All Norms on a Finite Dimensional Real Vector Space V Determine the Same Topology on V? In this instance, a norm determines a metric by $d(x,y) = |x-y|$.
I am not entirely what "determines" fully means. I am thinking it is interchangeable with "can be described by" but I am not sure. I was told to first consider $\mathbb{R}^n$ and then use the restriction of the norm onto the unit sphere.
Does it make sense to show that if $||\cdot||$ is a norm, then it is induced by the standard metric on $\mathbb{R}^n$ in terms of open balls? I loosely understand what the problem wants, but I'm not sure how to sew everything together and bridge the gap. Perhaps if $||\cdot||_{1}$ and $||\cdot||_{2}$ are norms, then they're equivalent if they induce the same topology on $\mathbb{R}^n$? In any case I want to show this for any real finite dimensional vector space.
 A: A relatively detailed answer, so maybe just read it after you have tried to solve it with the given hints:
Any finite-dimensional real vector space equipped with a norm $\lVert \cdot \rVert_V$ is isometrically isomorphic to $(\mathbb R^n, \lVert \cdot \rVert)$ with a suitable norm on $\mathbb R^n$. This can be seen by taking a basis $(v_1, \dots, v_n)$ and mapping
$$\phi: (V, \lVert \cdot \rVert_V) \to (\mathbb R^n, \lVert \cdot \rVert), \quad v_i \mapsto e_i$$
where $(e_1, \dots, e_n)$ is the standard basis of $\mathbb R^n$. This map is clearly linear and bijective. We choose as a norm on $\mathbb R^n$ the norm
$$\lVert \sum_{i=1}^n a_ie_i \rVert =\lVert \sum_{i=1}^n a_iv_i \rVert_V$$
which makes $\phi$ into an isometry. All properties follow from the fact that $\lVert \cdot \rVert_V$ is a norm on $V$ and that $\phi$ is bijective.
This means that to show that all norms on a finite-dimensional vector space are equivalent you only have to worry about norms on $\mathbb R^n$. Two norms $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$ are said to be isometric if there are constants $A,B>0$ such that
$$A\lVert x\rVert_1 \le \lVert x\rVert_2 \le B\lVert x\rVert_1$$
for all $x \in \mathbb R^n$.
A standard proof is to fix one norm on $\mathbb R^n$ (e.g. thenorm $\lVert x \rVert_1 = \sum_{i=1}^n|x_i|$) and to show that any other norm $\lVert \cdot \rVert$ on $\mathbb R^n$ is equivalent to $\lVert \cdot \rVert_1$. Then you can use that the equivalence of norms is transitive to show that any two norms on $\mathbb R^n$ are equivalent.
(To show the equivalence of $\lVert \cdot \rVert$ and $\lVert \cdot \rVert_\infty$ for one inequality we use the fact that we have a finite basis $(v_1, \dots, v_n)$ and for the other inequality we use that a norm $\lVert \cdot \rVert$ is always a continuous function and it obtains a minimum on $\{ \lVert x \rVert_1 = 1\}$ - a compact set)
Then you have to see that two normed spaces have the same topology (induced by their norms) if and only if the two norms are equivalent. Having the same topology means having the same open and closed sets. Now if two norms $\lVert \cdot \rVert_1, \lVert \cdot \rVert_2$ on a space are equivalent ($C \lVert \cdot \rVert_2 \le \lVert \cdot \rVert_1 \le D \lVert \cdot \rVert_2$) and a set $A$ is open in $(X, \lVert \cdot \rVert_1)$ that means for any $x \in A$ there is a ball $B_\delta(x)$ w.r.t. $\lVert \cdot \rVert_1$ still in $A$.
But that means for all $y$ s.t. $\lVert y - x \rVert_1 < \delta$ we have $y \in A$ which implies that for all $y$ such that $\lVert y-x \rVert_2 < \frac{\delta}{c}$ we have $y \in A$ so $A$ is also open w.r.t. $\lVert \cdot \rVert_2$. By switching the roles of $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$ you get that the two equivalent norms determine the same topology.
If on the other hand two topologies induced by norms are the same then we have
$$B^{\lVert \cdot \rVert_1}_1(0) \supset B_r^{\lVert \cdot \rVert_2}(0)$$
for some $r>0$ because the open set $B^{\lVert \cdot \rVert_1}_1(0)$ is also open w.r.t. $\lVert \cdot \rVert_2$, so it has to contain a $\lVert \cdot \rVert_2$-neighborhood of $0$. But this means for any $x \ne 0$ that by writing $y = \frac{rx}{\lVert x \rVert_2 2}$ we have $\lVert y \rVert_2 < r$ which implies $\lVert y \rVert_1 <1$ and this means
$$\lVert x \rVert_1 \le \frac 2r \lVert x \rVert_2$$
Again by symmetry you get the other direction and this shows that the norms are equivalent.
A: 
I am not entirely what "determines" fully means.

If you have a norm, then you always have a metric. (Conversely, if you have a metric, you don't necessarily have a norm, most obviously because metric spaces aren't necessarily vector spaces.)
(Furthermore, if you have an inner product, then you always have a norm. However, you didn't ask about this, so I won't elaborate further.)
A roadmap to show that "All Norms on a Finite Dimensional Real Vector Space V Determine the Same Topology on V" could be as follows: first, show that the topologies induced by two norms are equivalent if and only if the two norms are equivalent. Second, show that all norms on a finite-dimensional vector space over $\mathbb C$ are equivalent.
A: 
In this instance, a norm determines a metric by (,)=|−|
I am not entirely what "determines" fully means. I am thinking it is
interchangeable with "can be described by" but I am not sure.  [...]

Usually, it is said "a norm induces a distance". But 'induces' is a vague term that must be made precise from a mathematical point of view.
The question is that way:
If we have a  normed vector space, equipped with the norm $||\cdot||$,  $X$ is a metric space with the distance $d$ given by $d(x,y)= ||x-y||,  \; x,y \in X$.
That is, the distance induced by the norm is the norm of the difference $x-y$.
Of course, it must be proved that $d(x,y)= ||x-y||$ is actually a distance.
This can be done resorting to the definitions of norm and distance.
Remember the properties of a norm $||\cdot||$ on a (real or complex) vector space $X$:

*

*$||x||\geq0$ and $||x||=0 \Leftrightarrow x=0$


*$||\alpha x||=|\alpha|||x||$ with $\alpha\in \mathbb{R}$ or $\mathbb{C}$


*$||x+y||\leq||x||+||y|| \;\;\;\; \;\;(Triangle \; inequality)$.
Properties of a distance $d$ on a set $Y$:
1') $d(x,y)\geq 0$ and $d(x,y)=0 \Leftrightarrow x=y$,
2') $d(x,y)=d(y,x)$,
3') $d(x,y)\leq d(x,z)+ d(z,y), \,\,\,\,\,(Triangle \;\;inequality)$
$\forall x,y,z\in Y$
Using  properties $1,2,3$ of a norm, one can easily prove that $||x-y|| $ satisfies the properties  $1'),2'), 3')$ of a distance.
Hence, normed spaces are metric spaces. But the reverse is not true: not every metric on a vector space is associated to a norm. It can be proved that a distance is induced by a norm if and only if it is translation invariant and homogeneous .

Does it make sense to show that if ||⋅|| is a norm, then it is induced
by the standard metric on ℝ in terms of open balls?

In general, the answer is 'no', a norm is not necessarily induced by a metric, as we said.
But if you are in $\mathbb{R^n}$, equipped with the Euclidean norm, the norm of a point $x$ can be seen (in fact is) as the distance from the origin: $d(x,0)=||x-0||=||x||.$
