Does $\kappa \lt \lambda$ implie $cf\kappa \lt cf\lambda$ for singular cardinals?

Question

In T. Jech's set theory, the part which he tries to prove a part of silver's theorem on the singular cardinal hypothesis, namely,
"If the singular cardinal hypothesis holds for all singular cardinals of cofinality $\omega$, then it holds for all singular cardinals."
Is confusing me a little bit. He assumes that $\kappa$ is a singular cardinal of uncountable cofinality and he uses induction on the cofinality of $\kappa$ to prove the statement. The part which i am struggling to understand is where he says that for each $\lambda \lt \kappa$ we have $\lambda^{cf\kappa} \lt \kappa$. I know that assuming the singular cardinal hypothesis $\lambda^{cf\kappa}$ is either $2^{cf\kappa}$ or $\lambda$ or $\lambda^{+}$ based on the continuum and cofinality functions. So if induction on the cofinality of $\kappa$ is the same as induction on singular cardinals then i can assume that the singular cardinal hypothesis holds bellow $\kappa$ and i'm fine. So the question is:
Does $\kappa \lt \lambda$ implie $cf\kappa \lt cf\lambda$ for singular cardinals?

Answer 1

No. Recall that if $\delta$ is a limit ordinal, then $\operatorname{cf}(\aleph_\delta)=\operatorname{cf}(\delta)$.

Now take $\kappa=\aleph_{\omega_1}$ and $\lambda=\aleph_{\omega_1+\omega}$.

---

As for the proof of Silver's theorem, note that if $\lambda<\kappa$, by induction on $\lambda$, and using both your assumption and Hausdorff's formula to obtain that $\lambda^{\operatorname{cf}(\kappa)}<\kappa$.