Why would the diameter of a subset of a metric space not be finite? The diameter of a subset $A$ of metric space $(X,d)$ is defined as
$\delta(A)=\sup_{x,y\in A}d(x,y)$
and the set $A$ is defined as bounded if $\delta(A)<\infty$. But, since the first axiom of metric space says that the metric should be finite. Does not it mean that every metric space should be bounded?
Seeing the comments below, I think my understanding of the concept is supremum and boundedness is not quite right. I am reading functional analysis from Kreyszig's 'Introductory functional analysis'. I don't have a background in real analysis. I will be helpful if someone can answer this questions with some explanation of these concepts.
 A: Indeed, any specific distance $d(x,y)$ must be finite. This follows from the definition of $d$ as a function from $X\times X$ to $\mathbb{R}^+$ ($+\infty \notin \mathbb{R}^+$).
But the diameter of a set $A$ is a supremum of infinitely many real numbers (if $A$ is infinite), namely all distances between the different points from $A$. And a supremum of a subset of $\mathbb{R}^+$ can be $+\infty$: e.g. in the reals, usual metric, if $A = \mathbb{N}$ for $\delta(A)$ we have to find the supremum of $\{d(n,m); n,m \in \mathbb{N}\}$ which contains $\{d(0,n)= n: n \in \mathbb{N}\} = \mathbb{N}$. And $\sup(\mathbb{N}) = +\infty$ as there is no upperbound in the reals for the set $\mathbb{N}$.
So $\delta(A) = +\infty$ is just saying that there is no upperbound $B \in \mathbb{R}$ such that $d(a,a') \le B$ for all $a,a' \in A$. 
A supremum is often not a maximum: if $A = (0,1)$, then $\delta(A) = 1$ even when there are no $x,y \in A$ with $d(x,y) = 1$. But we take the sup of real numbers like $d(\frac{1}{n},1-\frac{1}{n}) = 1-\frac{2}{n}$ which can approach $1$ as closely as you like, taking larger and larger $n$. And it's certainly not a maximum for trivial reasons if $\delta(A) = +\infty$, as we saw.
