On Expansive Mappings

When finding an original proof to a known result describing expansive mappings on compact metric spaces as surjective isometries, we reveal that relaxing the condition of compactness to total boundedness preserves the isometry property and nearly so that of surjectivity. While a counterexample is found showing that the converse to the above descriptions do not hold, we are able to characterize boundedness in terms of specific expansions we call anticontractions.


Introduction
We take a close look at the nature of expansive mappings on certain metric spaces (compact, totally bounded, and bounded), provide a finer classification for such mappings, and use them to characterize boundedness.
When finding an original proof to a known result describing all expansive mappings on compact metric spaces as surjective isometries [1, Problem X.5.13 * ], we reveal that relaxing the condition of compactness to total boundedness still preserves the isometry property and nearly so that of surjectivity.
We provide a counterexample of a not totally bounded metric space, on which the only expansion is the identity mapping, demonstrating that the converse to the above descriptions do not hold.
Various examples for different types of expansions are furnished, in particular the one of a nonsurjective expansion on a totally bounded "dial set" in the complex plane which allows to better understand the essence of the latter.
(1) An expansion T : X → X such that ∀ x, y ∈ X : d(T x, T y) = d(x, y) is called an isometry, which is the weakest form of expansive mappings.
(2) An expansion T : X → X such that we call a proper expansion.
we call a strict expansion.
(4) Finally, an expansion T : X → X such that we call an anticontraction with expansion constant E.
Remark 3.1. Clearly, any anticontraction is necessarily a strict expansion, which in turn is also a proper expansion. However, as the following examples demonstrate, the converse statements are not true. 1. On C with the standard metric, the mapping i.e., the counterclockwise rotation by one radian, is an isometry which is not a proper expansion.
2. On the space ℓ ∞ of all real-or complex-termed bounded sequences with its standard supremum metric the right shift mapping is also an isometry which is not a proper expansion.
3. On ℓ ∞ , the mapping  ( χ · (·) is the characteristic function of a set), which is a subset of the unit sphere For any m, n ∈ N with n > m, we have: The map T f n := f kn , n ∈ N, with an arbitrary fixed k ∈ N is an isometry on {f n } n∈N since, for any m, n ∈ N with n > m, On the other hand, the map Sf n := f n 2 , n ∈ N, is a strict expansion on {f n } n∈N since, for any m, n ∈ N with n > m, which is not an anticontraction since 5. On R with the standard metric, the mapping is an anticontraction with expansion constant E = 2. However, the same mapping, when considered on R equipped with the metric turning R into a bounded space (see, e.g., [3]), is merely a strict expansion, which is not an anticontraction since Proof. For an arbitrary point x ∈ X, and an increasing sequence (n(k)) k∈N of natural numbers, consider the sequence

Expansions on Compact Metric Spaces
Since the space (X, d) is compact, there exists a convergent subsequence x n(k(j)) j∈N , which is necessarily fundamental.
Remark 4.1. Subsequently, we use only the fundamentality, and not the convergence of the subsequence, and hence, only the total boundedness and not the compactness of the underlying space (Remark 2.2).
By the fundamentality of x n(k(j)) j∈N , without loss of generality, we can regard the indices n(k(j)), j ∈ N, chosen sparsely enough so that Since T is an expansion, We thus conclude that T (X) = X.
Now, let x, y ∈ X be arbitrary. Then, for the sequence (x n := T n x) n∈N , we can, by the above argument, select a subsequence x n(k) k∈N such that and then, in turn, for the sequence y n(k) := T n(k) y k∈N , we choose a subsequence y n(k(j)) j∈N for which y n(k(j)) → y, j → ∞.
Since x n(k(j)) j∈N is a subsequence of x n(k) k∈N , we also have: Then, in view of the expansiveness of T , for any j ∈ N, x, T n(k(j)) y) = d(x n(k(j)) , y n(k(j)) ).
Whence, passing to the limit as j → ∞, by joint continuity of metric, we arrive at which implies that ∀ x, y ∈ X : d(T x, T y) = d(x, y), i.e., T is an isometry. Being an isometry, the mapping T is continuous, whence, since X is compact, we infer that the image T (X) is compact as well, and therefore closed in (X, d) (see, e.g., [3]).
In view of the denseness and the closedness of T (X), we conclude that i.e., T is also a surjection, as desired, which completes the proof. Remark 4.3. For the surjectivity of T , the requirement of the compactness of the underlying space is essential, as we rely on the fact the continuous image of a compact set is compact. Example 5.1 demonstrates that this requirement cannot be relaxed even to total boundedness.

Expansions on Totally Bounded Metric Spaces
We proceed now to demonstrate that relaxing the condition of the compactness of the underlying space to total boundedness yields a slightly weaker result, in which expansions emerge as "presurjective" isometries.
Proof. As is shown in the corresponding part of the proof of Theorem 4.1 (see Remarks 4.1 and 4.2), the image T (X) is dense in (X, d), i.e., T (X) = X, and T is an isometry.
As is mentioned in Remark 4.3, the compactness of the underlying space is essential for the surjectivity of expensions, the following example demonstrating that, when compactness is relaxed to total boundedness, surjectivity is not guaranteed.

Example 5.1 (Dial Set). Let
D := e in n∈Z+ ⊂ {z ∈ C | |z| = 1} (Z + is the set of nonnegative integers) be a dial set in the complex plane C with the usual distance, which is bounded in C, and hence, totally bounded (see, e.g., [3]), and D ∋ e in → T e in := e i(n+1) ∈ D, n ∈ Z + , be the counterclockwise rotation by one radian, which is, clearly, an isometry (see Examples 3.1) but not a surjection on D since, as is easily seen, Remarks 5.1.
• This, in particular, implies that, by Theorem 4.1, the dial set D is not compact, and hence, not closed, in C (see, e.g., [3]).
• Thus, on a totally bounded, in particular compact, metric space, any expansion is not proper but is an isometry which may fall a little short of being surjective.
By Theorem 5.1, the range T (D) is dense in the dial set D, which is not closed, relative to the usual distance. This allows us to "turn the tables" on the dial set and derive the following rather interesting immediate corollary. Then, (1) for an arbitrary n ∈ Z + , there exists an increasing sequence (n(k)) k∈N of natural numbers such that e in(k) → e in , k → ∞; (2) there exists a θ ∈ R\Z + for which there is an increasing sequence (n(k)) k∈N of natural numbers such that e in(k) → e iθ , k → ∞. Proof.
(1) Part (1) immediately follows from the fact that, by Theorem 5.1, the range T (D) = e in n∈N is dense in D.
(2) Part (2) follows from the fact that the set D, being not closed (see Remarks 5.1), has at least one limit point not belonging to D, which, by continuity of metric, is located on the unit circle {z ∈ C | |z| = 1}, i.e., is of the form e iθ with some θ ∈ R \ Z + .
Remark 5.2. If posed as a problem, the prior statement, although simply stated, might be quite challenging to be proved exclusively via the techniques of classical analysis.
6. Are the Converse Statements True?
Now, there are two natural questions to ask.
• If every expansive map T on a metric space (X, d) is a surjective isometry, is the space compact?
• If every expansive map T on a metric space (X, d) is a presurjective isometry (see Theorem 5.1), is the space totally bounded?
In other words, do the converse statements to Theorems 4.1 and 5.1 hold?
The following example answers both questions in the negative. Then T x 1 = x k with some k ∈ N, k ≥ 2. Let n ∈ N, n ≥ 2, be arbitrary.
There are two possibilities: either In the first case, we have: contradicting the expansiveness of T .
In the second case, for any m ∈ N, m = n, by the injectivity of T , and hence, which again contradicts the expansiveness of T .
The obtained contradictions making assumption (6.1) false, we conclude that Therefore, by the injectivity of T , we can restrict the expansion T to the subset {x n } n≥2 . Applying the same argument, one can show that Continuing inductively, we see that ∀ n ∈ N : T x n = x n , i.e. T is the identity map, which is both a surjection and an isometry, even though the set {x n } n∈N is not totally bounded, let alone compact (see Remark 2.2), as ∀ m, n ∈ N, m = n : d ∞ (x m , x n ) > 1.
Remark 6.1. Thus, a metric space with the property that every expansion on it is a presurjective isometry need not be totally bounded. Such spaces, which, by Theorems 4.1 and 5.1, encompass compact and totally bounded, can be called nonexpansive.

A Characterization of Boundedness
Although bounded sets support strict expansions (see Examples 3. 1 4, 5). Any attempt to produce an anticontraction on a bounded set would be futile, the following characterization explaining why.
Theorem 7.1 (Anticontraction Characterization of Boundedness). A metric space (X, d) is bounded iff no subset of X supports an anticontraction.
Proof. The case of a singleton being trivial, suppose that X consists of at least two distinct elements.
"Only if" part. We proceed by contradiction, assuming that X is bounded and there exists a subset A ⊆ X supporting an anticontraction T : A → A with expansion constant E. Then ∀ x, y ∈ A, x = y ∀ n ∈ N : T n x, T n y ∈ A, which implies Hence, A is unbounded, and since A ⊆ X, this contradicts the boundedness of X, the obtained contradiction proving the "only if " part.
"If" part. Here, we proceed by contrapositive assuming X to be unbounded and showing that there exists a subset of X which supports an anti-contraction.
Since X is unbounded, we can select two distinct points x 1 , x 2 ∈ X, and subsequently pick x 3 so that min 1≤i≤2 d(x 3 , x i ) > 2 max S ∋ x n → T x n := x n+1 ∈ S, n ∈ N.
Then, for any m, n ∈ N with n > m, d(T x n , T x m ) = d(x n+1 , x m+1 ) ≥ min 1≤i≤n d(x n+1 , x i ) > 2 max 1≤i,j≤n d(x i , x j ) ≥ 2d(x n , x m ), which implies that T is an anti-contraction with expansion constant E = 2 on S ⊆ X completing the proof of the "if " part and the entire statement.
Reformulating equivalently, we arrive at Theorem 7.2 (Anticontraction Characterization of Unboundedness). A metric space (X, d) is unbounded iff there exists a subset of X which supports an anticontraction.

Acknowledgments
The authors would like to express their appreciation to Drs. Michael Bishop, Przemyslaw Kajetanowicz, and other members of the Functional Analysis and Mathematical Physics Interdepartmental Research Group (FAMP) of California State University, Fresno for insightful questions and stimulating discussions.