If A and B are subspaces of a vector space, then dim A < dimB, implies A is contained in B. Is this statement True? I think that the statement is true but NO.
Can someone state the reason, Please.
 A: Consider, in $\mathbb{R}^3$, the subspace $A = \{(a,0,0)| a \in \mathbb{R}\}$ and the subspace $B = \{(0,b,c)| (b,c) \in \mathbb{R}^2\}$
Is it true for $A$ and $B$?
A: No. It is false. Consider three dimensional space. Take x-y plane and z axis. Dimension of x-y plane is 2.
Dimension of z axis is 1. But z axis is not contained in x-y plane
A: Let $A$ be the set of all vectors in $\mathbb{R}^{3}$ whose second and third coordinates are $0$, and $B$ the set whose first coordinate is $0$. The set $A$ has dimension $1$, and $B$ dimension $2$, but $A \cap B$ contains only the zero vector.
Edit: We would write $A = \mathbb{R} \times \{0\}^2 = \{ (x, 0, 0) : x \in \mathbb{R}$, and $B = \{0\} \times \mathbb{R}^2 = \{ (0, y, z) : y, z \in \mathbb{R} \}$. To visualize, we have a line $A$ and a plane $B$ perpendicular to that line, so $A$ passes through $B$ at exactly one point, in this case $\{0, 0, 0\}$.
Edit 2: Someone pointed out that all answers so far have given the same counterexample. I'd like to take a moment to comment on why that might be. Firstly, the claim is actually correct if we stick to $\mathbb{R}^2$, but for uninteresting reasons. We can either say
\begin{align*}
\dim A & = 0 & \dim B & = 1 , \\
\dim A & = 0 & \dim B & = 2, \\
\dim A & = 1 & \dim B & = 2 .
\end{align*}
The first two are trivial, as the $0$-dimensional vector space is just the zero vector, and is thus contained in all vector spaces. The final case is what we did, however if these were both subspaces of $\mathbb{R}^2$ then the counterexample wouldn't work. This is because if $W$ is an $n$-dimensional subspace of an $n$-dimensional space $V$, then in fact $V = W$ (this must suppose that $n$ is finite, or it fails). So we have to pass to $\mathbb{R}^3$, where we get the above example.
A: $L(\vec i$x$\vec j)$ and $L(\vec k)$ are both subspaces of $L(\vec i$x$\vec j$x$\vec k)$ and $L(\vec k)$ is not contained in $L(\vec i$x$\vec j)$
