# (rudin theorem 3.17) Struggling understanding a part of the proof

I need help understanding a part of the proof of Theorem 3.17 in Rudin Real Analysis

The full theorem can be found here: Is Rudin being redundant in this proof?

The problem is on this part

If $$s^* = -\infty$$, then $$E$$ contains only one element, namely $$-\infty$$, and there is no subsequential limit. Hence, for any real $$M$$, $$s_n > M$$ for at most a finite number of values of $$n$$, so that $$s_n \rightarrow -\infty$$.

The answer to the question in the link was basically this: " if E were empty, then you would still have that its supremum is −∞", which makes sense. The part I am struggling with is how this proves that $$-\infty$$ is in the set.

My understanding: If $$s^* = -\infty$$ then there are no subsequential limits(either the set $$E$$ is empty or that it has only the $$-\infty$$ element). Thus for any real $$M$$, $$s_n > M$$ for at most a finite number of values of $$n$$. This holds as if it were infinite number of values of $$n$$, then we could construct a subsequence that $$s_{n_k} \rightarrow \infty$$, which will be a contradiction to $$s^* = -\infty$$. now this allows us to have finite\infinite number of values of $$n$$ for all real numbers $$M$$ such that $$s_n \leqslant M$$ thus $$s_n = -\infty$$ ,but as we are dealing with sequences which contains infinite terms, thus $$s_n = -\infty$$ for infinitely many $$n \in \mathbb{N}$$ thus we can form a sequence of $$-\infty$$ terms thus $$s_n \rightarrow -\infty$$

Does my reasoning work? I would really appreciate any help you can offer.

$$s_n \leq M$$ for all but finitely many $$n$$ does not give you $$s_n =-\infty$$. $$(s_n)$$ is a sequence of real numbers so no $$s_n$$ can be $$\infty$$ or $$-\infty$$.
What you have to notice is $$s_n \leq M$$ holds for $$n \geq n_0$$ with $$n_0$$ depending on $$M$$. Though $$M$$ is arbitrary $$n_0$$ keeps varying with $$M$$ and you will never get $$s_n=-\infty$$.