Prove that the set of singleton subsets $\{\{a_1\},\ldots, \{a_n\}\}$ is a basis for the power set of $U$. Question:

Let $U = \{a_1,a_2,\dots,a_n\}$ be a finite set and consider the power set $\mathcal P(U)$ as a vector space over the field $\Bbb Z_2 = \{0,1\}$ with addition defined by
$$
A+B = (A \cup B) \setminus (A \cap B)
$$
and scalar multiplication defined by $0A = \emptyset$, $1A = A$ for all $A,B \subseteq U$. Prove that the set of singleton subsets $\{a_1\},\{a_2\},\dots,\{a_n\}\}$ is a basis for $\mathcal P(U)$.

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Solution:

For any $A \in \mathcal P(U)$. we can write
$$
A = \sum_{a \in A}\{a\},
$$
and so the set of singleton subsets spans $\mathcal P(U)$.
Furthermore, the set of singleton subsets is linearly independent since any linear combination of them is equal to $\emptyset$ (the zero vector in this space) if and only if all the coefficients are zero.

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The links are to the question and solution respectively.
The solution to this question, given my professor, is quite empty and not very helpful.
I'm confused by how the solution defined $A$ and why it's defined the way it is. im also confused by how defining $A$ like that implies that the set of singelton subsets spans the power set of $U$. Also, why is any linear combination of the set of singelton subsets equal to the empty set?
Any help is appreciated, thank you.
Joe
 A: $A$ is not "defined" that way. Instead, in this proof $A$ is given to you as an arbitrary element of $\mathcal P(U)$.
But then comes the statement "We can write $A = \sum_{a_i \in A} \{a_i\}$", and that statement is simply an assertion that this equation is true.
If you read this statement out loud to yourself, it sounds like this: "$A$ is the sum of its singleton subsets", which perhaps gives you an intuitive reason why it is true.
Of course this is a proof and so you might not be satisfied with an intuitive reason for this equation. If you want to prove this equation rigorously, you can prove it by induction on $A$; proving the induction step will be where your proof uses the actual definition of the sum operation.
You have also asked does this implies that the collection of singleton subsets spans $\mathcal P(U)$, and that's because you can rewrite the above equation in the form
$$A = \sum_{a_i \in A} 1 \cdot \{a_i\} + \sum_{a_i \in U-A} 0 \cdot \{a_i\}
$$
For your final question regarding linear independence of the set of singleton sets, it looks like you have misunderstood the suggested formulation of linear independence. But that suggestion is not very clearly written, so your confusion is not too surprising. Here is a rewording of that suggested proof, so you can try proving this:

For any given linear combination of the singleton sets, the coefficients of that linear combination are $0$ if and only if that linear combination is equal to the empty set.

A: (1). You can show by induction on $k,$ for $1\le k\le n,$ that if $A_1,...,A_k$ are $k$ distinct members of $P(U)$, then any $x\in\sum_{j=1}^kA_j$ iff $x$ belongs to exactly one of $A_1,...,A_k.$
This   implies that if   $\emptyset\ne A\in P(U)$ then $\sum_{a\in A}\{a\}=A.$
And $0x=0=\emptyset$ for any $x\in P(U).$
So the linear span of the set $C= \{\{a\}:a\in U\}$ is all of  $P(U).$
(2). If $k>0$ and if $\{a_1\},...,\{a_k\}$ are any $k$ distinct members of $C$ and if $s_1,...,s_k$ are scalars then, with the usual convention that the sum of the members of $\emptyset$ is $0,$ (because some of the following sums may be empty), we have $$\sum_{j=1}^ks_j\{a_j\}=\sum_{s_j=0}s_j\{a_j\}+\sum_{s_j=1}s_j\{a_j\}=$$ $$=0+\sum_{s_j=1}\{a_j\}=$$ $$=\{a_j: s_j\ne 0\}.$$ This last value cannot be $0,$ (i.e. cannot be $\emptyset$), unless $\{j:s_j\ne 0\}$ is empty, that is, unless every $s_j=0.$ So $C$ is a linearly independent set.
