Does the $2$-dimensional real vector space have infinitely many subspaces? Can anyone provide links to a concrete proof? Intuitively, the two-dimensional real space is infinite. so there should be infinitely many subspaces. But how do I go about a proof?
 A: Let $V$ be a $2-$dimensional vector space over $\mathbf{R}$. Fix a basis $v_1,v_2$ for $V$ so that $V$ can be identified with $\mathbf{R}^2$. Then we can see that the subspaces 
$$X_\theta=\{\lambda(\cos\theta, \sin\theta):\lambda \in \mathbf{R}\}$$
are distinct $1-$dimensional vector spaces for each $0\le \theta< \pi$. Visually, each of the vectors $(\cos\theta,\sin\theta)$ describes a unit vector in the unit circle of $\mathbf{R}^2$. The subspaces $X_\theta$ are then the lines through the origin making a counter-clockwise angle of $\theta$ with the positive $x-$axis. In particular, there are uncountably many such subspaces.
A: Take $v_\epsilon=(\epsilon,1)$, for $\epsilon \in [0,1]$. Then $\lambda v_\epsilon + \mu v_\nu=0$ implies $\lambda \epsilon + \nu \nu =0$ and $\lambda + \nu =0$, hence $\lambda (\epsilon -\nu)=0$. This shows that for $\epsilon \neq \nu$, $v_\epsilon$ and $v_\nu$ are linearly independent and therefore span different subspaces in $\mathbb{R}^2$. Hence we get an injective map from $[0,1]$ into the set of all subspaces of $\mathbb{R}^2$. Since $[0,1]$ is clearly infinite, we must have infinitely many subspaces in $\mathbb{R}^2$.
A: You should use your intuition from Euclidean geometry.
If you fix a point $O$ in a geometrical ($2$-dimensional!) plane, and identify every geometrical vector $v=\vec{OA}$ with its ending point $A$, then:


*

*The whole $2$-dimensional vector space is identified with the whole plane;

*Any $1$-dimensional subspace is identified with a line through $O$. (That is why it is obviously infinitely many of them.)

*The unique $0$ dimensional subspace is identified with the $1$-element set consisting of $O$ as its only element.


Hope this clears the matters a bit.
