Quick question about inaccessible cardinals

I am trying to find a source that states, it is consistent with ZFC that weakly inaccessible cardinal does not exist. Can I please get some sources? It was quite hard for me to find such.

Indeed, I could find in Jech's textbook that "one cannot prove the existence of weakly inaccessible cardinals within ZFC." Does this imply that ZFC+"there does not exist weakly inaccessible cardinals" are consistent? If so, why?

Solution 1:

If a first-order theory (e.g. ZFC) cannot prove statement $\lnot S$ from axioms $\mathcal A$ (which itself assumes $\mathcal A$ is consistent, otherwise it could prove any statement!), then $\mathcal A\cup\{S\}$ is consistent. This should be common knowledge.

Formal proof: as per Deduction theorem, if $\mathcal A\cup\{S\}$ is inconsistent (proves $\bot$), then $\mathcal A$ proves $S\to\bot$, which is the same as proving $\lnot S$.

Solution 2:

Adding to the other answers, one can even find a specific model of $\mbox{ZFC}$ in which there is no inaccessible cardinal:

Let $\kappa$ be the least inaccessible cardinal, and look at $M=V_\kappa$.

Assume that $M$ think there is an inaccessible cardinal. For instance, $M\models \lambda \text{ is inaccessible}$ for some $\lambda<\kappa$.

Since $\kappa$ is inaccessible, $V_\kappa$ computes correctly the powerset of every set $X\in V_\kappa$, and also the regularity of any ordinal $\alpha<\kappa$. It follows that if $M$ thinks that $\lambda$ is inaccessible, then it is really inaccessible in $V$, which isn't possible, since $\kappa$ was the least inaccessible.

Remark:

Taking the least inaccessible assumes that there is one to begin with. If there are'nt any, then we are done: $V$ itself is a model of $\mbox{ZFC}$ with no inaccessible cardinals.