In Euclidean space interior of finite set is empty? I seem to be somewhat lost in certain concepts. I'm asked to prove that if $W\subset\mathbb{R}^n$ is a linear subspace (a vector subspace) of $\mathbb{R}^n$, with $W\ne \mathbb{R}^n$, then the interior of $W$ is the empty set.
But how is this possible? Take, for example, $\mathbb{R}^2\subset \mathbb{R}^3$. Consider an open ball $B(0;r)$ of radius $r$ around $0$ in $\mathbb{R}^2$. Don't we then have that the interior of $B(0;r)$ is $B(0;r)$ itself? Also, for every $r>0$ and every $x\in\mathbb{R}^2$, $B(x;r)\in\mathbb{R}^2$, so that the interior of $\mathbb{R}^2$ is $\mathbb{R}^2$.
Also, in Wikipedia it is said that the interior of any finite subset of a Eucledian space is empty. Again, I don't see how this is possible.
What is it that I'm missing here?
 A: Let's abbreviate your $B(0;r)$ as just $B$.
The interior of your $B$, as a subset of the topological space $\mathbb R^2$, is $B$ itself, because $B$ is open in the topology of $\mathbb R^2$.
But $B$ is not open in the topology of $\mathbb R^3$. In fact, $B$ does not contain any open ball in $\mathbb R^3$. So the interior of $B$ as a subset of the topological space $\mathbb R^3$ is empty.
The interior operation doesn't just operate on a set all by itself; it operates on a subset of a topological space. If you change the ambient space, you get a different interior.
A: A subset $W$ of $\Bbb R^n$ is open in it provided for each point $x\in W$ the set $W$ also contains a ball $B(x,r)$ (from $\Bbb R^n$) of some radius $r$ centered at $x$. Balls are $n$-dimensional bodies, not "flat". For instance, 

Take, for example, $\mathbb{R}^2\subset \mathbb{R}^3$. Consider an open ball $B(0;r)$ of radius $r$ around $0$ in $\mathbb{R}^2$. 

$B(0,r)=\{x\in\Bbb R^3:\|x\|<r\}$. If $x=(x_1,x_2,x_3)$ then $\|x\|=\sqrt{x_1^2+x_2^2+x^2_3}$, so all points $(r/2,0,0)$, $(0,r/2,0)$, and $(0,0,r/2)$ belong to $B(0,r)$.
A: In order for any set $S$ in a topological space to have a non-empty interior, it must contain an open set $U$. Explicitly, $U \subseteq S$. If you are working with the base of a topology this is equivalent to saying that there is a basic open set which is contained in $S$.
You should be able to see that the only way a finite set can have interior is if there is at least one finite set in your space which is open. This certainly can't happen in the Euclidean topology.
When you look at $\mathbb {R}^2$ as a subset of $\mathbb {R}^3$ then you should hopefully be able to convince yourself that you cannot fit a (three dimensional) ball inside the plane, so $\mathbb {R}^2$ contains no basic open sets in $\mathbb {R}^3$ and thus has empty interior.
A: Suppose that $B(x, r)\subset W$ for some $x\in \mathbb{R}^n$ and $r > 0$.
Then, $x\in W$, hence $-x\in W$. If follows that $-x + B(x, r) = B(0, r)\subset W$.
Now, take an arbitrary $y\in\mathbb{R}^n$. Define $\alpha = \frac{r}{2||y||}$ and $y_0 = \alpha y$. Since $||y_0|| <r$, we have $y_0\in B(0, r)\subset W$. But this means that also $\frac{1}{\alpha}y_0 = y\in W$. Since  $y$ was arbitrary, $W = \mathbb{R}^n$.
