Sequence of events $\{C_n\}_n\in\mathbb{N}$: why $\mathbb{P}\left(\bigcup_{m\ge n}C_m\right)\ge\sup_{m\ge n} \mathbb{P}\left(C_m\right)$?

Question

Let us start from probability space $(\Omega,\mathcal{C},\mathbb{P})$ and a sequence of events $\{C_n\}$. I know that: $$\mathbb{P}\left(\bigcup_{m\ge n}C_m\right)\ge\mathbb{P}\left(C_m\right),\text{ for each $m\ge n$}\tag{1}$$

At this point, why could I state that: $$\mathbb{P}\left(\bigcup_{m\ge n}C_m\right)\ge\sup_{m\ge n} \mathbb{P}\left(C_m\right)\tag{2}$$ ?

---

Maybe it is extremely more trivial than I am thinking now. My doubt arises from the fact that, by def., supremum of a sequence does not necessarily belong to the sequence itself.

Answer 1

In general, if $a_{n}\leq M$, it means $M$ is an upper bound of $a_{n}$.
By definition, $\sup(a_{n})$ is the smallest upper bound, so $M\geq \sup(a_{n})$. Same applies here.

Answer 2

$$A\subseteq B\implies\mathbb{P}(A)\le \mathbb{P}(B)$$ Now $C_m\subseteq \bigcup_{k\ge n}C_k$ for all $m\ge n$, hence $\mathbb{P}(C_m)\le\mathbb{P}(\bigcup_{k\ge n}C_k)$ for all $m\ge n$. $\sup_{m\ge n}\mathbb{P}(C_m)$ is the smallest number that is $\ge\mathbb{P}(C_m)$ for all $m\ge n$. Any number that is $\ge\mathbb{P}(C_m)$ for all $m\ge n$, has to be greater than $\sup_{m\ge n}\mathbb{P}(C_m)$.