Integrating the two-views of lim sup and lim inf Preliminaries
Let $(x_n)$  be a bounded real number sequence and $ (x_n )_{n≥k}  $be a subsequence of $x_n$  which only takes the values of the sequence starting from the k−th term.
Let {$x_n $}  and {$x_n$ }$_{n≥k}$ denote a subset of R that contains all values of the corresponding sequences $(x_n )$  and $(x_n )_{n≥k}.$
Initial Considerations
Define two sequences ($\alpha_k$) and ($\beta_k$) out of the original ($x_n$) sequence :
($α_k$)= sup{$x_n$}$_{n≥k}$
($\beta_k$)= inf{$x_n$}$_{n≥k}$
Call a number $c \in R$ a subsequential limit of the real number sequence ($x_n$) iff there exists a subsequence of ($x_n$) that converges to c.
Define C $\in R$ as the set of all subsequential limits of ($x_n$).
Definition of Lim sup and Lim inf
1 - Define lim sup $x_n$ as the supremum of C and define lim inf $x_n$ as the infimum of C.
2 - Define lim sup $x_n$ as the infimum of {$\alpha_k$}  and lim inf $x_n$ as the supremum of {$\beta_k$}.
My doubt
Why the supremum of C is the same as the infimum of {$\alpha_k$}  and the infimum of C is the same as the supremum of {$\beta_k$} ?
In another words, what's special about the construction of the ($\alpha_k$) and ($\beta_k$) sequences that make their limits ( their infimums/supremums because they are bounded monotone decreasing/increasing, respectively ) to be largest/smallest possible subsequential limits of the original sequence sequence ($x_n$) ?
Can i understand the equivalence between these two definitions ( preferably both intuitively and rigorously ) without touching on anything related to topology ?
The reason is that my real analysis teacher ( who has not taught us anything about topology yet ) is teaching us sequences using the second definition of lim sup and lim inf and many of the theorems we are going through are better understood ( in my view ) if one understands both definitions of lim sup and lim inf ( and obviously, their equivalence ).
Thanks a lot in advance.
 A: Well, I find the relation (limsup by def. 1) $\le$ (limsup by def. 2) pretty intuitive - if there was a $c \in C$ such that inf {$\alpha_k$} $<$ c, that would mean that there is a k such that sup{$x_n$}$_{n \ge k} < c$, but since a subsequence of $x_n$ gets arbitrarily close to $c$, that must be a contradiction.
To see that (limsup by def. 2) = $\alpha$ $\in C$ you just have to consider that for any $k$ there are elements of {$x_n$}$_{n \ge k}$ abritrarily close to $\alpha$, and so you can "pick" a subsequence of $x_n$ that will tend to $\alpha$.
Of course this is all very hand-wavy and you might not find it intuitive or convincing at all. :) We did prove this in class rigorously without using topology, but the proof was rather long and technical (for my level of knowledge at least), so I'm hoping that someone else can provide a simpler proof.
A: I can show you a proof that if you define $\limsup x_n:=\lim_{n\to \infty} \sup\{x_k\}_{k\ge n}$, then $\limsup x_n=C$, where $C$ is the supremum of all subsequential limits. First, see this question for proofs that $\limsup x_n\ge C$.
To show that $\limsup x_n\le C$, it suffices to show there is a subsequence $(x_{n_m})$ that converges to $\limsup x_n$. This subsequence can be constructed inductively. Put $x_{n_1}=x_1$. Let $\alpha_k:=\sup \{x_k\}_{k\ge n}$. Now, suppose $x_{n_1},\ldots x_{n_{m-1}}$ have been defined. There must be some $k_0\ge n_{m-1}$ such that $\alpha_{n_{m-1}}-\frac1m\le x_{k_0}\le \alpha_{n_{m-1}}$. Put $n_m:=k_0$. This completes the construction of $(x_{n_m})$. Now, it follows from the construction that $|x_{n_m}-\alpha_{n_{m-1}}|<\frac1m$ for all $m\in \mathbb{N}$. Since $\lim_{m\to \infty}\alpha_{n_{m-1}}=\limsup x_n$, it follows that $\lim_{n\to \infty} x_{n_m}=\limsup x_n$.
