Difference between metric space and normed space What is the difference between a normed linear space and a metric space?
When I was going through definition I saw that a metric space is defined on a set. But a normed space  is defined on a vector space. 
Is there any metric space where the metric is defined on a set but not on a vector space?
 A: Take a normed space and remove the origin. Now you have a metric space which is not a normed space. Alternatively, consider any non-contractible Riemannian manifold. For example, take a sphere with its standard round metric. This will be a metric space which is certainly not a normed space.
A: The only rules a metric has to satisfy are positivity, symmetry, and the triangle inequality. Given any set, finite or infinite, you can define the trivial metric: $d(a,b) = 1$ if $a\neq b$, $d(a,b) = 0$ if $a = b$. E.g. this metric can be used on the set $S = \{a, b, c\}$ to create a metric space, but $S$ is clearly not a vectorspace!
A: A necessary condition for a metric space $(X,d)$ to be normed is that the metric must be translation-invariant, i.e., we must have $d(x,y)=d(x+a, y+a)$ for all $a \in X$, and the metric must be homogeneous, i.e. $d(cx,cy)=cd(x,y)$. This, of course, requires additional  structure to be able to add/subtract points , which we can't always do in a generic Metric space.
Notice $$||x||:=\sqrt {\Sigma x_i^2}$$ is a norm that generates the metric $$d(x,y)=\Sigma_{i=1}^n (x_i-y_i)^2 $$ in Euclidean $n$-space. To recover the norm from the metric, just set ||x||=d(x,0). And notice $$|| x||=d(x,0)$$ in Euclidean space returns the generating norm $$||x||:= \sqrt { \Sigma_{i=1}^n x_i^2}$$ from above .
You can verify the same is the case for $L^p$ spaces, where the metric  $d(x,y):= \int |f-g|^p d\mu $ returns the norm $$||f||_p := \int |f|^p d\mu $$ as $d(x,0)$.
