By Julian Lowell Coolidge

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If we have V(x, y) 2 0, then V(g(x),g(y)) 2 E for some E > 0 since V(g(x), g(y)) > V(x, y). Since T2 is compact, there exists 6 > 0 such that for u, v E T2 d(u, v) < 6 =$ V(S(U),g(v)) < E, from which we have d(x, y) 6. Since V(g2(x),g2(y)) > V ( g ( ~ ) , g ( ~ ) ) E , we have d(g(x), g(y)) 2 6 and by induction > > Define Y = min{V(g(u), g(v)) - V(U,v) : (u, v ) E T2 satisfying 6 5 d(u, v) Ie}. Then we have and thus V(gn(s), gn(y)) > nu + V(x, Y) L nv, n L 0. However, we have max{V(u,v) : (u,v) E T2}< 00,thus contradicting.

Every linearly ordered set of 0 has the least element. Thus 0 has a minimum element by Zorn's lemma. This element is a minimal set with respect to f . 26. There exists no expansive homeomorphism of the unit circle S' . Proof, Suppose f : S' -, S1 is expansive. 23, f has no periodic points. Indeed, suppose f has distinct periodic points x and y. Then fk(x) = x and fk((y = y for some L > 0. For simplicity put g = f k . There exists a closed arc C in S1 such that g(C) = C, or g2(C) = C. If g(C) = C, then g p : C -, C is expansive.

Thus we have E = X. Conversely, for x E X,cl(Of (3)) is a closed f-invariant set and thus cl(Of(x)) = X. Figure 4 A closed subset E which is f-invariant is called a minimal set with respect to f if f l E is minimal. 25. Every homeomorphism f : X -, X has a minimal set. Proof. Let 0 denote the collection of all closed nonempty f-invariant subsets of X. 0 is a partially ordered set under the inclusion. Every linearly ordered set of 0 has the least element. Thus 0 has a minimum element by Zorn's lemma.

### A History of Geometrical Methods by Julian Lowell Coolidge

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