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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!

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2 Answers 2

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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 $I=\text{ linear closure of }S$

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  • $\begingroup$ @JustDroppedln But is my reasoning correct? $\endgroup$
    – wessi
    Mar 27, 2020 at 22:00
  • $\begingroup$ @jacques99 as far as I understand, yes, you are correct. $\endgroup$ Mar 27, 2020 at 22:02
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Comments on your argument:

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.

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