1 Introduction

Let \(\ G\ \) be a group, \(\ X\ \) an arbitrary set. A map \(\ F: G\rightarrow 2^X\ \) is called subadditive if

$$\begin{aligned} F(gh) \subset F(g)\cup F(h) \qquad \text { for all }g,h\in G. \end{aligned}$$
(1.1)

This notion was introduced in [5, Theorem 2.2], where the following result was proved:

Theorem 1.1

([5, Theorem 2.2]) Let \(\ F\) be a subadditive map from a group G into the power set \(2^X\) of a set X. Let

$$\begin{aligned} \nu (F):= \sup _{g\in G} |F(g)|, \end{aligned}$$
(1.2)

where |A| denotes the cardinality of a set A. If \(\ \nu (F)<\infty \) then the cardinality of the global image \(F(G) = \cup _{g\in G}F(g)\) is finite and admits the estimate

$$\begin{aligned} |F(G)| < 4\nu (F). \end{aligned}$$
(1.3)

Moreover, if one assumes additionally that \(F(g)=F(g^{-1})\), for each \(g\in G\), then

$$\begin{aligned} |\cup _{g\in G}F(g)| < 2\nu (F). \end{aligned}$$
(1.4)

A trivial example of a subadditive map is the map with constant values \(F(g) = X\), for any \(g\in G\); the above result shows that every subadditive map with finite values is majorized by a trivial subadditive map with finite image. Among other examples the most immediate is the following one: \(\ G\) is a free group, X is the set of generators, and F(g), for each \(g\in G\), is the set of all \(x\in X\) that participate in the word representing g. With a minor modification this construction can be extended to an arbitrary group with a fixed set of generators.

Special examples of subadditive maps were constructed and used by the author in the study of multivariable Levi-Civita equations on a group [5]. In this case the role of X is played by the spectrum (the set of all irreducible representations up to the unitary equivalence) of G.

Another class of examples arises as follows. Let X be a set of subgroups of a group G. For any \(g\in G\), set

$$\begin{aligned} F(g) = \{H\in X: g\notin H\}. \end{aligned}$$
(1.5)

Then it is not difficult to show that the map \(F: G\rightarrow X\) is subadditive.

Applying Theorem 1.1 to this situation one can obtain the following statement.

Proposition 1.2

If \(\ H_1, H_2, \ \ldots , \ H_k\) are proper subgroups of G, then there is an element \(g\in G\) that belongs to fewer than k/2 of these subgroups.

An extension of Theorem 1.1, also presented in [5], provides a similar estimate for coverings of a group by semigroups.

Proposition 1.3

If a group is covered by k proper subsemigroups then there is an element belonging to fewer than 3k/4 of them.

In the literature (see e.g. [3, 6]) the term “subadditive mapping” is usually used for maps between groups, satisfying the condition

$$\begin{aligned} F(gh)\subset F(g)+F(h). \end{aligned}$$

This notion (as well as the results of the works listed above) is not directly related to our one, but both of them can be considered as special cases of the general notion of a subadditive map from a group to a lattice. Another special case of this general concept is the notion of a subadditive map from a group to the lattice of subspaces of a Banach space. In this case subadditivity means that

$$\begin{aligned} F(gh) \subset F(g)+F(h), \;\;\; \text {for all }\;g,h \in G. \end{aligned}$$

A version of Theorem 1.1 for this case gives, under some additional assumptions, an estimation of the dimension of \(\sum _{g\in G}F(g)\) via \(\sup _{g\in G} \dim (F(g))\); for the precise formulation see [5, Theorem 4.2].

Let us return to our definition. The above result on subadditive maps with finite values (Theorem 1.1) was extended in [5, Theorem 3.1] to subadditive set-valued maps from a group G to a measure space \((X,\mu )\) – with an estimate of \(\mu (\cup _{g\in G}F(g))\) via \(\sup _{g\in G} \mu (F(g))\).

It was shown in [5, Theorem 3.2] that the described phenomenon possesses a kind of stability resembling the Hyers-Ulam stability (see [1, 2] for the general discussion of the notion): if (1.1) holds “up to a set of small measure” then the estimate of \(\mu (\cup _{g\in G}F(g))\) also holds with a small correction.

Here we will consider set-valued maps from groups to metric spaces and prove a metric version of the stability.

2 Formulation of the main results

Let (X, d) be a metric space. For \(Y\subset X\) and \(\delta >0\), we denote by \(O_{\delta }(Y)\) the \(\delta \)-neighborhood of Y:

$$\begin{aligned} O_{\delta }(Y) = \{x\in X: d(x,Y)<\delta \}. \end{aligned}$$

A map \(F: G\rightarrow 2^X\) is called \(\delta \)-subadditive if

$$\begin{aligned} F(gh) \subset O_{\delta }(F(g)\cup F(h)) \qquad \text { for all }g,h\in G. \end{aligned}$$
(2.1)

Our aim is to prove the following statement:

Theorem 2.1

For any \(n\in \mathbb {N}\) and \(\varepsilon >0\), there is a \(\delta = \delta (n,\varepsilon )\) such that if \(F: G\rightarrow 2^X\) is \(\delta \)-subadditive and \(\nu (F) \le n\), then

$$\begin{aligned} \cup _{g\in G}F(g) \subset O_{\varepsilon }(P), \end{aligned}$$
(2.2)

where P is a subset of X with \(|P|\le n(n+3)/2\). Moreover, one can take \(\delta = \varepsilon /(3\cdot 6^{n-1}-1).\)

To illustrate this result, let us look at a very simple case: \(G = \mathbb {Z}\), \(\ X = \mathbb {R}\) and \(\nu (F) = 1\). This means that \(F(n) = \{a_n\}\), for any \(n\in \mathbb {Z}\), where \(a_n\in \mathbb {R}\). The condition of \(\delta \)-subadditivity takes the form

$$\begin{aligned} \min \{|a_{m+n}-a_m|, |a_{m+n}-a_n|\} < \delta , \text { for all } m,n\in \mathbb {Z}. \end{aligned}$$
(2.3)

So Theorem 2.1 implies that, for every two-sided sequence \(\{a_n\}_{n\in \mathbb {Z}}\) satisfying (2.3), there are numbers \(a,b\in \mathbb {R}\) such that

$$\begin{aligned} \min \{|a_n - a|, |a_n - b|\} < 2\delta , \qquad \forall n\in \mathbb {Z}. \end{aligned}$$
(2.4)

Note that the same statement for one-sided sequences fails. To see this it suffices to consider the sequence \(a_n = \log n\): it satisfies (2.3) with \(\delta = \log 2\) but nothing similar to (2.4) because \(a_n\rightarrow \infty \).

Modifying this example one can show that Theorem 2.1,unlike [5, Theorem 2.2], by no means extends even to the simplest of semigroups (that are not groups). In fact if \(G = \mathbb {N}\) then, for any \(\delta >0\), one can find a \(\delta \)-subadditive map \(F: G \rightarrow \mathbb {R}\) with \(\nu (F) = 1\) (e.g. \(F(n) = \{\delta \log n\}\)) such that the set \(\cup _{g\in G}F(g)\) is unbounded and therefore not contained in a neighborhood of a finite set.

The proof of Theorem 2.1 will be obtained as a consequence of a more general (and somewhat more technical) statement.

Given a subset \(S\subset X\) and a number \(\delta >0\), we call a map \(F: G \rightarrow 2^{X} (S,\delta )\)-subadditive if

$$\begin{aligned} F(gh) \subset O_{\delta }(F(g)\cup F(h)\cup S) \qquad \text { for all }g,h\in G. \end{aligned}$$
(2.5)

Theorem 2.2

For any \(n\in \mathbb {N}\) and any \(\delta > 0\), there is \(k_n > 0\) such that if \(F: G\rightarrow 2^X\) is an \((S,\delta )\)-subadditive map and \(\nu (F) \le n\), then

$$\begin{aligned} \cup _{g\in G}F(g) \subset O_{k_n \delta }(S\cup P) \end{aligned}$$
(2.6)

where P is a subset of X with \(|P|\le n(n+3)/2\). Moreover, one can take \(k_n = 3\cdot 6^{n-1}-1.\)

3 Proofs

Our aim is to prove Theorem 2.2 so in what follows we assume that S is a fixed subset of a metric space (X, d) and \(F: G\rightarrow 2^X\) is an \((S,\delta )\)-subadditive map with \(\nu (F)<\infty \).

As usual we denote by d(x, A) the distance from \(x\in X\) and \(A\subset X\):

$$\begin{aligned} d(x,A) = \inf \{d(x,y): y\in A\}. \end{aligned}$$

Furthermore

$$d(A_1,A_2):= \inf \{d(x_1,x_2): x_i\in A_i\},$$

for any \( A_1,A_2\subset X.\)

For \(g\in G\), we say that the set F(g) is \((S,\delta )\)-big if

  1. a)

    \(|F(g)| = \nu (F)\),

  2. b)

    \(F(g)\cap O_{\delta }(S) = \emptyset \)

    and

  3. c)

    \(d(x,y) > 2\delta \), for all \(x,y \in F(g)\), (\(x\ne y\)).

Let us say that sets \(F(g_1)\) and \(F(g_2)\) are \((S,\delta )\)-far if

$$\begin{aligned} d(x_1,x_2) > 2\delta , \text { for all }x_i\in F(g_i)\setminus O_{\delta }(S), \qquad (i=1,2). \end{aligned}$$

In other words \(F(g_1)\) and \(F(g_2)\) are \((S,\delta )\)-far if

$$\begin{aligned} d(F(g_1)\setminus O_{\delta }(S),F(g_2)\setminus O_{\delta }(S))>2\delta . \end{aligned}$$

Lemma 3.1

Suppose that the set F(m) is \((S,\delta )\)-big, for some \(m \in G\). Then for any two sets F(g) and F(h) which are \((S,\delta )\)-far from F(m), there is a subset K of X such that \(|K| \le \nu (F) \) and

$$\begin{aligned} F(g) \cup F(h) \subset O_{\delta }(S\cup K). \end{aligned}$$

Proof

Since F(m) is \((S,\delta )\)-big, it does not intersect \(O_{\delta }(S)\). Hence (2.5) implies

$$\begin{aligned} F(m) \subset O_{\delta }(F(mg^{-1}))\cup O_{\delta }(F(g)). \end{aligned}$$
(3.1)

Taking into account that F(g) is \((S,\delta )\)-far from F(m), we get that

$$\begin{aligned} F(m) \subset O_{\delta }(F(mg^{-1})). \end{aligned}$$
(3.2)

It can be deduced from (3.2) that

$$\begin{aligned} F(mg^{-1}) \subset O_{\delta }(F(m)). \end{aligned}$$
(3.3)

Indeed, let us choose, for each \(x\in F(m)\), an element \(y = y(x)\in F(mg^{-1})\) with \(d(x,y) < {\delta }\). Then the map \(x\mapsto y(x)\) is injective: if \(x_1\ne x_2\) and \(y(x_1) = y(x_2)\) then \(d(x_1,x_2) < 2 \delta \), a contradiction. Since \(|F(m)| = \nu (F) \ge |F(mg^{-1})|\), each \(y\in F(mg^{-1})\) is of the form y(x). So \(d(y,F(m)) < \delta \) for each \(y\in F(mg^{-1})\).

Changing g for h in (3.3), we get also that

$$\begin{aligned} F(mh^{-1}) \subset O_{\delta }(F(m)). \end{aligned}$$
(3.4)

Since

$$\begin{aligned}{} & {} F(g) \subset O_{\delta }(F(ghm^{-1}))\cup O_{\delta }(F(mh^{-1}))\cup O_{\delta }(S) \\{} & {} \subset O_{\delta }(F(ghm^{-1}))\cup O_{2\delta }(F(m))\cup O_{\delta }(S) \end{aligned}$$

and \(F(g)\cap O_{2\delta }(F(m)) = \emptyset \), we obtain that

$$\begin{aligned} F(g) \subset O_{\delta }(F(hgm^{-1}))\cup O_{\delta }(S). \end{aligned}$$
(3.5)

Similarly

$$\begin{aligned} F(h) \subset O_{\delta }(F(hgm^{-1}))\cup O_{\delta }(S). \end{aligned}$$
(3.6)

Thus we proved that \(F(h)\cup F(g)\) is contained in \(O_{\delta }(S\cup K)\) where \(K =F(hgm^{-1})\). Clearly \(|K| \le \nu (F)\) \(\square \)

Proof of Theorem 2.2

We use induction on \(\nu (F)\).

Let \(\nu (F) =1\). In this case F(g) is \((S,\delta )\)-big if and only if it is not contained in \(O_{\delta }(S)\). If there are no \((S,\delta )\)-big sets then there is nothing to prove. So, let us fix \(m\in G\) such that F(m) is \(\delta \)-big, and denote by \(G_1\) the set of all \(g\in G\) with the property that F(g) is not (\(S,\delta \))-far from F(m). Then clearly

$$\cup _{g\in G_1}F(g) \subset O_{\delta }(S)\cup O_{2\delta }(F(m)).$$

Let \(G_2 = G\setminus G_1\), then no F(g), for \(g\in G_2\), intersects \(O_{\delta }(S)\) or \(O_{2\delta }(F(m))\). Fix \(h\in G_2\). If \(g\in G_2\) is such that F(g) is \(\delta \)-far from F(h) then, by Lemma 3.1, \(F(g)\cup F(h) \subset O_{\delta }(S\cup K)\) with \(|K| = 1\). Therefore \(F(g)\cup F(h) \subset O_{\delta }(K)\) whence \(d(F(g),F(h)) < 2\delta \), a contradiction. Thus \(\cup _{g\in G_2}F(g) \subset O_{2\delta }(F(h))\) and

$$\begin{aligned} \cup _{g\in G}F(g) \subset O_{\delta }(S)\cup O_{2\delta }(F(m)\cup F(h)) \subset O_{2\delta }(S\cup (F(m)\cup F(h))). \end{aligned}$$

Thus we may take \(k_1 = 2\).

Suppose now that the statement holds for \(\nu (F)< n\). Let F be an \((S,\delta )\)-subadditive map with \(\nu (F) = n\). \(\square \)

Claim 3.2

If no F(g) is \((S,\delta )\)-big then

$$\begin{aligned} \cup _{g\in G}F(g) \subset O_{C\delta }(S\cup E), \end{aligned}$$
(3.7)

where \(|E| = (n-1)(n+2)/2\) and \(C = 3k_{n-1}+2\).

Proof of the claim

In each F(g) with \(|F(g)| = n\) one can choose \(x = x_g\) such that

$$\begin{aligned} \text {either } d(x_g,S)< \delta \text { or } d(x_g,F(g)\setminus \{x_g\}) < 2\delta . \end{aligned}$$

Now we define a new map \(F'\) by setting \(F'(g) = F(g){\setminus } \{x_g\}\) for such g, and \(F'(g) = F(g)\) for g with \(|F(g)| < n.\)

The map \(F'\) is \((S,3\delta )\)-subadditive, because

$$\begin{aligned} F'(gh)\subset F(gh)\subset O_{\delta }(F(g)\cup F(h)\cup S) \end{aligned}$$

and

$$\begin{aligned} F(g)\subset O_{\delta }(S)\cup O_{2\delta }(F'(g)), \;\;F(h)\subset O_{\delta }(S)\cup O_{2\delta }(F'(h)). \end{aligned}$$

Moreover \(\nu (F') \le n-1\) whence, by the induction hypothesis, there is \(E \subset X\) with \(|E| \le (n-1)(n+2)/2\) such that

$$\begin{aligned} \cup _{g\in G} F'(g) \subset O_{k_{n-1}3\delta }(S\cup E). \end{aligned}$$

Again using the condition \(F(g) \subset O_{2\delta }(F'(g)\cup S)\) we get

$$\begin{aligned} \cup _{g\in G}F(g) \subset O_{2\delta }(O_{3k_{n-1}\delta }(S\cup E)) = O_{(2\delta + 3k_{n-1}\delta )}(S\cup E). \end{aligned}$$

The claim is proved. \(\square \)

We may assume now that there is \(m\in G\) such that F(m) is \((S,\delta )\)-big.

Case 1::

No \((S,\delta )\)-big set F(g) is \((S,\delta )\)-far from F(m).

In such a case we set \(\widetilde{S} = S\cup F(m)\) and define a map \(\widetilde{F}: G\rightarrow 2^X\) by the formula

$$\begin{aligned} \widetilde{F}(g) = F(g) \setminus O_{\delta }(\widetilde{S}), \qquad g \in G. \end{aligned}$$

Using the \((S,\delta )\)-subbadditivity of F and the inclusion

$$\begin{aligned} F(g) \subset O_{\delta }(\widetilde{F}(g)\cup \widetilde{S}), \end{aligned}$$
(3.8)

it is easy to see (as in the proof of Claim 3.2) that the map \(\widetilde{F}\) is \((\widetilde{S},2\delta )\)-subadditive.

Since no \((S,\delta )\)-big set is \((S,\delta )\)-far from F(m), the map \(\widetilde{F}\) has no \((\widetilde{S},2\delta )\)-big values. Indeed if \(\widetilde{F}(g)\) is \((\widetilde{S},2\delta )\)-big then by definition \(d(\widetilde{F}(g),\widetilde{S})> 2\delta \) and \(\widetilde{F}(g) = F(g)\). So \(d(F(g),S) > 2\delta \) and \(d(F(g),F(m))> 2\delta \), a contradiction.

Applying Claim 3.2 we get that

$$\begin{aligned} \cup _{g\in G}\widetilde{F}(g)\subset O_{C2\delta }(\widetilde{S}\cup E), \end{aligned}$$

where \(|E| = (n-1)(n+2)/2\) and \(C = 3k_{n-1}+2\). By (3.8), this implies

$$\begin{aligned}{} & {} \cup _{g\in G}F(g)\subset O_{\delta }( O_{2C\delta }(\widetilde{S}\cup E)) = O_{(2C+1)\delta }(\widetilde{S}\cup E) \\{} & {} = O_{(2C+1)\delta }(S\cup P) = O_{C_1\delta }(S\cup P), \end{aligned}$$

where \(\ P = E \cup F(m)\) and \(\ C_1 = 2C+1 = 6k_{n-1}+5\). Clearly \(\ |P| \le n + (n-1)(n+2)/2 < n(n+3)/2\).

Case 2::

There is \(g_0 \in G\) such that \(F(g_0)\) is \((S,\delta )\)-big and \((S,\delta )\)-far from F(m).

Choosing a point \(x_0\in F(g_0)\) we set \(S_1 = F(m)\cup \{x_0\}\cup S\) and define a map \(F_1: G\rightarrow 2^X\) by

$$\begin{aligned} F_1(g) = F(g) \setminus O_{\delta }(S_1). \end{aligned}$$

Since \(F(g)\subset F_1(g)\cup O_{\delta }(S_1)\), one checks as above that the map \(F_1\) is \((S_1,2\delta )\)-subadditive.

Let us show that \(F_1\) has no \((S_1,2\delta )\)-big values. Indeed, let \( F_1(g)\) be \((S_1,2\delta )\)-big, then \(F_1(g) = F(g)\) and \(d(F_1(g), S_1) > 2\delta \). It follows that \(d(F(g),F(m))> 2\delta \) and, since F(g) and \(F(g_0)\) do not intersect \(O_{\delta }(S)\), they both are \((S,\delta )\)-far from F(m). Applying Lemma 3.1 to the sets F(g) and \(F(g_0)\) we get that \(F(g)\cup F(g_0) \subset O_{\delta }(S \cup K)\), where \(K= \{k_1,...,k_r\}\) has cardinality at most n. Since F(g) and \(F(g_0)\) do not intersect \(O_{\delta }(S)\) we obtain

$$\begin{aligned} F(g)\cup F(g_0) \subset O_{\delta }(K). \end{aligned}$$

Note that if two elements \(x_1,x_2\) of F(g) were \(\delta \)-close to the same \(k_i\) then \(d(x_1,x_2) < 2\delta \), in contradiction to the condition that \(F(g) = F_1(g)\) is \(\delta \)-big (recall that it is even \(2\delta \)-big). Therefore \(|K|=n\) and F(g) is not contained in \(O_{\delta }(K_1)\) for any proper subset \(K_1\) of K. Thus for each \(k_i\) there is a \(\ y = y(i)\in F(g)\) with \(d(y(i),k_i)< \delta \). On the other hand, since \(F(g_0) \subset O_{\delta }(K)\), there is a \(k_i\) with \(d(x_0,k_i) < \delta \). Therefore \(d(x_0,y(i)) < 2\delta \) in contradiction with our assumption that \(d(F(g),S_1)> 2\delta \).

Applying Claim 3.2 to the map \(F_1\) we obtain

$$\begin{aligned} \cup _{g\in G}F_1(g) \subset O_{C 2\delta }(S_1\cup E), \end{aligned}$$

where \(|E| = (n-1)(n+2)/2\) and \(C = 3k_{n-1}+2\). Therefore

$$\begin{aligned}{} & {} \cup _{g\in G}F(g) \subset O_{\delta }(S_1) \cup O_{2C \delta }(S_1\cup E) = O_{3C\delta }(S_1\cup E) \\{} & {} \quad = O_{2C \delta }(S\cup F(m) \cup \{x_0\}\cup E) = O_{C_2 \delta }(S\cup P), \end{aligned}$$

where \(|P| \le |E|+n+1 \le (n-1)(n+2)/2+n+1 = n(n+3)/2\), \(C_2 = 6k_{n-1}+4\).

So the result will be proved if we choose a sequence \(k_n\) with \(k_1 \ge 2\) and \(k_{n} \ge 6k_{n-1}+5\) for all n. Clearly the sequence \(k_n =3\cdot 6^{n-1}-1\) satisfies these conditions.

Now to deduce Theorem 2.1 it suffices to take \(S = \emptyset \) and \(\delta = \frac{\varepsilon }{(3\cdot 6^{n-1}-1)}\) in Theorem 2.2. \(\square \)

It would be interesting to find a precise formula, or at least a precise asymptotic for \(\delta \) (as a function of n) and a sharp estimate for |P|.

We will finish the paper with a simple application of Theorem 2.1 to subadditive maps.

Corollary 3.3

Let G be a group, \((\mathcal {X}, d)\) a metric space and let \(F: G\rightarrow 2^X\) be a subadditive map. If, for each \(g \in G\), the diameter of F(g) does not exceed some \(D>0\), then the entire image F(G) can be covered by two balls of radius 3D.

Proof

For each \(g \in G\), let us fix an element \(x_g\in F(g)\). Then the function \(\tilde{F}: G\rightarrow 2^X\) that assigns to \(g\in G\) the singleton \(\{x_g\}\) is D-subadditive. Indeed,

$$\begin{aligned} \tilde{F}(gh){} & {} =\{x_{gh}\} \in F(gh) \subset F(g)\cup F(h) \subset O_{D}(\{x_g\})\cup O_{D}(\{x_h\}) \\{} & {} = O_{D}(\tilde{F}(g)\cup \tilde{F}(h). \end{aligned}$$

We are going to apply Theorem 2.1 with \(\varepsilon = 2D\). Since \(n=1\), we get that \(|P|=n(n+3)/2 = 2\), P consists of two elements: \(P = \{a,b\}\). Furthermore \(2D/(3\cdot 6^{n-1}-1) = D\), so Theorem 2.1 shows that

$$\begin{aligned} {\tilde{F}}(G) \subset O_{2D}(P) = B(a,2D)\cup B(b,2D), \end{aligned}$$

where by B(x, R) we denote the ball with center x and radius R. Thus \({\tilde{F}}(G)\) is contained in the union of two balls. Since \(F(g)\subset O_D({\tilde{F}}(g))\), we conclude that \(F(G)\subset B(a,3D)\cup B(b,3D)\). \(\square \)