Why does $\operatorname{Con}(\operatorname{Con}(\Gamma)) = \Gamma$?

Question

I'll understand if it was

$\Gamma \subseteq \operatorname{Con}(\operatorname{Con}(\Gamma))$

but why equals? There should be other formulas in $\operatorname{Con}(\operatorname{Con}(\Gamma$)) which aren't part of $\Gamma$.

Edit (lecture notes):

Answer 1

I am certain that this is a typo, and should read "$Con(Con(\Gamma))=Con(\Gamma)$". This is because a closure operator $c$ on a family of sets $\mathcal{F}$ is one satisfying the following three properties:

• $X\subseteq c(X)$,

• $X\subseteq Y$ implies $c(X)\subseteq c(Y)$, and

• $c(c(X))=c(X)$

for all sets $X, Y\in\mathcal{F}$.

In particular, note the "that is" between the statement of the lemma and the three properties: the three properties listed are meant to define what a closure operator is.

---

As further evidence for this being a typo, note that as written the properties imply that $Con(X)=X$ for all $X$:

• We have $X\subseteq Con(X)$ and $Con(X)\subseteq Con(Con(X))$ by property $1$.

• But $Con(Con(X))=X$, by property $3$ as your professor wrote it.

• So $X\subseteq Con(X)\subseteq X$, that is, $X=Con(X)$.

So clearly your professor didn't mean to write that, since otherwise $Con$ is trivial!