Proving that $[\mathcal F, \mathcal F]$ is an ideal. Let $\mathcal{F}$ be a Lie algebra over a field $k.$ A lie subalgebra of $\mathcal{F}$ is a vector subspace $S \subset \mathcal{F}$ such that $[S,S] \subset S.$ An ideal of $\mathcal{F}$ is a vector subspace $I \subset \mathcal{F}$ such that $[I, \mathcal{F}] \subset I$.
Now, I want to show that
Show that $[\mathcal F, \mathcal F] = \{\sum_{i}^n[x_i,y_i]| x_i, y_i \in \mathcal{F}$} is an ideal of $\mathcal{F}.$
My questions are:
-1- (closure under Lie bracket)My intuition says that  $[\mathcal F, \mathcal F] \in \mathcal F,$ but I am unable to give an appropriate wording because I do not understand what is [,] exactly. (any help with that wording will be appreciated!)
0- I have a question, does this [,] here always means the commutator or just the Lie bracket?
1-Also, I have a question, can I say $[0,0] = 0 \in F$? my intuition say no (can the commutator by any means be zero?), but how can I show that the additive identity is in $\mathcal F$? Also, do I have to show that
the additive identity is in $\mathcal F$?
2- Showing that it is closed under addition (I know that the given definition is giving it to me as the set of finite linear combinations of the commutator. ):
so assume $a,b \in [\mathcal F, \mathcal F],$ we want to show that $a+b \in [\mathcal F, \mathcal F],$ i.e., it is a set of finite linear combinations of the commutator.
Since $a \in [\mathcal F, \mathcal F],$  then $a = \{\sum_{i}^n[x_i,y_i]| x_i, y_i \in \mathcal{L}\},$ Since $b \in [\mathcal F, \mathcal F],$  then $b = \{\sum_{j}^m[x_j,y_j]| x_j, y_j \in \mathcal{L}\},$
But then how can I add 2 sets?
Should I use induction here? if so, how?
3-  Showing that it is closed under scalar multiplication with elements from the field $k.$
Let $c \in k, a \in [\mathcal F, \mathcal F],$ we want to show that  $ca \in [\mathcal F, \mathcal F].$
How multiplying a lie bracket (or a commutator ) by a scalar look like ? do I have to multiply the first coordinate only or the two coordinates inside [,]?
4- Then showing $[[\mathcal F, \mathcal F], \mathcal F] \in [\mathcal F, \mathcal F]$ is extremely hard to me, Any help in that part will be greatly appreciated!
Can anyone help me answer all the above questions so that I can have a peaceful, alert mind instead of the confusion I am having now?
thanks in advance!
 A: Q: I have a question, does this [,] here always means the commutator or just the Lie bracket?A: It is an alternating, bilinear map sending a pair $(a, b)$ to some element $[a, b]$ that is required to satisfy the Jacobi identity $[[a, b], c] + [[b, c], a] + [[c, a], b] =0$.
Q: (closure under Lie bracket)A: I didn't tell you the domain and codomain of $[-, -]$ above, but I will do it now. It is a map $\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$, and in your situation $\mathfrak{g}=\mathcal{F}$.
Q: $[0, 0]=0$A: It is bilinear, so it is linear in both variables. $[0, 0]=[0+0, 0]=[0, 0]+[0,0]$. This gives you what you are after.
Q: Showing that it is closed under additionA: The ideal is defined as sums of Lie brackets. It is, by definition, closed under addition. Also, the bracket is alternating (which means $[a, b]=-[b, a]$), so it also contains the additive inverses. You are also misinterpreting the notation $[X, Y]$. It does not mean that you literally take two sets and somehow do the bracket thing to them (whatever that would even mean!), but that you take one element from each, bracket those and your $[X, Y]$ then consists of all finite sums of brackets $[x, y], x\in X, Y\in Y$.
Q: Showing that it is closed under scalar multiplication with elements from the field $k$.A: Again, bilinearity. $[ra, b] = r[a, b] = [a, rb]$ for all $r\in k, a, b\in\mathcal{F}$.
Q: $[[\mathcal{F}, \mathcal{F}], \mathcal{F}]\subseteq[\mathcal{F}, \mathcal{F}]$A: Once again, we need the bilinearity here. If we look at the elements the claimed subset, those are of the form $\sum\limits_{i=1}^m[\sum\limits_{j=1}^n[a_{ij}, b_{ij}], c_i]$. I'll let you figure out what happens to them!
