Measure Theory – Proof that the intersection of Sigma algebras is a Sigma algebra

For $ A subset 2 ^ { Omega} $,

let $ sigma (A): = bigcap _ { mathcal {F} in {F}} mathcal {F} $,

or $ F = { mathcal {F} subset 2 ^ { Omega}: {F} $ is a $ sigma $-Algebra and $ A subset mathcal {F} }. $

The goal is to prove that $ sigma ({A}) $ is a $ sigma $-algebra.

It seems to me that we can define $ sigma {A} $ as $ bigcap _ { mathcal {F} _ {k} in F} mathcal {F} _ {k} forall k in K $.

Then if $ A in mathcal {F} $, $ A in mathcal {F} _ {k} $ for everyone k $And the same is true $ A_ {c} $.

So, to show that $ sigma ({A}) $ is a sigma algebra, we simply show the following: that if $ A_ {n} in sigma ({A}) $ for everyone $ n = 1,2, dots $,

then $ A_ {n} in mathcal {F} _ {k} $ for everyone $ n, k $, and

$ bigcup_ {n = 1} A_ {n} in mathcal {F} _ {k} forall k $,

which implies $ bigcup_ {n = 1} A_ {n} in mathcal {F} $.

The only thing that puts me off, is that we have $ A subset mathcal {F} $, do not $ A in mathcal {F} $.

Is it true that if A, F are both $ subset $ $ 2 ^ { Omega} $, and $ A subset F $, this $ A in F $? I think that's true, but does it have to be proven?