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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 fβˆ’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.

Definition 14.8.

Two topological spaces (X,dX) and (Y,dY) are topologically equivalent if there is a continuous bijection f:Xβ†’Y such that fβˆ’1 is also continuous.
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.
The function f (or fβˆ’1) in Definition 14.8 is called a homeomorphism.

Definition 14.9.

Let (X,Ο„X) and (Y,Ο„Y) be topological spaces. A function f:Xβ†’Y is a homeomorphism if f is a continuous bijection such that fβˆ’1 is also continuous.
If there is a homeomorphism from (X,Ο„X) to (Y,Ο„Y) we say that the spaces (X,Ο„X) to (Y,Ο„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 (R2,dE) to the metric space (R2,dM), the continuity of f ensures a smooth deformation from R2 to R2. In terms of the metrics, this means that distances cannot get distorted too much β€” in fact, the amount distances are distorted should be bounded. In other words, we might expect that there is a constant K so that dE(x,y)≀KdM(f(x),f(y)) for any x,y∈R2. The next theorem tells us that this is a sufficient condition for topological equivalence when we work in the same underlying space.

Proof.

Let X be a set on which two metrics d and dβ€² are defined. Suppose there exist positive constants K and Kβ€² so that
dβ€²(x,y)≀Kd(x,y)d(x,y)≀Kβ€²dβ€²(x,y)
for all x,y∈X. Let iX:(X,d)β†’(X,dβ€²) be the identity mapping. That is, iX(x)=x for all x∈X. We will prove that iX is a homeomorphism. We know that iX is a bijection, so we only need verify that iX and iXβˆ’1 are continuous. Let Ο΅>0 be given, and let a∈X. Let Ξ΄=Ο΅K. Suppose x∈X so that d(x,a)<Ξ΄. Then
dβ€²(iX(x),iX(a))=dβ€²(x,a)≀Kd(x,a)<KΞ΄=K(Ο΅K)=Ο΅.
Thus, iX is continuous. The same argument shows that iXβˆ’1 is also continuous. Therefore, iX is a homeomorphism between (X,d) and (X,dβ€²).

Activity 14.4.

(c)

Do you expect that (R2,dE) and (R2,dM) are topologically equivalent. Explain without doing any calculations or comparisons.