Define $\{s_n\}$ to be a sequence of real numbers. Let $E$ be the set of all numbers $x$ (in the extended real number system such that $s_{n_k} \rightarrow x$ for some subsequence $\{s_{n_k}\}$.

Walter Rudin defines $s^*=\sup E$ and $s_*=\inf E$. Now, a following theorem states that if $y>s^*$, then there is an $N \in \mathbb{N}$ such that $\forall n \ge N \implies s_n<y$

There is a part on the proof of this statement that I do not understand. (Bold and italicized)

The proof goes as follows:

Suppose there is a number $y>s^*$ such that $s_n \ge y$ for infinitely many values of $n$. In that case, there is a number $z \in E$ such that $z\ge y>s^*$, contradicting the definition of $s^*$.

Can anyone tell me why this statement is true?


Let $\;\{s_{n_k}\}\;$ be the infinite number of elements of $\;\{s_n\}\;$ s.t. $\;s_{n_k}\ge y\;$ . Then either this subsequence is bounded and thus by Bolzano-Weierstrass theorem it has a partial (finite) limit, or it is unbounded and then it has a subsequence that covnerges to $\;\infty\;$ . Either way, there's an element $\;z\in E\;$ which is a partial limit of the original limit, an by cosntruction $\;z>s^*\;$ , contradicting the definition of $\;s^*\;$


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