As before, let \(E\) be the set of all subsequential limits of \(\{p_n\}\) and let \(q\) be a limit point of \(E\). We have to show that \(q \in E\). To show this we will construct a subsequence of \(\{p_n\}\) which converges to \(q.\)

Choose \(n_1\) so that \(p_{n_1} \neq q\). If no such \(n_1\) exists, then \(E\) has only one element, namely \(q = p_1 = p_2 = \cdots\), and there is nothing to prove. Define \(\delta = d(q, p_{n_1})\). Suppose \(n_1, \ldots, n_{i-1}\) for some fixed \(i \ge 2\) are chosen. Since \(q\) is a limit point of \(E\), there is a point \(x \in E\) with \(d(q, x) < \delta /2^i\). Since \(x \in E\), there is an \(n_i > n_{i-1}\) such that \(d(x, p_{n_i}) < \delta/2^i\). Thus

This implies \(p_{n_k} \to q\), showing \(q \in E\).

Assume for the sake of the contradiction that \(\alpha \notin F\). Then since \(F^\mathsf{c}\) is an open set (because \(F\) is closed) there exists an \(\varepsilon > 0\) such that \((\alpha - \varepsilon, \alpha + \varepsilon) \subset F^\mathsf{c}\). But this implies \(\alpha - \frac{\varepsilon}{2}\) is an upper bound for \(F\) which is smaller than \(\alpha\). This gives us our required contradiction.

We start by showing the two properties.

1. We divide it into three cases depending on what value \(s^*\) takes:

   1. If \(s^* = +\infty\), then \(E\) is not bounded above, hence \(\{s_n\}\) is not bounded above, and thus there is a subsequence \(\{s_{n_k}\}\) such that \(s_{n_k} \to +\infty.\) Therefore, \(+\infty \in E\), or in other words \(s^* \in E\).

   2. If \(s^*\) is real, then \(E\) is bounded above, and at least one subsequential limit exists by the definition of \(\sup\). **Therefore, \(s^* \in E\) follows from the Lemmas 1 and 2, and the fact that \(s^* = \sup E.\)

   3. Lastly, if \(s^* = -\infty\), then \(E\) contains only one element, namely \(-\infty\), and there is no subsequential limit. Thus, \(s^* \in E\).

2. Suppose for the sake of contradiction that there is a number \(x > s^*\) such that \(s_n \geq x\) for infinitely many values of \(n\). Let's denote this set of \(n\)'s with \(\mathcal{K}\) and let \(\{s_k\}_{k \in \mathcal{K}}\) be this subsequence. If \(\{s_k\}_{k \in \mathcal{K}}\) is unbounded then \(s^* = +\infty\) contradicting the fact that there exists an \(x > s^*.\) And if \(\{s_k\}_{k \in \mathcal{K}}\) is bounded that it contains a convergent subsequence (by the Bolzano–Weierstrass theorem). Suppose this convergent subsequence converges to \(y\). Then \(y \geq x > s^*\). This contradicts the definition of \(s^*.\)

To show the uniqueness, suppose there are two distinct numbers, \(p\) and \(q\), which satisfy the two properties, and suppose \(p < q\). Choose \(x\) such that \(p < x < q\). Since \(p\) satisfies the second property, there exists an integer \(N\) such that \(s_n < x\) for \(n \geq N\). But then \(q\) cannot satisfy the first property.

Let \(s^* = \limsup_{n \to \infty} s_n\) and \(t^* = \limsup_{n \to \infty} t_n\), as before. Suppose for the sake of contradiction \(t^* < s^*\). Choose \(x\) such that \(t^* < x < s^*\). Then by the second property of Theorem 1 there is an integer \(N_1\) such that \(n \geq N_1\) implies \(t_n < x\). Also by the first property of Theorem 1 there exists a subsequence \(\{s_{n_k}\}\) such that \(s_{n_k} \to s^*\). This implies that there exists an integer \(N_2\) such that \(n \geq N_2\) implies \(x < s_n\). But then for \(n \geq \max\{N_1, N_2\}\) we have \(t_n < x < s_n\). This gives us our required contradiction.

A similar argument can be made for the \(\liminf\) case.

We divide the analysis into three cases.

1. First, let \(s \in \mathbb R\). Then if \(s^* = s_* = s\), Theorem 1 implies that for any \(\varepsilon > 0\) we have \(s_n \in (s-\varepsilon, s+\varepsilon)\) for all but finitely many \(n\), which means \(s_n \to s.\) On the other hand if \(s_n \to s\) then every subsequence \(\{s_{n_k}\}\) must converge to \(s\) and hence \(s^* = s_* = s\).

2. Now let \(s = +\infty\). Then \(s_n \to s\) means that for every \(M \in \mathbb R\) there is an integer \(N\) such that \(n \geq N\) implies \(s_n \geq M.\) This is same as saying \(s_* = +\infty\), which gives \(s^* = +\infty\) curtsy of Theorem 2.

3. Lastly, let \(s = -\infty\). Then \(s_n \to s\) means that for every \(M \in \mathbb R\) there is an integer \(N\) such that \(n \geq N\) implies \(s_n \leq M.\) This is same as saying \(s^* = -\infty,\) which gives \(s_* = -\infty\) curtsy of Theorem 2.

We will prove the first equation. The proof for the second is similar. We prove the equation in two steps.

Let \(S = \{ s_n : n \in \mathbb N \}\). Let's show that \(\sup S = +\infty\) if and only if \(s^* = + \infty.\) Suppose first that \(\sup S = + \infty.\) Then we construct a subsequence \(\{s_{n_k}\}\) as follows. We let \(n_1 = 1.\) Suppose \(n_1, \ldots, n_{k}\) for some fixed \(k \ge 1\) have been chosen and let

Notice that \(S_k\) is infinite as otherwise we can find an \(M \in \mathbb R\) such that \(s_n \leq M\) for all \(n \geq 1\), contradicting the fact that \(\sup S = + \infty\). We pick \(n_{k+1}\) to be the smallest element of \(S_k\) which is bigger than \(n_k.\) The resulting subsequence satisfies the condition that \(s_{n_k} \geq k\) for \(k \geq 2\) and thus we conclude that \(s_{n_k} \to +\infty\) which gives \(s^* = +\infty.\) Now suppose that \(s^* = +\infty.\) From Theorem 1 we can conclude that there exists a subsequence \(\{s_{n_k}\}\) such that \(s_{n_k} \to +\infty.\) This immediately implies \(\sup S = +\infty.\)

For the second step, suppose \(\sup S < + \infty.\) Denote

Notice that \(a_1 < +\infty\) and \(\{a_n\}\) is a monotonically decreasing sequence. Therefore, we have that either \(\{a_n\}\) is lower bounded, in which case it converges to, say, \(a,\) or it is not, in which case \(a_n \to -\infty.\) Since \(s_n \leq a_n,\) by Theorem 2 we can conclude where the last equality follows from Theorem 3. We will now show that which will give us our required equality. If \(\lim_{n \to \infty} a_n = -\infty\), there is nothing to prove and so we assume that \(a > -\infty.\) Let \(\varepsilon > 0\) be given and define We claim that \(B\) is infinite. Indeed, if \(B\) were finite we could find \(N \in \mathbb N\) so that \(N \geq \max B.\) This will imply that \(s_n \le a - \varepsilon\) for all \(n \geq N\) and so \(a_n \leq a - \varepsilon\) for \(n \geq N.\) But then by Theorem 2 we would conclude \(a = \lim a_n \leq a-\varepsilon.\) Thus \(B\) is infinite and we let \(\{s_{n_k}\}\) be a subsequence of \(\{s_n\}\) with \(n_k \in B.\) Notice that \(\limsup_{k \to \infty} s_{n_k} \leq s^*,\) since any subsequential limit of \(\{s_{n_k}\}\) is also a subsequential limit of \(\{s_n\}.\) This along with Theorem 2 give Since \(\varepsilon\) was arbitrary we conclude that \(s^* \ge a\), which is what we wanted.

If for a set \(S\) of real numbers, we denote by \(-S\) the set \(\{-s : s \in S\},\) then the theorem follows immediately from the simple observation that and Theorem 4.

If \(s^* = +\infty\) or \(t^* = +\infty\) or \(s^* = -\infty\) or \(t^* = -\infty\), it is easily checked that the inequality is true, and so we may assume \(s^*, t^* \in \mathbb R.\)

For two sets \(S\) and \(T\) of real numbers satisfying \(S \subseteq T\), we know that \(\sup S \le \sup T\). This fact along with the observation

for any \(n \in \mathbb N\), imply Taking limit as \(n \to \infty\) now gives us our desired inequality.

Similar to the proof of Theorem 6.