1.5 Equivalent Metrics
In the previous examples, we have seen that it is possible to have more than one metric defined on a set. It is desirable to have some notion of equivalence of two metrics defined on the same set. Before we give the definition of equivalent metrics, we start by defining a convergent sequence in a metric space.
Definition 1.5.1. A sequence \((x_n)\) in a metric space \((X,d)\) is said to converge to a point \(x\in X\) if for every \(\epsilon >0\) there exists \(n_0\in \mathbb {N}\) such that \[d(x_n,x)<\epsilon \quad \text {for all } n\geq n_0.\] The point \(x\) is called the limit of \((x_n)\), and we write \(\lim \limits _{n\rightarrow \infty } x_n=x\) or \(x_n\rightarrow x\) as \(n\rightarrow \infty \).
Remark. Equivalently, \(x_n\rightarrow x\) in \((X,d)\) if and only if \(d(x_n,x)\rightarrow 0\) as a sequence of real numbers, so convergence in any metric space reduces to convergence in \(\mathbb {R}\).
A sequence has at most one limit, which is what justifies calling \(x\) the limit. If \(x_n\rightarrow x\) and \(x_n\rightarrow y\), then for every \(n\), \[0\leq d(x,y)\leq d(x,x_n)+d(x_n,y),\] and the right-hand side tends to \(0\). Hence \(d(x,y)=0\), and so \(x=y\).
Definition 1.5.2. Two metrics \(d_1\) and \(d_2\) defined on a set \(X\) are said to be equivalent if there exist two real numbers \(\alpha >0\) and \(\beta >0\) such that for all \(x,y\in X \), \[\alpha d_1(x,y)\leq d_2(x,y)\leq \beta d_1(x,y).\]
Theorem 1.5.3. Let \(d_1\) and \(d_2\) be metrics defined on a set \(X\). If \(d_1\) and \(d_2\) are equivalent, then every sequence \((x_n)\) in \(X\) that converges to \(x\) with respect to \(d_1\) will also converge to \(x\) with respect to \(d_2\).
Proof. Suppose that \(d_1\) and \(d_2\) are equivalent. Then there exist real numbers \(\alpha >0\) and \(\beta >0\) such that \(\alpha d_1(x,y)\leq d_2(x,y)\leq \beta d_1(x,y)\) for all \(x,y\in X\). If \((x_n)\) is a sequence in \(X\) which converges to x with respect to \(d_1\), then \(d_1(x_n,x)\rightarrow 0\) as \(n\rightarrow \infty \). It follows that \(d_2(x_n,x)\leq \beta d_1(x_n,x)\rightarrow 0\) as \(n\rightarrow \infty \), so that \((x_n)\) converges to \(x\) with respect to \(d_2\). □
Note that the contrapositive of the previous theorem says that if a sequence \((x_n)\) in \(X\) does not converge to the same limit with respects to \(d_1\) and \(d_2\) are not equivalent.
Example 1.5.4. The metrics \(d_1(x,y)=\sqrt {(x_1-y_1)^2+(x_2-y_2)^2}\),
\(d_2(x,y)=|x_1-y_1|+|x_2-y_2|\) and \(d_3=\max \{|x_1-y_1|,|x_2-y_2|\}\) are equivalent.
Proof. The equivalence of \(d_1\) and \(d_3\) follows from \(d_3(x,y)\leq d_1(x,y)\leq \sqrt {2}d_3(x,y)\).
The equivalence of \(d_2\) and \(d_3\) follows from \(d_3(x,y)\leq d_2(x,y)\leq 2d_3(x,y)\).
The equivalence of \(d_1\) and \(d_2\) follows from \(d_1(x,y)\leq d_2(x,y)\leq \sqrt {2}d_1(x,y)\). □
Example 1.5.5. Let \(C\) be the set of all convergent sequences of real numbers and \(d_1\) and \(d_2\) be metrics on \(C\) defined as follows for \(x=(x_n)\) and \(y=(y_n)\) in \(C\). \(d_1(x,y)=\sup \limits _{n\geq 1} |x_n-y_n|\) and \(d_2(x,y)=\sum \limits ^{\infty }_{n=1}\dfrac {1}{2^n}\,\dfrac {|x_n-y_n|}{1+|x_n-y_n|}\). Then \(d_1\) and \(d_2\) are not equivalent.
Proof. Consider the sequences \(e_1=(1,0,0,\ldots )\), \(e_2=(0,1,0,\ldots )\) and, in general, \(e_n=(0,\ldots ,0,1,0,0,\ldots )\) with 1 in the \(n^{th}\) place and zeroes elsewhere. Take \(e_0=(0,0,\ldots )\). Then \(d_1(e_n,e_0)=1\) for all \(n\in \mathbb {N}\), so \(d_1(e_n,e_0)\) does not converge to 0 as \(n\rightarrow \infty \). This means that the sequence \((e_n)\) does not converge to \(e_0\) with respect to \(d_1\). With respect to \(d_2\), only the \(n^{th}\) term of the sum is non-zero, so \[d_2(e_n,e_0)=\frac {1}{2^n}\,\frac {1}{1+1}=\frac {1}{2^{n+1}} \rightarrow 0 \quad \text {as } n\rightarrow \infty .\] Thus the sequence \((e_n)\) converges to \(e_0\) with respect to \(d_2\). By the previous theorem \(d_1\) and \(d_2\) are not equivalent. □
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