About inner products, norms and metrics Do these three kinds of vector spaces, those with an inner-product, those with a norm and those with a metric, are the same sets of vector spaces? At least for finite dimensional vector spaces all of these coincide? 
It would be great to know finite dimensional counter-examples if they exist any and if anyone can link to some lecture notes explaining this point. 

Like are there examples of inner product spaces which cannot have a metric or metric spaces which cannot have a norm or various other such possible conflicts between these 3 properties. 
 A: Even in finite dimension, the converses are not true.
For instance, the $p$-norm ($p≠2$) on $\Bbb R^n$ doesn't come from an inner product. This can be proven by contradiction, using the parallelogram law or the polarization identity. The polarization identity
$$2 \|x\|^2+2\|y\|^2=\|x+y\|^2+\|x-y\|^2 \qquad \forall x,y \in V$$
 precisely gives a necessary and sufficient condition for a normed vector space $V$ to have an inner product that induces that norm.
The discrete metric on $\Bbb R$ doesn't come from a norm, because it takes only $0$ and $1$ as values.
If $d$ is a metric such that there is a norm $\|\cdot\|$ on $V$ with $d(x,y)=\|x-y\|, \; \forall x,y \in V$, then two necessary conditions are that
$$d(ax,0)=|a|d(x,0) \tag 1 $$ 
$$d(x,y)=d(x+z,y+z) \tag 2$$
 for every $x,y,z \in V$ and $a \in \Bbb R$.
Conversely, if the metric $d$ satisfies $(1)$ and $(2)$, then $\|x\|:=d(x,0)$ yields a norm on $V$.
You can also see this question.
A: Take $X=\mathbb R^n$ and equip it with the topology induced by the the distance function
$$
{\rm d}(x,y) = \begin{cases} 0 &\text{if } x=y\\1&\text{if }x\neq y\end{cases}
$$
Then $X$ is a metric space, specifically the discrete metric space (in which all points are isolated)
However, $(X,{\rm d})$ fails to be a normed space in that ${\rm d}$ does not induce any norm, as you can easily verify:
$$
  {\rm d}(\lambda x,\lambda y) \neq |\lambda|{\rm d}(x,y)
\quad\text{whenever $x\neq y$ and $\lambda\neq\pm1$}
$$
Let now
$$
\|x\|_p = \left(\sum_{i=1}^n|x_i|^p\right)^{\frac1p}
\quad\text{for some $p>0$}
$$
Then you can easily verify that $\|{}\cdot{}\|_p$ is a norm for any $p>0$, hence $(X,\|{}\cdot{}\|_p)$ is a normed space, but for $p\neq 2$ it is not induced by a scalar product.
Therefore, for $p\neq 2$ $(X,\|{}\cdot{}\|_p)$ is a normed space which is not induced by an inner product.
A: You can always use an inner product to define a norm and you can always use a norm to define a metric.  Specifically, if $<a,b>$ is the inner product of $a$ and $b$, then $\sqrt{<a,a>}$ is a norm, i.e. you can say $\Vert a\Vert = \sqrt{<a,a>}$ and it has all the properties required of a norm.  If you have a norm $\Vert \cdot \Vert$, then you can define a metric by $d(a, b)=\Vert a-b\Vert$ and that metric has all the properties required.
However, as pointed out in the comments, the reverse is not true.  I.e. having a metric does not imply that you have an inner product.
A: I don't know if this helps, but a metric space is not necessarily a vector space.
"The set of positive real numbers, ℝ+=(0,∞), with the metric given by d(x,y):=|x−y| is a metric space, but it is not a linear space, since it contains neither an additive identity (0) nor additive inverses (−x)."
Source: https://wiki.math.ntnu.no/users/ehrnstro/teaching/linearmethods/metricspaces
A: In the plane the ordinary Euclidean metric comes from the norm that comes from the usual inner product.
In the taxicab metric the distance from $(a,b)$ to $(c,d)$ is $|a-c| + |b-d|$. That metric does come from a norm, but the norm doesn't come from an inner product.
The metric for which the distance between any two distinct points is $1$ doesn't come from a norm.
