1.5 Equivalent Metrics
In the previous examples, we have seen that it is possible to have more than on 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 given \(\epsilon >0\),
there exists a natural number \(n_0\) such that \(d(x_n,x)<\epsilon \) for all \(n\geq n_0\).
The point \(x\) is said to be the limit of the sequence \((x_n)\) and \((x_n)\) is said to converge to \(x\) as \(n\) approaches
infinity. This is denoted by \(\lim \limits _{n\rightarrow \infty } x_n=x\) or \(x_n \rightarrow x\) as \(n\rightarrow \infty \).
From the definition, it is clear that \(x_n \rightarrow x\) as \(n\rightarrow \infty \) if and only if \(d(x_n,x)\rightarrow 0\) as \(n\rightarrow \infty \).
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.
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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 \(d_1(x,y)\leq d_2(x,y)\leq \sqrt {2}d_1(x,y)\).
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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 \(e_1=(1,0,0,......),e_2=(0,1,0,......),................,\\e_n=(0,0,0,........,1,0,0,..........)\) with 1 in \(n^{th}\) place and zeroes elsewhere.
Take \(e_0=(0,0,.......)\). Then \(d_1(e_n,e_0)=1\) for all \(n\in \mathbb {N}\) and 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 with respect to \(d_1\), with respect to \(d_2\) we have
\(d_2(e_n,e_0)=\dfrac {1}{2^n}\,\dfrac {1}{1+1}=\dfrac {1}{2^{n+1}} \rightarrow 0\) as \(n\rightarrow \infty \). Thus the sequence \((e_n)\) converges to 0 with respect to \(d_2\). By the previous theorem \(d_1\) and \(d_2\) are not
equivalent.
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