A basic principle in tame geometry is that there are no pathological definable functions in o-minimal structures.\u00a0 \u00a0One precise sense in which this principle is true is given by the Monotonicity Theorem.<\/p>\n
Monotonicity Theorem<\/b> :\u00a0 Let $f: R \\to R$ be a definable function in some o-minimal structure $\\mathfrak{R}$.\u00a0 Then $f$ is piecewise continuous and is piecewise constant or strictly monotone.\u00a0 That is, we can find $-\\infty = a_0 < a_1 < \\ldots < a_m = \\infty$ so that for $0 \\leq i < m$ the function $f \\upharpoonright (a_i,a_{i+1})$ is continuous and constant, strictly increasing, or strictly decreasing.<\/p>\n
<\/p>\n
We will follow the outline of the proof presented by van den Dries.<\/p>\n
Fix for now an o-minimal structure $\\mathfrak{R}$ so that “definable” will mean “definable in $\\mathfrak{R}$.\u00a0 The Monotonicity Theorem is proven as a consequence of three lemmas.<\/p>\n
Lemma 1<\/b>:\u00a0 If $f:I \\to R$ is a definable function with domain an interval $I$, then there is a subinterval $J \\subseteq I$ for which either $f \\upharpoonright J$ is contant or it is injective.<\/p>\n
Lemma 2<\/b>: If $f:I \\to R$ is a definable, injective function with domain an interval $I$, then there is a subinterval $J \\subseteq I$ on which $f$ is strictly monotone.<\/p>\n
Lemma 3<\/b>: If $f:I \\to R$ is a definable, strictly monotone function with domain an interval $I$, then $f$ is piecewise continuous.<\/p>\n
Let us check first that the Monotonicity Theorem follows from Lemmas 1, 2, and 3.<\/p>\n
Proof<\/b> (of the Monotonicity Theorem given Lemmas 1, 2, and 3):\u00a0 Let $f:R \\to R$ be a definable function.\u00a0 \u00a0Let us note that the set $C$ of points at which $f$ is continuous is definable.\u00a0 Indeed, the usual $\\epsilon$-$\\delta$-definition of continuity (slightly modified as we are not assuming that $\\mathfrak{R}$ expands an ordered group) gives a formal definition of $C$:<\/p>\n
$$C = \\{ x \\in R :\u00a0 \\forall \\epsilon_1 \\forall \\epsilon_2 (\\epsilon_1 < f(x) < \\epsilon_2 \\to \\exists \\delta_1 \\exists \\delta_2 [\\delta_1 < x < \\delta_2 ~\\&~ \\forall u (\\delta_1 < u < \\delta_2 \\to \\epsilon_1 < f(u) < \\epsilon_2)])$$<\/p>\n
By o-minimality, if $R \\smallsetminus C$ were infinite, then there would be an interval $I \\subseteq R \\smallsetminus C$.\u00a0 By Lemma 1, there would be some subinterval $J \\subseteq I$ on which $f$ is contant or injective.\u00a0 We cannot be in the first case as constant functions are continuous.\u00a0 Thus, $f$ is injective on $J$ which implies by Lemma 2 that there is a subinterval $K \\subseteq J$ on which $f$ is strictly monotone from which Lemma 3 gives a subinterval $L \\subseteq K$ on which $f$ is continuous, again contradicting the fact that $K \\cap C = \\varnothing$.\u00a0 Therefore, $f$ is piecewise continuous.<\/p>\n
Let $B_0 := \\{ x \\in R : f $ is locally constant at $x \\}$, where to say that $f$ is locally constant at $x$ means $\\exists \\epsilon_1 \\exists \\epsilon_2\u00a0 (\\epsilon_1 < x < \\epsilon_2 ~\\&~ \\forall y [\\epsilon_1 < y < \\epsilon_2 \\to f(y) = f(x)])$.\u00a0 \u00a0Likewise, let $B_+ := \\{ x \\in R : f $ is locally increasing at $x \\}$ and $B_- := \\{ x \\in R : f $ is locally decreasing\u00a0 at $x \\}$.\u00a0 Each of these sets is definable.\u00a0 Arguing as we did for proving piecewise continuity, the complement of $B_0 \\cup B_+ \\cup B_-$ must be finite, for otherwise it would contain an interval $I$ with by Lemma 1 would have a subinterval on which $f$ is constant (which would contradict $I \\cap B_0 = \\varnothing$) or on which $f$ is injective, and then by Lemma 2, we would find a subinterval on which $f$ is strictly monotone, now contradicting disjointness from $B_+ \\cup B_-$.<\/p>\n
Finally, we observe that if $f$ is continuous and locally constant (respectively, locally monotone)\u00a0 on some interval, then it is actually constant (respectively, monotone).\u00a0 <\/span>That is, we assume that $I$ is an interval, $f$ is continuous on $I$, and $f$ is locally constant, increasing, or decreasing on $I$.\u00a0 <\/span>Let us write the proof in the case that $f$ is locally constant. The other cases are similar.\u00a0 <\/span>Let $x \\in I$.\u00a0 <\/span>Let $A := \\{ y \\in I : y < x $ and $ f(y) \\neq f(x) \\}$ and $B := \\{ z \\in I : x < z $ and $ f(z) \\neq f(x) \\}$.\u00a0 <\/span>We wish to show that both $A$ and $B$ are empty. \u00a0 <\/span>Suppose that $A \\neq \\varnothing$.\u00a0 <\/span>Let $\\alpha := \\sup A$.\u00a0 <\/span>(Recall that in an o-minimal structure each nonempty definable subset has a supremum \u2014 possibly $\\infty$ \u2014 and an infimum \u2014possibly $-\\infty$).\u00a0 <\/span>Since $f$ is locally constant at $x$, there is some $\\delta < x$ so that for all $z \\in (\\delta,x)$ we have $f(z) = f(x)$. \u00a0 <\/span>Thus, $\\alpha \\leq \\delta$. \u00a0 <\/span>Since $f$ is locally constant at $\\alpha$, there are $\\beta < \\alpha < \\gamma$ so that $f$ is constant on $(\\beta,\\gamma)$.\u00a0 <\/span>Consider $u$ and $v$ with $\\beta < u \\leq \\alpha < v < \\min \\{ \\gamma, x \\}$.\u00a0 <\/span>Then $f(v) = f(x)$ as $v > \\sup A$ and $v < x$ while $f(u) = f(v)$ as $f$ is constant on $(\\beta,\\gamma)$.\u00a0 <\/span>Thus, $(\\beta,x) \\cap A = \\varnothing$ which would imply that $\\alpha \\leq \\beta < \\alpha$ which is absurd. $\\Box$<\/p>\n Let us now prove the lemmas.<\/p>\n Proof<\/b> of Lemma 1: If there is some $b \\in f(I)$ with $f^{-1} \\{ b \\}$ infinite, then because $f^{-1} \\{ b \\}$ is definable, there would be some interval $J \\subseteq f^{-1} \\{ b \\}$ meaning that $f$ would be constant on $J$.\u00a0 <\/span>So, we may assume that for every $b \\in f(I)$ the preimage $f^{-1} \\{ b \\}$ is finite (and nonempty as $b$ is in the range).\u00a0 <\/span>Define $g:f(I) \\to I$ by $y \\mapsto \\min f^{-1} \\{ y \\}$.\u00a0 <\/span>Being a right inverse of $f$, $g$ is itself injective.\u00a0 <\/span>The function $g$ is a left inverse to $f \\upharpoonright g(f(I))$.\u00a0 <\/span>Hence, $f \\upharpoonright g(f(I))$ is injective. Since $f$ is finite to one and $I$ is infinite, the definable set $g(f(I))$ is also infinite and therefore contains an interval $J$ on which $f$ is injective.\u00a0 <\/span>$\\Box$<\/p>\n Of the three lemmas, Lemma 2 is the most complicated.\u00a0 <\/span>We intend to show that definable functions are not pathological.\u00a0 <\/span>We will achieve this by showing using o-minimality to argue that if there are pathologies, then they appear everywhere on an interval and moreover the pathologies feed on themselves.\u00a0 <\/span>Ultimately, we will produce such a pathological situation\u00a0 <\/span>that it cannot exist.<\/p>\n Proof<\/b> of Lemma 2: \u00a0 <\/span>Since $f$ is injective on $I$, for any point $x \\in I$ the definable sets $L_{+}(x) := \\{ y \\in I : y < x$ and $f(y) > f(x) \\}$ and $L_-(x) := \\{ y \\in I : y < x$ and $f(y) < f(x) \\}$ partition the set $I \\cap (-\\infty,x)$. By o-minimality, each of $L_+(x)$ and $L_-(y)$ is a finite union of points and intervals.\u00a0 <\/span>Thus, one or the other contains an interval of the form $(z,x)$ for some $z \\in I$. \u00a0 <\/span>Likewise, one or the other of $R_+(x) :-= \\{ y \\in I : y > x$ and $ f(y) > f(x) \\}$ or $R_-(x) := \\{ y \\in I : y > x$ and $f(y) < f(x) \\}$ contains an interval of the form $(x,z)$.\u00a0 <\/span>From these observations we see that $I$ may be partitioned into the following four definable sets.<\/p>\n $$X_{++} := \\{ x \\in I : (\\exists \\alpha < x)(\\exists \\beta > x)(\\forall y)(\\forall z)(\\alpha < y < x < z < \\beta \\to [f(y) > f(x) ~\\&~ f(z) > f(x)])$$<\/p>\n $$X_{+-} := \\{ x \\in I : (\\exists \\alpha < x)(\\exists \\beta > x)(\\forall y)(\\forall z)(\\alpha < y < x < z < \\beta \\to [f(y) > f(x) ~\\&~ f(z) < f(x)])$$<\/p>\n $$X_{-+} := \\{ x \\in I : (\\exists \\alpha < x)(\\exists \\beta > x)(\\forall y)(\\forall z)(\\alpha < y < x < z < \\beta \\to [f(y) < f(x) ~\\&~ f(z) > f(x)])$$<\/p>\n $$X_{- -} := \\{ x \\in I : (\\exists \\alpha < x)(\\exists \\beta > x)(\\forall y)(\\forall z)(\\alpha < y < x < z < \\beta \\to [f(y) < f(x) ~\\&~ f(z) < f(x)])$$<\/p>\n By o-minimality, we may break $I$ into finitely many subintervals and points so that each such piece is contained in exactly one of these sets.\u00a0 <\/span>Working with each of these pieces separately, we may assume that $I$ itself is equal to one of the sets $X_{++}$, $X_{+-}$, $X_{-+}$, or $X_{- -}$.<\/p>\n Claim<\/b> 1: If $I = X_{+-}$, then $f$ is strictly decreasing.<\/p>\n Proof<\/b> of Claim 1:\u00a0 <\/span>Let us note that for $x \\in I$ the set $B(x) :=\\{ y \\in I : x < y ~\\&~ f(y) \\geq f(x) \\}$ is always empty.\u00a0 <\/span>Indeed, suppose that $B(x) \\neq \\varnothing$. As $x \\in X_{+-}$ there is some $b > x$ for which $(x,b) \\cap B(x) = \\varnothing$.\u00a0 <\/span>Therefore $\\beta := \\inf B(x) \\geq b > x$.\u00a0 <\/span>Since $\\beta \\in I$ there some $\\gamma < \\beta$ so that for all $y \\in (\\gamma,\\beta)$ we have $f(y) > f(\\beta) \\geq f(x)$.\u00a0 <\/span>Hence, $\\max \\{x,\\gamma\\},b] \\subseteq B(x)$ which implies that $\\beta = \\inf B(x) \\leq \\max \\{ x, \\gamma \\} < \\beta$, which is impossible.<\/p>\n To say that for every $x \\in I$ we have that $B(x) = \\varnothing$ says that for $x < y$ from $I$ we have $f(x) < f(y)$.\u00a0 <\/span>That is, $f$ is strictly decreasing on $I$.\u00a0 <\/span>$\\maltese$<\/p>\n Reversing the inequalities we see dually that if $I = X_{-+}$, then $f$ is strictly increasing.<\/p>\n We are left with showing that $X_{++} = X_{- -} = \\varnothing$.\u00a0 <\/span>Since the arguments for $X_{- -}$ are essentially the same as those for $X_{++}$ with the inequalities reversed, we will write out the details only for $X_{++}$.<\/p>\n Claim<\/b> 2: For each $x \\in I$ there is some $y \\in I$ with $x < y$ and $f(y) < f(x)$.<\/p>\n Proof <\/b>of Claim 2:\u00a0 <\/span>If not, then, because $f$ is injective on $I$, for $x < y$ in $I$ we have $f(x) < f(y)$.\u00a0 <\/span>That is, $f$ would be strictly increasing on $I$\u00a0 <\/span>contrary to the hypothesis that for $y \\in I$ there is $a < y$ so that for $a < x < y$ we have $f(x) > f(y)$. $\\maltese$<\/p>\n Define $\\beta:I \\to I$ by $x \\mapsto \\inf \\{ y \\in I : y > x ~\\&~ f(y) < f(x) \\}$.\u00a0 <\/span>Since for each $x \\in I$ there is some $b > x$ for which $x < y < b$ implies $f(y) > f(x)$, we have $x <\u00a0 <\/span>b \\leq \\beta(x)$. \u00a0 <\/span>Thus, the range of $\\beta$ is infinite as for any $x \\in I$ it contains the infinite sequence of points $\\beta(x) < \\beta^2(x) < \\ldots < \\beta^n(x) < \\ldots$.\u00a0 \u00a0In particular, this shows that for every $x \\in I$ the definable set $\\{ y \\in I : y > x ~\\&~ f(y) < f(x) \\}$ is infinite, and, hence, contains an interval.\u00a0 Define $H(x) := \\{ y \\in I : (\\exists b)[b > y ~\\&~ (\\forall t) (y < t < b \\to f(t) < f(x) )]\u00a0 \\}$ and let $\\gamma:I \\to I$ be given by\u00a0 $x \\mapsto \\inf H(x)$.\u00a0 Note that because $I \\subseteq X_{++}$, $\\gamma(x) \\in H(x)$.\u00a0 \u00a0Since the range of $\\beta$ is cofinal in $I$, so is the range of $\\gamma$.\u00a0 <\/span>\u00a0\u00a0<\/span>Hence, the range of $\\gamma$ contains an interval $J$ which is cofinal in $I$.<\/p>\n Let us define $$Y_{+-} := \\{ y \\in I : (\\exists z)(\\exists a < y)(\\exists b > y)(\\forall u)(\\forall v)(a < u < y < v < b \\to [f(u) > z ~\\&~ f(v) < z]) \\}$$<\/p>\n <\/p>\n Claim<\/b> 3:\u00a0 <\/span>$J \\subseteq Y_{+-}$<\/p>\n Proof<\/b> of Claim 3: \u00a0 <\/span>If $y \\in J$, then there is some $x \\in I$ with $y = \\gamma(x)$.\u00a0 <\/span>Let $z = f(x)$.\u00a0 Let $b > y$ witness the existential quantifier showing that $y \\in H(x)$.\u00a0 Since $y = \\inf\u00a0 H(x)$ and $f(t) > f(x)$ for $t$ just to the right of $x$, we have $y > x$.\u00a0 Since $y$ is the least point at which there is an interval to the right of $y$ on which $f$ is smaller than $f(x)$,\u00a0 there is some $a < y$ so that for $u \\in (a,y)$ we have $f(u) > f(x)$.\u00a0 These choices of $a$ and $b$ witness that $y \\in Y_{+-}$. $\\maltese$<\/p>\n We now reverse the above analysis.\u00a0 That is, again possibly restricting to a coinitial interval we may assume to for each $x \\in J$ there is some $y \\in J$ with $y < x$ and $f(y) < f(x)$.\u00a0 \u00a0As above, it follows that we may define $$G(x) := \\{ y \\in J : y < x ~\\&~ (\\exists a < y)(\\forall u) [a < u < y \\to f(u) < f(x) ] \\}$$ always obtaining a nonempty definable set.\u00a0 We set $\\alpha(x) := \\sup G(x)$ and see that there is a cointial interval $K$ in the range of $K$.\u00a0 As with Claim 3, we see that\u00a0 $K \\subseteq Y_{-+}$ where $$Y_{-+} := \\{ y \\in I : (\\exists z)(\\exists a < y)(\\exists b > y)(\\forall u)(\\forall v)(a < u < y < v < b \\to [f(u) < z ~\\&~ f(v) > z]) \\}$$<\/p>\n However, $K \\subseteq Y_{+-}$ and $Y_{+-} \\cap Y_{-+} = \\varnothing$.\u00a0 \u00a0 With this contradiction we conclude. $\\Box$<\/p>\n <\/p>\n To finish the proof of the Monotonicity Theorem we need only prove Lemma 3.<\/p>\n Proof<\/strong> of Lemma 3:\u00a0 Without loss of generality, we may assume that $f$ is strictly increasing.\u00a0 The argument for the case where $f$ is strictly decreasing is essentially the same with the inequalities reverses.<\/p>\n Since $f:I \\to R$ is strictly increasing, it is, in particular, injective and its range is definable and infinite, and, thus, contains an interval $J$.\u00a0 Let $x \\in f^{-1} J$ and consider $\\epsilon_1 < f(x) < \\epsilon_2$.\u00a0 \u00a0Let $\\epsilon_1′ \\in J$ and $\\epsilon_2′ \\in J$ with $\\epsilon_1 < \\epsilon_1′ < f(x) < \\epsilon_2′ < \\epsilon_2$.\u00a0 Let $\\delta_1 := f^{-1} (\\epsilon_1′)$ and $\\delta_2 := f^{-1} (\\epsilon_2′)$.\u00a0 Then for $u \\in (\\delta_1,\\delta_2)$ because $f$ is strictly increasing, we have $\\epsilon_1 < \\epsilon_1′ = f(\\delta_1) < f(u) < f(\\delta_2) = \\epsilon_2′ < \\epsilon_2$.\u00a0 Thus, $f$ is continuous on $f^{-1} J$, which being an infinite definable subset of $R$ contains an interval.\u00a0 (In fact, it is an interval itself.) $\\maltese$<\/p>\n <\/p>\n <\/p>\n <\/p>\n \n","protected":false},"excerpt":{"rendered":" A basic principle in tame geometry is that there are no pathological definable functions in o-minimal structures.\u00a0 \u00a0One precise sense in which this principle is true is given by the Monotonicity Theorem. Monotonicity Theorem :\u00a0 Let $f: R \\to R$ … Continue reading