Trouble on proof step of: $\lim_{n\rightarrow+\infty}a_n=+\infty \Longrightarrow \liminf_{n\rightarrow+\infty}a_n=+\infty$ I'm proving that:
$$
    \lim_{n\rightarrow+\infty}a_n=+\infty \Longrightarrow \liminf_{n\rightarrow+\infty}a_n=+\infty 
$$
By definition:
$$
\lim_{n\rightarrow+\infty}a_n=+\infty \iff \forall M\in\mathbb{R} \quad \exists \nu\in\mathbb{N}:\forall 
n\in\mathbb{N}, n \ge \nu \Rightarrow a_n > M.
$$
In particular we have that:
$$
c_{\nu} := \inf_{h\ge \nu} a_h \ge M   \tag{1}
$$
Now I know that $(c_{\nu})$ is  non-decreasing. From other post on this site I've read that the next step is that we have:
$$
\lim_{\nu\rightarrow+\infty} c_{\nu} \geq M \tag{2}
$$
Since $M$ was arbitrary, it follows that:
$$\liminf_{n\rightarrow+\infty}a_n=+\infty$$
My problem:
I'm stuck thinking based on what is it possible to go from $(1)$ to $(2)$.
In particular:
There is a theorem according to which, given two convergent sequences $(a_n), (b_n)$ with $a_n \rightarrow l$, $b_n \rightarrow m$ and $l,m\in\mathbb{R}$.
If $a_n\le b_n \forall n\in \mathbb{N}$ then $l\le m$.
In my case if $\lim_{\nu}c_{\nu}$ is finite then it's possible to pass from $(1)$ to $(2)$ thanks to the previous theorem.
However since $(c_{\nu})$ is a monotonically non-decreasing sequence its limit could also be $+\infty$:
what would be the line of reasoning in this case?
 A: Explaining the logic from (1) to (2): Because $a_n\to \infty$, for every $M\in \mathbb{R}$ there exists an $N$ such that for all $n>N$ we have $a_n>M$. For $v>N$, then, $c_v =\inf_{n\geq v}a_n$ is greater than or equal to $M$ because for all $a_n\in \{a_n:n>N\}$ we have $a_n>M$.
This actually explains why the inequality still holds when taking the limit as $v\to \infty$. Suppose for contradiction that $c_v\geq M$ but $\lim_{v\to\infty}c_v\not\geq M$. Then there exists a $v'>N$ such that $c_{v'}<M$. If $c_{v'}<M$ then there is an $n\geq v'>N$ such that $a_n<M$. But this is a contradiction of our previous discussion.
In the question it was stated "why, if we take the limit of both members in (1), the inequality $\geq$ still holds?" I'm not sure what is meant by "both members". If this means the limit of the left and right side as $n\to \infty$, the limit of the constant $M$ as $n\to \infty$ is $M$, because it's a constant. At this point in the argument we're not letting $M$ go to infinity, we just want to prove that the limit $\lim_{v\to\infty}c_v\geq M$. But our $M$ was arbitrary, and this means $\lim_{v\to\infty}c_v\geq M$ for all $M\in\mathbb{R}$.
A: I think I've overcome my doubt.
By taking the supremum of $c_{\nu}$, follows:
$$
c_{\nu} := \inf_{h\ge \nu} a_h \ge M \Rightarrow \sup_{\nu\in \mathbb{N}}c_{\nu}  \ge M \iff \lim_{\nu \rightarrow +\infty}c_{\nu} \ge M \Rightarrow \liminf_{n\rightarrow+\infty}a_n=+\infty
$$
since $c_{\nu}$ is non-decreasing then $\lim_{\nu \rightarrow +\infty}c_{\nu}= \sup_{\nu}c_{\nu}$
