Long sequences in singular cardinals can't be cofinal

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My attempts to understand a solution to the Kunen exercise (two well orderings on a cardinal $\kappa$ necessarily agree on a set of size $\kappa$) have foundered on the following claim. If someone would be kind enough to confirm it's correct and provide a proof, I'd be grateful.

Lemma: Assume $\operatorname{cf} \kappa \lt \lambda \lt \kappa$, where $\kappa$ and $\lambda$ are cardinals with $\lambda$ regular. Let $S=\{\alpha_i~\vert ~ i \lt \lambda \}$ be an increasing sequence of ordinals in $\kappa$. Then $S$ is not cofinal in $\kappa$.

This makes sense to me. I certainly can't envision a strictly increasing cofinal map $f: \aleph_1 \to \aleph_\omega$. But I haven't been able to prove no such map can exist.

Thanks for any help. I've been bogged down on this exercise for literally months. (I'm a lawyer who's been out of the field for nearly $30$ years. I'm engaging in self-study for my own entertainment.) I'd really like to move on to chapter $2$ of the book.

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Fix a cofinal increasing sequence $(\beta_j)_{j<\operatorname{cf}\kappa}$ below $\kappa$. Define $f:\operatorname{cf}\kappa\to\lambda$ by $f(j)=i$ for the least $i$ such that $\alpha_i\geq \beta_j$ (such an $i$ always exists if $S$ is cofinal in $\kappa$). Then $f$ is cofinal, since the $\beta_j$ are eventually above any fixed $\alpha_i$. But this is a contradiction since $\lambda$ is regular and $\operatorname{cf}\kappa<\lambda$.