I'm reading about inacessible cardinals. I don't know anything about them, I just know basic set theory, cardinal operations etc., but I am too curious to leave this question that now I'm (probably) not able to totally understand...
So, a (strongly) inacessible cardinar is a regular uncountable cardinal $\lambda$ such that $\kappa < \lambda \implies 2^{\kappa}<\lambda$. I know that if the hypothesis of being uncountable is dropped then $\aleph_0$ satisfy the other 2 properties (regular and $\kappa < \aleph_0\implies 2^{\kappa}<\aleph_0)$. But I can't think what happens if we drop the hypothesis of $\lambda$ being regular. Does exist a uncountable singular cardinal $\lambda$ such that $\kappa < \lambda \implies 2^{\kappa}<\lambda$? Why?
Are you familiar with $\beth$ (beth) notation? It is similar to $\aleph$ (aleph) notation, except that the exponential function $\kappa\mapsto2^\kappa$ replaces the successor function $\kappa\mapsto\kappa^+$. That is:
$$\beth_0=\aleph_0;$$ $$\beth_\lambda=\sup\{\beth_\alpha:\alpha\lt\lambda\}\text{ if }\lambda\text{ is a limit ordinal};$$ $$\beth_{\alpha+1}=2^{\beth_\alpha}.$$ Now the cardinal $$\beth_\omega=\aleph_0+2^{\aleph_0}+2^{2^{\aleph_0}}+2^{2^{2^{\aleph_0}}}+\cdots$$ is a singular cardinal of cofinality $\omega$, and it is a strong limit cardinal, that is, if $\kappa\lt\beth_\omega$, then $\kappa\lt\beth_n$ for some $n\lt\omega$, and so $$2^\kappa\le2^{\beth_n}=\beth_{n+1}\lt\beth_\omega.$$