Consistent and independence of ZFC

Question

My question is very simple but I want to make sure. when we say the statement is independent of ZFC means, we can not this statement true or false by using only the axioms of ZFC. Is that right?

The second thing: when we say that is consistent with ZFC follows from, e.g., CH statement true means there is a module ZFC+CH for which the statement is true. Is that right?

Basically, I am looking for the easiest way to understand these concepts without going to more complicated definitions.

Answer 1

Jech describes the relationship between these concepts thusly:

Let $T$ be a mathematical theory, and let $A$ be an additional axiom. We say that $T+A$ is consistent relative to $T$ (or that $A$ is consistent with $T$) if the following implication holds: if $T$ is consistent, then so is $T+A$. If both $A$ and $\neg A$ are both consistent with $T$, we say that $A$ is independent of $T$. The question whether $A$ is consistent with $T$ is equivalent to the question whether the negation of $A$ is provable in $T$ (provided $T$ is consistent); this is because $T+A$ is consistent if and only if $\neg A$ is not provable in $T$. The way to show that an axiom $A$ is consistent with $ZF$ ($ZFC$) is to use models. For assume that we have a model $\mathfrak{M}$ of $ZF$ such that $\mathfrak{M}\models A$. Then $A$ is consistent with $ZF$: If it were not, then $\neg A$ would be provable in $ZF$, and since $\mathfrak{M}$ is a model of $ZF$, $\mathfrak{M}$ would satisfy $\neg A$. (Here we tacitly use the fact that if $\mathfrak{M}$ is a model of a theory $T$ and $\sigma$ is a sentence provable in $T$, then $\mathfrak{M}$ models $\sigma$.) However, $\mathfrak{M}\models \neg A$ contradicts $\mathfrak{M} \models A$ since $\mathfrak{M} \models \neg A$ iff $\mathfrak{M} \not\models A$.

("Set Theory", 1978 p. 79)

To summarize: yes to your first question, your second question has the right idea but the wording is a little off.