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A Sequence Converges on the Extended Reals iff the number of Complete Upcrossings is Finite

Last updated Nov 1, 2022

# Statement

Let $(x_{n}){n=0}^{\infty} \subset \mathbb{R}$. Then $(x{n})$ converges in $\bar{\mathbb{R}}$ If and Only If $U_{\infty}^{*}[a, b] < \infty$ $\forall a, b \in \mathbb{Q}$.

# Proof

$(\Leftarrow)$ Suppose $(x_{n})$ does not converge in $\bar{\mathbb{R}}$. Recall that The Extended Reals is Sequentially Compact, so there exists $(x_{n_{k}}) \subset (x_{n})$ that converges in $\bar{\mathbb{R}}$ to limit $L$. Because $(x_{n})$ does not converge, there must be a $(x_{n_{l}}) \subset (x_{n}) \setminus (x_{n_{k}})$ so that $x_{n_{l}}$ converges to $R \neq L$. Without Loss of Generality suppose $R > L$. Then, because Rationals are Dense in the Reals, there exists $a,b \in \mathbb{Q}$ so that $R > b > a > L$ TODO

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