Here's Theorem 3.17 in the book Principles of Mathematical Analysis by Walter Rudin, third edition.
Let $\{s_n\}$ be a sequence of real numbers. Let $E$ denote the set of all the subsequential limits of $\{s_n \}$ (in the extended real number system). Let $s^* = \sup E$. Then $s^*$ has the following two properties.
(a) $s^* \in E$.
(b) If $x > s^*$, there is an integer $N$ such that $n \geq N$ implies $s_n < x$.
Moreover, $s^*$ is the only number with the properties (a) and (b).
I've given a my reading of Rudin's proof. Here's the link. Theorem 3.17 in Baby Rudin: Infinite Limits and Upper and Lower Limits of Real Sequences
Now I would like to state and prove the analogous result which Rudin refers to after the statement.
So here's the statement.
Let $\{s_n \}$ be a sequence of real numbers. Let $E$ denote the set of all the subsequential limits of $\{s_n \}$ (in the extended real number system), and let $s_* = \inf E$. Then
(a) $s_* \in E$.
(b) If $x< s_*$, there is an integer $N$ such that $n \geq N$ implies $s_n > x$.
Moreover, $s_8$ is the only number with the properties (a) and (b).
And, here's my proof.
If $s_* = - \infty$, then, for every real number $M$, there is an element $x_M \in E$ such that $x_M < M$, and since this element $x_M$ is a subsequential limit of $\{s_n\}$, there is a subsequence of $\{s_n \}$ all but finitely many of whose terms are less than $M$. Thus, there is a natural natural number $n_1$ such that $s_{n_1} < -1$. And, for each $k \in \mathbb{N}$ such that $k > 1$, we can find a natural number $n_k$ such that that $n_k > n_{k-1}$ and $s_{n_k} < -k$. And, the subsequence $\{s_{n_k}\}$ diverges to $- \infty$. So $-\infty \in E$.
If $s_*$ is finite (i.e. real), then the set $E$ is bounded below in $\mathbb{R}$, and for every $\delta > 0$, we can find an element $x_\delta \in E$ such that $s_* \leq x_\delta < s_* + \delta$. Now as $x_\delta$ is a subsequential limit of $\{s_n \}$, there are infinitely many terms of $\{s_n\}$ in the half open interval $[x_\delta, s_* + \delta ) \subset [s_* , s_* + \delta)$. So for $\delta = 1$, there is some natural number $n_1$ such that $s_* \leq s_{n_1} < s_* + 1$, and, for any $k \in \mathbb{N}$ such that $k > 1$, we can find a natural number $n_k$ such that $n_k > n_{k-1}$ and $s_* \leq s_{n_k} < s_* + \frac 1 k$. Then the subsequence $\{s_{n_k}\}$ converges to $s_*$, showing that $s_* \in E$.
If $s_* = +\infty$, then no real number or $- \infty$ can be a seubseuqntial limit of $\{s_n \}$. So, for every real number $M$, there is a positive real number $\delta_M$ such that the segment $(M-\delta_M, M+\delta_M)$ contains only finitely many terms of the sequence $\{s_n\}$. And there is a real number $t$ such that there are only finitely many terms of $\{s_n\}$ in the infinite open interval $(-\infty, t)$. Moreover, if there were some segment (i.e. finite open interval) $(a, b)$ containing infinitely many terms of $\{s_n\}$, then some subsequence of $\{s_n\}$ would be bounded and thus would have a subsequence converging to a real number in the interval $[a,b]$, and that real number would thus be a subsequential limit of $\{s_n \}$, contradicting the definition of $s_*$. So, for every real number $M$, there would be at most finitely many terms in $(-\infty, M + 1 ]$. Thus there would be a natural number $n_1$ such that $s_{n_1} > 1$, and for each $k \in \mathbb{N}$ such that $k > 1$ there would be a natural number $n_k$ such that $n_k > n_{k-1}$ and $s_{n_k} > k$. Therefore the subsequence $\{s_{n_k}\}$ diverges to $+ \infty$ and hence $+\infty \in E$.
This establishes (a) in all cases.
If $x < s_*$ and if there are infinitely many $s_n \leq x$, then the range of some subsequence of $\{s_n\}$ would be contained in $(-\infty, x]$, and this subsequence (depending on whether it is unbounded below or bounded) would in turn have a subsequence which either diverges to $-\infty$ or converges to a real number $\leq x$; in either case we would have an element $y \in E$ such that $y \leq x < s_*$, a contradiction.
Thus $s_*$ satisfies (a) and (b).
To show the uniqueness, suppose there are two numbers $p$ and $q$ which satisfy (a) and (b) and suppose that $p < q$. Choose a real number $x$ such that $p < x < q$. As $q$ satisfies (b), there is an integer $N$ such that $n \ge N$ implies that $s_n > x$. Thus, for $\delta \in (0, x-p)$, the neighborhood $(p-\delta, p+\delta)$ of $p$ contains only finitely many terms of $\{s_n\}$, which implies that $p$ cannot be a subsequential limit (in the extended real number system) of $\{s_n\}$, which contradicts (a).
Is this statement correct? And if so, is the proof correct? Or, are there any issues? How about my presentation?