Collection of independent sets equal to downset? I have been struggling with the following problem.

Every set here is supposed to be finite.

If we have a closure $\lambda$ on $X$, we define the collection of independent sets of $X$ as $$ I_\lambda = \{ A \subseteq X : x \notin\lambda(A\setminus \{ x \})\forall x\in A \} $$

Supose now we have a downset $I \subseteq 2^X$. Is there a closure $\lambda$ on $X$ such that $I_\lambda = I$? 

I first tried a pretty straightforward closure, by setting $\lambda(A) = A$ for elements in $I$ and $\lambda(A) = X$ for the rest. This of course does not work but only on a few sets, namely the ones that are not in $I$ but that any $A \setminus \{x\}$ is in $I$ for every $x \in A$, I tried to refine this idea to get rid of those cases but I haven't been able to do it.

This lead to my believe that it may not be possible and probably the trouble sets are those I just mentioned, but on a few small example I have been able to find the closure even when there are sets like those.

Edit: Now I know this is not possible, but I need help finding a counterexample, or even better, trying to understand when it exactly fails, and when it works.

Any help is appreciated

Definitons

A closure $\lambda$ on $X$ is a function $\lambda: 2^X \to 2^X$, such that is increasing (i.e $A \subseteq \lambda(A)$), monotone and idempotent.

An $I$ is said to be a downset if for any $A \in I$ and $B \subseteq A$, then $B \in I$
My first idea was to try
$$
\lambda(A)=\begin{cases}A,&amp;\text{if $A\cup\{x\}\in I$ for some $x\in X\setminus A$}\\X,&amp;\text{otherwise,}\end{cases}
$$
but this does not work (see alejopelaez's comment). Another idea I had was to try something like
$$
\lambda(A)=\{x\in X:\text{$x\in A$ or $A\cup\{x\}\notin I$}\},
$$
but this is not idempotent.

It is possible that an idea from matroids might work. For example, we could take
$$
\lambda(A)=\{x\in X:r(A)=r(A\cup\{x\})\},
$$
where
$$
r(A)=\max\{\lvert B\rvert:\text{$B\subseteq A$ and $B\in I$}\}
$$
is the size of the largest subset of $A$ that is in $I$. The idea is that $r$ is some kind of "rank" function, having the property that $r(A)=\lvert A\rvert$ if and only if $A\in I$.
Try the closure that have the exchange property.