Definable Continuous Induction on Ordered Abelian Groups

TL;DR

This paper develops a first-order definable version of continuous induction on densely ordered abelian groups, enabling a proof of a definable Heine-Borel theorem within this algebraic structure.

Contribution

It introduces a novel, first-order definable form of continuous induction for ordered abelian groups, extending the classical real analysis tools to algebraic structures.

Findings

Proves Heine-Borel theorem in densely ordered abelian groups.

Introduces pseudo finite sets for definability.

Establishes a definable version of continuous induction.

Abstract

As mathematical induction is applied to prove statements on natural numbers, {\it continuous induction} (or, {\it real induction}) is a tool to prove some statements in real analysis.(Although, this comparison is somehow an overstatement.) Here, we first consider it on densely ordered abelian groups to prove Heine-Borel theorem (every closed and bounded interval is compact with respect to order topology) in those structures. Then, using the recently introduced notion of pseudo finite sets, we introduce a first order definable version of {\it continuous induction} in the language of ordered groups and we use it to prove a definable version of Heine-Borel theorem on densely ordered abelian groups.

Equations

$$S=\{x\in [a,b)|[a,x]\mathrm{is}\mathrm{covered}\mathrm{by}\mathrm{a}\mathrm{finite}\mathrm{subset}\mathrm{of}\mathcal{U}\}.$$

$$S=\{x\in G|a<x\le b\to \exists t\in G,[a,x]\subseteq \bigcup _{u\in \psi (t,G)}\phi (u,G)\}.$$

Full text

1. Introduction and Preliminaries

As mathematical induction is a good proof technique for statements on natural numbers, some efforts have been made to provide a similar technique in the real field, see [1, 7, 13, 14, 8, 2]. Here, we define continuous induction on densely ordered abelian groups. Our motivation is to give a definable version of continuous induction in ordered structures such as the definable version of mathematical induction in Peano arithmetic.

A structure $(G,+,<)$ is an ordered abelian group if $(G,+)$ is an abelian group, $(G,<)$ is a linear ordered set, and for all $x,y,z$ in $G$, if $x<y$ then $x+z<y+z$. An ordered abelian group is dense if for all $x<y$ in $G$ there exists $z\in G$ such that $x<z<y$. If $G$ is not dense, then it is discrete, i.e. it contains a smallest positive member. Throughout the paper, let $G$ be a fixed arbitrary densely ordered abelian group. Intervals in $G$ are defined in common way like in the real field $\mathbb{R}$. Then, $G$ is equipped with the interval topology for which open intervals in $G$ constitute a topology base.

A nonempty subset $C$ of $G$ is a Dedekind cut if it is downward closed. A cut $C$ is proper if $C\neq G$. A proper cut $C$ is said to be a gap in $G$ if it does not have a least upper bound in $G$. An ordered abelian group $G$ which has no any gaps is said to be Dedekind complete. Clearly, $G$ is Dedekind complete if and only if every nonempty and bounded from above subset of $G$ has a least upper bound in $G$.

In the following, we formulate the continuous induction over an ordered abelian group.

Definition 1.1**.**

Continuous Induction (${\rm CI}$): Let $S\subseteq G$ satisfy the following conditions:

$i$) there exists at least one element $g\in G$ such that $(-\infty,g)\subseteq S$,

$ii$) for every element $x\in G$, if $(-\infty,x)\subseteq S$ then there exists $y>x$ in $G$ such that $(-\infty,y)\subseteq S$.

Then, $S=G$.

We refer the conditions $i$ and $ii$ above as the assumptions of ${\rm CI}$.

2. The Strength of Continuous Induction

In order to see the usefulness of proving by continuous induction, we refer to [1], [2], and [8]. Here, we use ${\rm CI}$ to prove that Heine-Borel theorem holds in densely ordered abelian groups. Our proof in somehow is different from ones in the mentioned references. It is such that to be adaptable for the definable version in the next section.

A linearly ordered set which is dense (in itself) and has no end points is Dedekid complete if and only if it satisfies the CI [7]. A similar statement is true for ordered (dense or non-dense) abelian groups.

Proposition 2.1**.**

An ordered abelian group $G$ is Dedekind complete if and only if it satisfies the ${\rm CI}$.

Proof.

Let $G$ be a Dedekind complete ordered abelian group, and $S\subset G$ satisfy the assumptions $i$ and $ii$ of ${\rm CI}$. Let $C=\{g\in G|\ (-\infty,g)\subseteq S\}$. Then, $C$ is a Dedekind cut in $G$. It follows from the condition $ii$ of the ${\rm CI}$ that $C$ does not have a least upper bound in $G$, and so by Dedekind completeness of $G$, $C=G$. From that we have $S=G$. Conversely, if $G$ is not Dedekind complete then it does have a gap, say $C$. The subset $C$ satisfies the conditions of ${\rm CI}$, but $C\neq G$. So $G$ does not satisfy the ${\rm CI}$. ∎

Therefore, every densely ordered abelian group $G$ which satisfies the ${\rm CI}$ is isomorphic to the ordered additive group of reals. For some practical purposes, we also define a bounded version of ${\rm CI}$, bounded continuous induction (${\rm BCI}$), on ordered abelian groups.

Definition 2.2**.**

For every $a<b$ in $G$ and every $S\subseteq[a,b)$, if

i) there exists $x>a$ such that $[a,x)\subseteq S$, and

ii) for every $x>a$ for which $[a,x)\subseteq S$, there exists $y>x$ such that $[a,y)\subseteq S$,

then $S=[a,b)$.

Proposition 2.3**.**

On densely ordered abelian groups, ${\rm CI}$ is equivalent to ${\rm BCI}$.

Proof.

Let $G$ be a densely ordered abelian group that satisfies ${\rm CI}$. Suppose that $S\subseteq[a,b)$, for $a<b$ in $G$, satisfies the two conditions of ${\rm BCI}$. If $S$ is not the whole of $[a,b)$, then $C=\{x\in G|\exists y\in S:\ x<y\}$ is a gap in $G$ which is Dedekind complete by Proposition 2.1. This is a contradiction.

Conversely, let $G$ satisfy ${\rm BCI}$, and the two conditions of ${\rm CI}$ hold for $S\subset G$. Then, there is an element $x\in G$ such that $(-\infty,x)\subset S$. Let $a<x$, and $b>x$ be arbitrary. Then, the subset $S^{\prime}=[a,b)\cap S$ of $G$ satisfies the conditions of ${\rm BCI}$. Thus, $S^{\prime}=[a,b)$ and so $[a,b)\subset S$. Therefore, $S=G$. ∎

In the following theorem, we give a proof for Heine-Borel theorem on densely ordered abelian groups by using ${\rm CI}$. We also show that any densely ordered abelian group which satisfies Heine-Borel theorem, holds ${\rm CI}$ too.

*Heine-Borel * ($\rm HB$) theorem: Every closed bounded interval $[a,b]$ is compact.111Here, we consider the order topology, i.e. a topology upon open intervals $(a,b)$ in $G$.

Theorem 2.4**.**

Any densely ordered abelian group satisfies ${\rm CI}$ if and only if it satisfies Heine-Borel theorem.

Proof.

Let $G$ be a densely ordered abelian group that satisfies ${\rm CI}$. Take arbitrary elements $a<b$ in $G$. We show that the closed and bounded interval $[a,b]$ is compact. Let $\mathcal{U}$ $=\{U_{i}\}_{i\in I}$ be an open cover for $[a,b]$, and let

[TABLE]

By Proposition 2.3, $G$ satisfies ${\rm BCI}$. In the following claim we show that the conditions $i$ and $ii$ of ${\rm BCI}$ on $[a,b)$ hold for $S$, so we have $S=[a,b)$. On the other hand, there exists $i_{n}\in I$ such that $b\in U_{i_{n}}$, and so $(b-\epsilon,\ b]\subseteq U_{i_{n}}$ for some $\epsilon\in G^{>0}$. Since $b-\epsilon\in S$, then there exists a finite subset of $\mathcal{U}$ which covers $[a,\ b-\epsilon]$. Consequently, $[a,b]$ is covered by a finite subset of $\mathcal{U}$.

Claim. The subset $S$ satisfies the conditions of ${\rm BCI}$ on $[a,b)$.

Proof of the claim. $i$) Since $\mathcal{U}$ covers $[a,b]$, then there exists $i\in I$ such that $a\in U_{i}$, so $[a,x_{0})\subseteq U_{i}$ for some $x_{0}\in(a,b)$. Hence, by the definition of $S$, $[a,x_{0})\subseteq S$.

$ii$) Let $a<x<b$, and $[a,x)\subseteq S$. There is an $i_{0}\in I$ such that $x\in U_{i_{0}}$, and so $(x-\epsilon,\ x+\epsilon)\subseteq U_{i_{0}}$ for some $\epsilon\in G^{>0}$. Since $x-\epsilon\in S$, then there is a finite subcover of $\mathcal{U}$ for $[a,\ x-\epsilon]$, say $\{U_{i_{1}},U_{i_{1}},...,U_{i_{n}}\}$. So, $\{U_{i_{0}},U_{i_{1}},U_{i_{1}},...,U_{i_{n}}\}$ is a finite subcover for $[a,\ x+\delta]$ for every $0<\delta<\epsilon$. Thus, $[a,\ x+\epsilon)\subseteq S$. End of the proof of the claim.

Conversely, suppose $G$ satisfies Heine-Borel theorem, i.e. all closed bounded intervals in $G$ are compact. If $G$ does not satisfy ${\rm CI}$, then by proposition 2.1, $G$ has a Dedekind gap, say $C$. Take $a$ in $C$ and b in $G\setminus C$. Then, $\{(-\infty,\ x)\ |\ x\in C\ and\ x>a\}\cup\{G\setminus C\}$ is an open cover for $[a,b]$ which does not have any finite subcover. ∎

In the real field $\mathbb{R}$, a consequence of ${\rm HB}$ theorem is that any continuous function on a closed bounded interval is uniformly continuous. In a similar method, one can easily conclude the following result in ordered abelian groups.

Corollary 2.5**.**

Let $G$ be an ordered abelian group that satisfies ${\rm CI}$. Then, if $f:[a,b]\rightarrow G$ is a continuous function, then it is uniformly continuous on $[a,b]$.

3. Definable Continuous Induction

Let $\mathcal{L}=\langle+,<,0,\dots\rangle$ be an expansion of the language of ordered abelian groups. A subset $D$ of an ordered abelian group $\langle G,+,<,0\rangle$ is said to be $\mathcal{L}$-definable (with parameters) if there exists an $\mathcal{L}$-formula $\varphi(x,\bar{g})$, with some parameters $\bar{g}$ from $G$, such that $D$ consists of all elements $a\in G$ for which the $\mathcal{L}$-structure $G$ satisfies $\varphi(a,\bar{g})$. $D=\{a\in G|G\vDash\varphi(a,\bar{g})\}$. Now, we can consider the first order version of the continuous induction in $\mathcal{L}$ by replacing the subset $S$ in the formulation of ${\rm CI}$ with a definable subset $D$. Let ${\rm DCI}$ denote the first order definable version of ${\rm CI}$. Indeed, ${\rm DCI}$ is a schema in the language $\mathcal{L}$.

Definition 3.1**.**

For any $\mathcal{L}$-formula $\varphi(v,\bar{w})$, let ${\rm DCI}_{\varphi}$ denote the $\mathcal{L}$-formula

$\forall\bar{w}((\exists s\forall v<s\varphi(v,\bar{w})\wedge\forall v(\forall s<v\varphi(s,\bar{w})\rightarrow\exists u>v\forall s<u\varphi(s,\bar{w})))\rightarrow\forall v\varphi(v,\bar{w}))$.

Now, ${\rm DCI}$ is the scheme $\{{\rm DCI}_{\varphi}|\ \varphi(v,\bar{w})\ \rm is\ an\ \mathcal{L}-formula\}$.

So, ${\rm DCI}$ is preserved under elementary equivalence relation between $\mathcal{L}$-structures, as well as under ultraproducts of those structures.

As in the second order case, it is easy to show that ${\rm DCI}$ is equivalent to its bounded version on intervals. Also, one can observe that ${\rm DCI}$ is indeed equivalent to the definable version of the Dedekind completeness in densely ordered abelian groups. We say that a densely ordered abelian group $\langle G,+,<,0\rangle$ is definably (Dedekind) complete if it has no any definable gap.

Proposition 3.2**.**

A densely ordered abelian group $G$ is definably complete if and only if it satisfies ${\rm DCI}$.

Proof.

Assume that $G$ is definably complete and $\phi(v,\bar{a})$ is a formula, with parameters $\bar{a}$, for which $G$ satisfies the induction hypotheses. If $G\nvDash\forall v\phi(v,\bar{a})$, then the formula $\psi(v,\bar{a})=\forall u<v\phi(u,\bar{a})$ defines a bounded from above nonempty subset, say $S$, of $G$. By definable completeness, the subset $S$ has a least upper bound in $G$. This contradicts the second induction hypothesis about $\phi(v,\bar{a})$.

Conversely, suppose that $G$ is an ordered abelian group that satisfies $DCI$. Let $C$ be a definable cut in $G$ which is defined by an $\mathcal{L}$-formula $\phi(v,\bar{a})$. If $C$ does not contain a least upper bound in $G$, then $G$ satisfies the hypotheses of $DCI$ for $\phi(v,\bar{a})$. Then, $G\vDash\forall v\phi(v,\bar{a})$ and therefor $C=G$. Hence, $G$ does not have any definable gap. ∎

By [10], definably complete ordered groups are divisible. So, we have the following.

Corollary 3.3**.**

If $G\vDash DCI$, then $G$ is divisible.

Remark 3.4**.**

According to [3] a linearly ordered structure $M$ is said to be o-minimal if any definable subset of $M$ is a finite union of intervals and points in $M$. Any o-minimal structure is definably complete, but the converse is not true for arbitrary ordered structures. (See [11], Proposition 1.2. and the following comments.) On the other hand, the theory of divisible ordered abelian groups in the language of ordered groups has quantifier elimination property (See [9]). So, every definable subset of a divisible ordered abelian group in the language of ordered groups is a finite union of intervals and points. Hence, those groups are o-minimal. The converse was proved in op. cit. (Theorem 2.1). So, every divisible ordered abelian group is definably complete and so satisfies ${\rm DCI}$. Note that there are models of ${\rm DCI}$ that are not not o-minimal. Here, we give an example. Let $\mathcal{F}$ be a non-principal ultrafilter on the set of natural numbers, $\mathbb{N}$. Let $\mathcal{G}$ be the ultra-product of the structures $\mathbb{Q}_{n}=(\mathbb{Q},+,0,<,\{0,\dots,n\})$ with respect to $\mathcal{F}$. Note that for each $n$, $\mathbb{Q}_{n}$ satisfies ${\rm DCI}$. So, $\mathcal{G}$ satisfies ${\rm DCI}$ too. But $\mathcal{G}$ is not o-minimal, as the ultra-product of the sets $\{0,\dots,n\})$ is an infinite definable discrete set in $\mathcal{G}$.

Definable compactness in ordered structures was introduced by [12] where it was proved that this concept is equivalent to closed and bounded for definable sets. Also, [5] contains a weak version this notion which is far from finiteness. Here, we use the concept of *pseudo-finite set * which was introduced in [6], to give an alternative definition for definable compactness in ordered abelian groups. Then, we provide a definable version of the Heine-Borel theorem in definably complete ordered groups using ${\rm DCI}$.

Let $G$ be an $\mathcal{L}$-structure, where $\mathcal{L}$ is an expansion of the language of ordered groups. Then, a definable subset $D\subseteq G$ is said to be pseudo-finite if it is discrete, closed, and bounded. Note that these are the only first order properties in the language $\mathcal{L}$ describing the finiteness for sets definable in $\mathcal{L}$-structures. It is easy to see that the union of two pseudo-finite sets is pseudo-finite. Even, the union of a pseudo-finite family of pseudo-finite sets is pseudo-finite (see [6]). In o-minimal structures, a definable set is pseudo-finite if and only if it is finite.

Let $\varphi(x,y)$ be an $\mathcal{L}$-formula. By $\varphi(a,G)$, where $a\in G$, we mean a set that is defined in $G$ by the formula $\varphi(a,y)$: $\{b\in G|G\vDash\varphi(a,b)\}$. Then, $\{\varphi(a,G)\}_{a\in G}$ is said to be a definable family of subsets of $G$. A definable open cover for a set $X\subseteq G$ is a definable family $\{\varphi(a,G)\}_{a\in G}$ which covers $X$ and for each $a\in G$, $\varphi(a,G)$ is an open subset of $G$. In the following definition we fix a directed definable family $\{\psi(t,G)\}_{t\in G}$ of pseudo-finite sets which covers $G$, that is $\psi(t_{1},G)\subseteq\psi(t_{2},G)$ for $t_{1}<t_{2}$, and $\bigcup_{t\in G}\psi(t,G)=G$.

Definition 3.5**.**

Let $G$ be an ordered abelian group. Then, a definable subset $D\subseteq G$ is said to be definably compact if for every definable open cover $\{\varphi(a,G)\}_{a\in G}$ of $D$, there exists a pseudo-finite sub-cover $\{\varphi(a,G)\}_{a\in P}$ for some pseudo-finite set $P=\psi(t_{0},G)$.

Theorem 3.6**.**

All closed and bounded intervals of definably complete ordered groups are definably compact.

Proof.

Let $\langle G,+,<,0\rangle$ be a definably complete ordered abelian group, and $\varphi(x,y)$ define an open cover for a closed and bounded interval $[a,b]$ in $G$. Let

[TABLE]

Let $\theta(x,a,b)$ be an $\mathcal{L}$-formula which defines the set $S$ in the $\mathcal{L}$-structure $G$. We show that $\theta(x,a,b)$ satisfies the conditions of the ${\rm DCI}_{\theta}$.

There exists $x_{0}\in G$ such that $a\in\varphi(x_{0},G)$. Since $\varphi(x_{0},G)$ is open, then there is $c\in(a,b)$ such that $[a,c)\subseteq\varphi(x_{0},G)$. On the other hand the family $\psi$ covers $G$, hence there is an element $t_{0}\in G$ such that $x_{0}\in\psi(t_{0},G)$. Thus, $G$ satisfies $\theta(g,a,b)$ for every $g\in(-\infty,c)$.

Now, let $h\in(a,b)$ and $G$ satisfy $\forall g<h\theta(g,a,b)$. There are $x_{1},h_{1},h_{2}\in G$ such that $a<h_{1}<h<h_{2}<b$ and $(h_{1},h_{2})\subset\varphi(x_{1},G)$. There is an element $t_{1}\in G$ with $x_{1}\in\psi(t_{1},G)$. By the assumption, $G$ satisfies $\theta(h_{1},a,b)$. So, there is a pseudo finite set $\psi(t_{2},G)$ such that $[a,h_{1}]\subseteq\bigcup_{u\in\psi(t_{2},G)}\varphi(u,G)$. Since the family $\psi$ is directed, then there is $t_{3}\in G$ such that $\psi(t_{1},G),\psi(t_{2},G)\subseteq\psi(t_{3},G)$. Hence, $[a,h_{2}]\subseteq\bigcup_{u\in\psi(t_{3},G)}\varphi(u,G)$.

Therefor the conditions of ${\rm DCI}_{\theta}$ hold in $G$. By Proposition 3.2, $G$ satisfies ${\rm DCI}_{\theta}$. Hence, $S=G$, and so there is a pseudo finite set $\psi(t,G)$ for some $t\in G$ such that $[a,b]\subseteq\bigcup_{u\in\psi(t,G)}\varphi(u,G)$.

∎