# Property of the linear closure of a vector space, Linear Algebra

I'm trying to test this curious property of the linear closure of a vector space( Property 6 of the attached image), I think we have to prove this equality:

let $$F\subset E$$ let's see that

$$E_{s}=\bigcap_{S\subset F} F$$

indeed: Let $$x\in E_s$$ and $$S\subset F$$, since $$x\in E_s$$ then $$x= \sum_{i\in \Lambda}\lambda_is_i$$, $$s_i \in S$$ so $$x\in S\subset F$$ hence $$x\in F$$.

On the other hand let $$x\in F$$ for all $$S\subset F$$ in particular since $$S\subset E$$ then $$x\in E$$

So Now I don't know how to finish!  First prove that an intersection of subspaces is a subspace. Therefore the intersection of all subspaces containing $$S$$ is a subspace and it contains $$S$$. Lets call this intersection $$I$$. Since the linear closure of $$S$$ is a subspace and it contains $$S$$, it is obvious that $$I$$ is a subset of the linear closure of $$S$$.
On the other hand, an element of the linear closure of $$S$$ is of the form $$\sum_{i=1}^nx_is_i$$ where $$s_1,\dots,s_n\in S$$ and $$x_i$$ are scalars. Therefore, this element belongs to any subspace containing $$S$$ (since it is a linear combination of elements of $$S$$). Therefore it is an element of the intersection of all subspaces containing $$S$$, which is exactly $$I$$. . Therefore $$I=\text{ linear closure of }S$$
In your sentence starting with "indeed" you cannot conclude $$x\in S$$. You can only conclude that $$x\in F$$. Why? Well, $$s_i\in S\subset E_S\subset F$$ so, since $$F$$ is a subspace and is therefore closed under linear combinations, we have $$x=\sum_i\lambda_is_i\in F$$. However, we do not know if $$S$$ is closed under linear combinations so $$x$$ might not lie in $$S$$. If $$S$$ were a subspace we would have $$x\in S$$ but we are not assuming $$S$$ is a subspace. The rest of your sentence is correct. And, to complete this part of the proof, knowing that $$x\in F$$ for any subspace $$F$$ such that $$S\subset F$$ means that $$x\in \cap_{S\subset F, F subspace}F$$. This shows that $$E_S\subset \cap_{S\subset F, F subspace}F$$.
The next part of the argument is to show the other inclusion: $$\cap_{S\subset F, F subspace}F \subset E_S$$. Since $$E_S$$ is a subspace containing $$S$$ it is one of the $$F$$'s appearing in the intersection $$\cap_{S\subset F, F subspace}F$$. Hence, the intersection is at most $$E_S$$, ie, $$\cap_{S\subset F, F subspace}F\subset E_S$$. I don't follow what you have done for this half of the proof.