Second direction:
Let \(\left(a_{n}\right)_{n=1}^{\infty}\) be a bounded sequence such that it has exactly one partial limit, \(L\in\mathbb{R}\).
Let us assume for the sake of contradiction that \(\left(a_{n}\right)_{n=1}^{\infty}\) doesn’t converge.
In particular, it doesn’t converge to \(L\), so there exists \(\varepsilon_{1}>0\) such that \(|a_{n}-L|>\varepsilon_{1}\) is a property that holds for infinitely many \(n\) (otherwise it would converge to \(L\)).
Since it’s a frequent property of \(\left(a_{n}\right)\), there exists a sub-sequence \(\left(a_{n_{j}}\right)_{j=1}^{\infty}\) of which all elements fulfill this property.
In particular, \(\left(a_{n_{j}}\right)_{j=1}^{\infty}\) inherits the boundedness, therefore by the Bolzano-Weierstrass theorem it has a converging sub-sequence \(\left(a_{n_{j_{w}}}\right)_{w=1}^{\infty}\) , i.e. there exists \(L'\in\mathbb{R}\) such that \(\underset{w\rightarrow\infty}{\lim}a_{n_{j_{w}}}=L'\).
However, every element of this sub-sequence fulfills the property of being far from \(L\) by at least \(\varepsilon_{1}\), and since limits weakly preserve sequence inequalities, \(|L'-L|\geq\varepsilon_{1}>0\), therefore \(L'\neq L\).
\(L'\) is also a partial limit of \(\left(a_{n}\right)\), since the sub-sequence of a sub-sequence is also a sub-sequence, which contradicts our assumption. ∎