Definition 14.8.
Two topological spaces \((X,d_X)\) and \((Y,d_Y)\) are topologically equivalent if there is a continuous bijection \(f : X \to Y\) such that \(f^{-1}\) is also continuous.
Section Topological Equivalence
When we can deform one set into another without poking holes in the set, we consider the two sets to be equivalent from a topological perspective. Such a deformation \(f\) has to be a bijection to ensure that the two sets contain the same number of elements, continuous so that the inverse images of open sets are open, and \(f^{-1}\) must be continuous so images of open sets are open. Such a function provides a one-to-one correspondence between open sets in the two spaces. This leads to the next definition.
Metric equivalence always implies topological equivalence (using the metric topologies), which is left for Exercise 3. So metric equivalence is a stronger condition than topological equivalence.
Definition 14.9.
Let \((X,\tau_X)\) and \((Y,\tau_Y)\) be topological spaces. A function \(f: X \to Y\) is a homeomorphism if \(f\) is a continuous bijection such that \(f^{-1}\) is also continuous.
If there is a homeomorphism from \((X,\tau_X)\) to \((Y,\tau_Y)\) we say that the spaces \((X,\tau_X)\) to \((Y,\tau_Y)\) are homeomorphic topological spaces.
It can be difficult to show directly that two metric spaces are homeomorphic, but there are ways to make the process easier in metric spaces. If \(f\) is a homeomorphism from the metric space \((\R^2, d_E)\) to the metric space \((\R^2, d_M)\text{,}\) the continuity of \(f\) ensures a smooth deformation from \(\R^2\) to \(\R^2\text{.}\) In terms of the metrics, this means that distances cannot get distorted too much — in fact, the amount distances are distorted should be bounded. In other words, we might expect that there is a constant \(K\) so that \(d_E(x,y) \leq K d_M(f(x),
f(y))\) for any \(x, y \in \R^2\text{.}\) The next theorem tells us that this is a sufficient condition for topological equivalence when we work in the same underlying space.
Theorem 14.10.
Let \(X\) be a set on which two metrics \(d\) and \(d'\) are defined. If there exist positive constants \(K\) and \(K'\) so that
\begin{align*}
d'(x,y) \amp \leq K d(x,y)\\
d(x,y) \amp \leq K' d'(x,y)
\end{align*}
for all \(x,y \in X\text{,}\) then \((X,d)\) is topologically equivalent to \((X,d')\text{.}\)
Proof.
Let \(X\) be a set on which two metrics \(d\) and \(d'\) are defined. Suppose there exist positive constants \(K\) and \(K'\) so that
\begin{align*}
d'(x,y) \amp \leq K d(x,y)\\
d(x,y) \amp \leq K' d'(x,y)
\end{align*}
for all \(x,y \in X\text{.}\) Let \(i_X : (X,d) \to (X,d')\) be the identity mapping. That is, \(i_X(x)=x\) for all \(x \in X\text{.}\) We will prove that \(i_X\) is a homeomorphism. We know that \(i_X\) is a bijection, so we only need verify that \(i_X\) and \(i_X^{-1}\) are continuous. Let \(\epsilon \gt 0\) be given, and let \(a \in X\text{.}\) Let \(\delta = \frac{\epsilon}{K}\text{.}\) Suppose \(x \in X\) so that \(d(x,a) \lt \delta\text{.}\) Then
\begin{equation*}
d'(i_X(x), i_X(a)) = d'(x,a) \leq Kd(x,a) \lt K\delta = K\left(\frac{\epsilon}{K}\right) = \epsilon\text{.}
\end{equation*}
Thus, \(i_X\) is continuous. The same argument shows that \(i_X^{-1}\) is also continuous. Therefore, \(i_X\) is a homeomorphism between \((X,d)\) and \((X,d')\text{.}\)
Activity 14.4.
(a)
Are \((\R^2,d_T)\) and \((\R^2, d_M)\) topologically equivalent? Explain.
(b)
Are \((\R^2,d_E)\) and \((\R^2, d_T)\) topologically equivalent? Explain.
(c)
Do you expect that \((\R^2,d_E)\) and \((\R^2, d_M)\) are topologically equivalent. Explain without doing any calculations or comparisons.
