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Topology: An Inquiry-Based Approach

Steven Schlicker

Section Metric Equivalence

We have seen that we can make a set into a metric space with different metrics. For example, the spaces \((\R^2, d_E)\text{,}\) \((\R^2, d_T)\text{,}\) \((\R^2, d_M)\text{,}\) and \((\R^2, d)\) are all metric spaces, where \(d_E\) is the Euclidean metric, \(d_T\) the taxicab metric, \(d_M\) the max metric, and \(d\) the discrete metric. But are these metric spaces really “different” metric spaces? What do we mean by “different”?

Activity 14.3.

We might consider two metric spaces \((X, d_X)\) and \((Y, d_Y)\) to be equivalent if we can find a bijection between the two sets \(X\) and \(Y\) that preserves the metric properties. That is, find a bijective function \(f : X \to Y\) such that \(d_X(a,b) = d_Y(f(a), f(b))\) for all \(a,b \in X\text{.}\) In other words, \(f\) preserves distances.

(a)

Let \(X = ((0,1), d_X)\) and \(Y = ((0,2), d_Y)\text{,}\) with both \(d_X\) and \(d_Y\) the Euclidean metric. Is it possible to find a bijection \(f : X \to Y\) that preserves the metric properties? Explain.

(b)

Now let \(X = ((0,1), d_X)\) and \(Y = ((0,2), d_Y)\text{,}\) where \(d_X\) is defined by \(d_X(a,b) = 2 | a-b |\) and \(d_Y = d_E\text{.}\) You may assume that \(d_X\) is a metric. Is it possible to find a bijection \(f : X \to Y\) that preserves the metric properties? Explain.
If there is a bijection between metric spaces that preserves distances, we say that the metric spaces are metrically equivalent.

Definition 14.5.

Two metric spaces \((X,d_X)\) and \((Y,d_Y)\) are metrically equivalent if there is a bijection \(f : X \to Y\) such that
\begin{equation*} d_X(x,y) = d_Y(f(x),f(y)) \end{equation*}
for all \(x,y \in X\text{.}\)
Because \(f\) is a bijection, it will also be the case in Definition 14.5 that
\begin{equation*} d_Y(u,v) = d_X(f^{-1}(u), f^{-1}(v)) \end{equation*}
for all \(u\) and \(v\) in \(Y\text{.}\) The proof is left for Exercise 1.
Any function \(f\) that preserves distances (like the one in Definition 14.5) is called an isometry.

Definition 14.6.

A function \(f\) from a metric space \((X,d_X)\) to a metric space \((Y, d_Y)\) is an isometry if \(f\) is a bijection and
\begin{equation} d_Y(f(a),f(b)) = d_X(a,b)\tag{14.1} \end{equation}
for all \(a, b \in X\text{.}\)
Metric equivalence is a very strong type of equivalence — the existence of an isometry does not allow for much flexibility since distances must be preserved. From a topological perspective, we are only concerned about the open sets — there are no distances. The open unit ball in \((\R^2, d_E)\) and the open ball in \((\R^2, d_M)\) (where \(d_E\) is the Euclidean metric and \(d_M\) is the max metric) are not that different as we can see in Figure 14.7. If we don’t worry about preserving distances, we can stretch the open ball \(B_E = B((0,0),1)\) in \((\R^2, d_E)\) along the lines \(y=x\) and \(y=-x\) uniformly in a way to mold it onto the unit ball \(B_M = B((0,0),1)\) in \((\R^2, d_M)\text{.}\) The important thing is that this stretching will preserve the open sets. This is a much more forgiving type of equivalence and maintains the central idea of topology that we have discussed — what properties of a space are not altered by stretching and bending the space. This type of equivalence that allows us to manipulate a space without fundamentally changing the open sets is called topological equivalence.
Figure 14.7. The open unit balls in \((\R^2, d_E)\) and \((\R^2, d_M)\text{.}\)
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