Does basis of $A$ in $4$ dimensions, make basis for $B$ Suppose that {$x_1, x_2, x_3, x_4$} is a basis for a vector space $A$ . Suppose that $B$ is a subspace of $A$, and that $x_1, x_2 ∈ B$ but $x_3 ∉ B$ and $x_4 ∉ B$. Is it true that {$x_1, x_2$} is a basis for $B$?
If this is true, then I need to write a proof but if it's not, then a counterexample should suffice right? I not sure if it's true, I thought it wasn't but I'm having a hard time coming up with a counterexample so now I'm thinking that it is true.
Here's what I have so far: Consider where $A = R^4$. So, the basis has $4$ vectors and the vectors in a basis must be linearly independent and span. Let the basis of $R^4$ be {$x_1, x_2, x_3, x_4$}  so $B$ is given by {$x_1,x_2$}, so $B$ has dimension of 2, so it's in $R^2$. If $B$ is $R^2$, then standard basis of $B$ is {$i,j$}, but, if the vectors from the basis of $R^4$ have $4$ components, each, $x_1$ and $x_2$ are not in $R^2$ since they have more than $2$ components. So it's not a basis of $B$.
I'm not really sure if this is the right way to go about the question and counterexample. If it's true, how could I do a proof for it and how do I know it's true? Thanks.
(I'm starting to think it's true but I'm pretty confused)
Edit: I just tried to answer this, so please refer to the answer section to give me feedback. Thanks!
 A: Consider $B = \operatorname{span}\{x_1,x_2,x_3+x_4\}$. Since $x_1$, $x_2$, $x_3$, and $x_4$ form a basis for $A$, they are all linearly independent and hence so are $x_1$, $x_2$, and $x_3+x_4$. The span $B$ is then a three-dimensional subspace of $A$.
Now, neither $x_3$ nor $x_4$ are in $B$. To show this by way of contradiction, assume $x_3\in B$. Then there exists $a_1$, $a_2$, and $a_3$ in the field such that
$$
x_3=a_1x_1+a_2x_2+a_3(x_3+x_4)\,.
$$
Rearranging, we can write this as
$$
0=a_1x_1+a_2x_2+(a_3-1)x_3+a_3x_4\,.
$$
Since the $x_i$'s are linearly independent, all of these coefficients must be zero, which means that $a_1=0$, $a_2=0$, $a_3-1=0$, and $a_3=0$. But this is a contradiction because the last two equalities say that $0=a_3=1$. Thus $x_3$ is not in the span of $x_1$, $x_2$, and $x_3+x_4$ and is therefore not in $B$. The same argument works for $x_4$.
Thus, $B$ is a proper subspace of $A$ containing neither $x_3$ nor $x_4$, but it is not spanned by $x_1$ and $x_2$.
A: Suppose that {$x_1, x_2, x_3, x_4$} is a basis for a vector space $A$ . Suppose that $B$ is a subspace of $A$, and that $x_1, x_2 ∈ B$ but $x_3 ∉ B$ and $x_4 ∉ B$. Is it true that {$x_1, x_2$} is a basis for $B$?
This is not true. Have that a set of vectors form a basis if they are linearly independent and span the vector space. Since {$x_1, x_2, x_3, x_4$} for $A$ are linearly independent, we know that {$x_1, x_2$} are also linearly independent for B. Now it remains to show that {$x_1, x_2$} are not a span for $B$. Consider the situation where $A=R^4$, with basis {$x_1, x_2, x_3, x_4$} and some subspace $B$. Let  {$x_1, x_2$} = {$(0,0,0,2), (0,0,2,0)$}. For this to span $B$, all of $B$ must be expressed as a linear combination of vectors {$(0,0,0,2), (0,0,2,0)$}. Now consider when $B$ is some subspace of $R^4$, where  $(2,0,0,2)$ is an element of the subspace $B$. $(2,0,0,2)$ can't be expressed as a linear combination of {$(0,0,0,2), (0,0,2,0)$}, so {$(0,0,0,2), (0,0,2,0)$} does not span $B$, so it's not a basis of $B$.
Can I get some feedback on this answer? I want to show a solid counterexample. Thanks!
