A metric space is an ordered pair \((X, d)\) where \(X\) is a set and \(d\) is a metric on \(X\), i.e., a function \[ d:X\times X \rightarrow \mathbb{R} \] satisfying the following axioms for all \(p,q,r\in X\):
(Nonnegativity): \(d(p,q)\ge 0\). Moreover, \(d(p,p)=0\) for all \(p\), and \(d(p,q)>0\) if \(p\ne q\).
(Symmetry): \(d(p,q)=d(q,p)\) for all \(p,q\).
(Triangle inequality): \(d(p,q)\le d(p,r)+d(r,q)\).
A metric space \((X,d)\) is bounded if there exists a real number \(B\) such that \(d(p,q)<B\) for all \(p,q\in X\).
Let \((X,d)\) be a metric space, and let \(E\) be a nonempty subset of \(X\). The followings are equivalent:
\((E,d)\), considered as a sub-metric space of \(X\), is bounded;
There exists \(p\in E\) and a real number \(M\) such that \(d(p,q)<M\) for all \(q\in E\);
There exists \(p\in X\) and a real number \(M\) such that \(d(p,q)<M\) for all \(q\in E\).
Proof: We will show \((1)\implies (2)\implies (3)\implies (1)\).
\((1)\implies (2)\) is clear.
\((2)\implies (3)\) can be showed by choosing the same \(M, p\) (you may look at the word exist).
\((3)\implies (1)\): Let \(B=2M\). For \(q,q'\in E\), we have \(d(q,q')\le d(p,q)+d(p,q')<M+M=2M=B\).
Finite metric space is bounded.
If \((X,d)\) is bounded, then so is any subspace.
If \((X,d_X)\), \((Y,d_Y)\) are bounded, so is \((X\times Y, d_{X\times Y})\), where for any \(x,x'\in X\) and \(y,y'\in Y\), \[ d_{X\times Y}((x,y),(x',y')):=(d_X(x,x'), d_Y(y,y')) \]
The metric space \((\mathbb{R},d)\) is not bounded.
Compare(cf.)
An inner product space is a vector space \(V\) over \(\mathbb{R}\) together with an inner product, that is a map \[\left<\cdot,\cdot\right>:V\times V\rightarrow \mathbb{R}\] which satisfies the following three properties for all \(\mathbf{x,y,z}\in V\) and \(a,b\in\mathbb{R}\) :
(Symmetry): \(\left<\mathbf{x},\mathbf{y}\right>=\left<\mathbf{y},\mathbf{x}\right>\);
(Linearity): \(\left<a\mathbf{x}+b\mathbf{y},\mathbf{z}\right>=a\left<\mathbf{x},\mathbf{z}\right>+b\left<\mathbf{y},\mathbf{z}\right>\);
(Positive-definiteness): If \(\mathbf{x}\ne \mathbf{0}\), then \(\left<\mathbf{x},\mathbf{x}\right>>0\).