Prove that if $M$ is a subspace of $\mathbb{R}^n$ such that $M \not= \mathbb{R}^n$ , then the interior of $M$ is empty.

Question

Prove that if $M$ is a subspace of $\mathbb{R}^n$ such that $M$ is not equal to $\mathbb{R}^n$ , then the interior of $M$ is empty.

Could someone help me with this problem?

Answer 1

Can you see why this is true intuitively? As an example, draw a line in $\mathbb{R}^2$. Or, draw a line or a plane in $\mathbb{R}^3$.

Formally, to show that the interior of $M$ is empty, pick any point $x$ in $M$ and show that any ball centered at $x$ will contain some point outside of $M$.

Answer 2

Hint:

By contrapositive. Suppose first there's a neighbourhood of $0$ entirely contained in $M$. We can suppose it's a ball centred at $0$. If $\mathcal C=(e_1, \dots,e_n)$ is the canonical basis of $\mathbf R^n$, $\lambda\mathcal C=(\lambda e_1,\dots, \lambda e_n)$ is also a basis of $\mathbf R^n$, and if $\lambda$ is small enough, it is contained in the ball, hence in $M$. Therefore $$\DeclareMathOperator\span{Span}M\supseteq\span(\lambda \mathcal C)=\mathbf R^n.$$

Answer 3

Assume that $M$ is a linear subspace of $\mathbb{R}^n$ with a nonempty interior.

Lemma. There exists an open ball around $0$ contained in $M$.

Proof. Since $M$ has a nonempty interior then there exists some open ball $B$ centered at $x\in M$ (not necessarily $0$) contained in $M$. Define $B'=\{v-x\ |\ v\in B\}$. You can easily check that $B'$ is an open ball around $0$ with the same radius as $B$. Since $M$ is a linear subspace then obviously $B'\subseteq M$ which completes the proof. $\Box$

So pick an open ball $B$ centered at $0$ of radius $r>0$ contained in $M$. Define $$r_i=(0,\ldots, \frac{r}{2},\ldots, 0)\in\mathbb{R}^n$$ where $\frac{r}{2}$ is on the $i$-th coordinate. Let

$$V=\mbox{span}(r_1,\ldots, r_n)$$

Obiously $\lVert r_i\rVert=\frac{r}{2}$ and thus $r_i\in B\subseteq M$. Therefore $V\subseteq M$.

On the other hand the set $\{r_1,\ldots, r_n\}$ is linearly independent (almost by definition). Therefore $\mbox{dim}(V)=n$ and since $\mbox{dim}(\mathbb{R}^n)=n$ then $V=\mathbb{R}^n$. Thus we have

$$\mathbb{R}^n=V\subseteq M\subseteq\mathbb{R}^n$$

which completes the proof.

Answer 4

A set with nonempty interior contains an open ball, which has positive Lebesgue measure, but any proper subspace is contained in an hyperplane, which has measure zero.