Metric Space Completeness is not Preserved by Homeomorphism

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Theorem

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $\phi: M_1 \to M_2$ be a homeomorphism.


If $M_1$ is complete then it is not necessarily the case that so is $M_2$.


Proof

Let $\Z_{>0}$ be the set of (strictly) positive integers.

Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.

Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the metric on $\Z_{>0}$ defined as:

$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$

Let $\tau_d$ denote the metric topology for $d$.

Let $\tau_\delta$ denote the metric topology for $\delta$.


From Topologies induced by Usual Metric and Scaled Euclidean Metric on Positive Integers are Homeomorphic:

$\tau_d$ and $\tau_\delta$ are homeomorphic.


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From Positive Integers under Usual Metric is Complete Metric Space:

$\struct {\Z_{>0}, d}$ is a complete metric space.

However, from Positive Integers under Scaled Euclidean Metric is not Complete Metric Space:

$\struct {\Z_{>0}, \delta}$ is not a complete metric space.


Hence the result.

$\blacksquare$


Sources

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