I have a problem that asks me to give a counterexample or prove that it is closed under addition/multiplication depending on if it forms a subspace or not.
$$S = \begin{bmatrix} |a-b|\\ |b-c| \\ a+b+c\end{bmatrix}$$
where S is the set of all vectors in R^3, a,b,c, in Real Numbers
a = 0, b = 0, c = 0 and results are 0 which is a real number so it contains the zero vector.
a = 1, b = 2, c = 3 results in 1, 1, 6 which are real numbers and a = 2, b = 1, c = 3 results in 1, 2, 6 which are real numbers and their sum a = 3, b =3, c = 6 results in 0, 0, 12 which are real numbers which means its closed under addition.
a = 1, b = 2, c = 3 results in 1, 1, 6 and you have an arbitrary constant, lets say 2, which gives a = 2, b = 4, c = 6. This results in 2, 2, 12 which are real numbers which means its closed under multiplication.
However, the answer given was that it was not a subspace. Where did I go wrong?
You say arbitrary but then you pick $2$; this doesn't prove it always works...!
Hint: the first two components are absolute values and thus non-negative. What about taking a negative scalar multiple, e.g. $-2$ instead of $2$? You found that $(1,1,6) \in S$, what about: $$-2 \cdot (1,1,6) = (-2,-2,-12)$$