# Proper subspaces of $R^n$

I'm trying to prove that proper subspaces of $$R^n$$ are closed and have empty interiors.

To prove that they are closed I'm trying to use the fact that inverse images of closed sets are closed sets by continuous functions. So I'm trying to find a linear map such that a proper subspace of $$R^n$$ is its Kernel but I don't know how to express it so that it would work on all dimensions.

Regarding the fact that they have empty interiors, I really don't know what to do.

Suppose $V \subset \mathbb{R}^n$ is a proper subspace, you can begin with a basis $\{v_1, v_2,\ldots, v_k\}$ of $V$ and extend it to a basis $\{v_1,v_2,\ldots, v_n\}$ of $\mathbb{R}^n$. Now for any $x\in \mathbb{R}^n$, we can write $x$ as $$x = v + \sum_{i=k+1}^n \alpha_i v_i$$ where $v \in V$. Now define the map $T: \mathbb{R}^n \to \mathbb{R}^n$ by $$T(x) = \sum_{i=k+1}^n \alpha_i v_i$$
Then $T$ is linear (by definition), and $\ker(T) = V$ (Why?)
Since $T$ is finite dimensional, $T$ is continuous (it is really just a matrix), and hence $$V = T^{-1}(\{0\})$$ is closed.
As for the empty interior : Suppose $V$ had interior, then there would be $x \in V$ and $\epsilon > 0$ such that $$B(x,\epsilon) \subset V$$ For any $y \in \mathbb{R}^n$, you can find $\alpha \in \mathbb{R}, \alpha\neq 0$ such that $$\|\alpha y\| < \epsilon$$ Now note that $w = x+\alpha y \in B(x,\epsilon) \subset V$. Since $x \in V$ and $V$ is a subspace, $$y = \frac{1}{\alpha}(w-x) \in V$$ This is true for any $y \in \mathbb{R}^n$, and hence $\mathbb{R}^n = V$, contradicting the fact that $V$ is a proper subspace.