Analysis in a Formal Predicative Set Theory

Abstract

We present correct and natural development of fundamental analysis in a predicative set theory we call ${\mathsf {PZF}^{\mathsf {U}}}$. This is done by using a delicate and careful choice of those Dedekind cuts that are adopted as real numbers. ${\mathsf {PZF}^{\mathsf {U}}}$ is based on ancestral logic rather than on first-order logic. Its key feature is that it is definitional in the sense that every object which is shown in it to exist is defined by some closed term of the theory. This allows for a very concrete, computationally-oriented model of it, and makes it very suitable for MKM (Mathematical Knowledge Management) and ITP (Interactive Theorem Proving). The development of analysis in ${\mathsf {PZF}^{\mathsf {U}}}$ does not involve coding, and the definitions it provides for the basic notions (like continuity) are the natural ones, almost the same as one can find in any standard analysis book.

A Appendix

Proof

(proof of Theorem 2). We prove the case of $\mathsf {PZF}$ (the case of ${\mathsf {PZF}^{\mathsf {U}}}$ is identical). In the following proof we argue in $\mathsf {PZF}$ .

Define first the formula $\varphi$ to be:

$$\begin{aligned}&\big (f=\emptyset \wedge g = \{\langle 0,t_i \rangle \} \big )\ \vee \\&\big ( f\ne \emptyset \wedge g = f \cup \{\langle {\mathsf {dom}} \langle f \rangle ,t_s\{{\mathsf {dom}} \langle f \rangle /n ,f({\mathsf {dom}} \langle f \rangle -1)/p\} \rangle \} \big ) \end{aligned}$$

Informally, we set g to be the finite sequence achieved from the finite sequence f by expanding the domain of f by 1 according to the “step function” $t_s$. Obviously $\varphi \succ _{{\mathsf {PZF}^{\mathsf {U}}}}\{g\}$. Let

$$ X=\{ g \mid g=\{\langle 0,t_i \rangle \} \vee \exists f. f = \emptyset \wedge ({\mathsf {TC}}_{f,g}\varphi )(f,g)\} $$

and define $t_\mathsf {rec}= \cup X$. Obviously $t_\mathsf {rec}$ is a term of $\mathsf {PZF}$.

Let $x_1 ,\ldots ,x_n$ be n sets. We observe first that for all $g \in X$ it holds that:

$$\begin{aligned} \mathsf {seqFin}[g] \wedge g(0)=t_i \wedge \forall n \in {\mathsf {dom}} \langle g \rangle . n>0 \rightarrow g(n)=t_s\{g(n-1)/p\} \end{aligned}$$

(5)

The proof is by a straightforward application of the induction rule for the ${\mathsf {TC}}$ operator.

Next we prove that ${\mathsf {func}} [t_\mathsf {rec}]$. Since every $g \in X$ is a function s.t. ${\mathsf {dom}} \langle g \rangle \subseteq {\mathbb {N}}$, it is obvious that ${\mathsf {rel}} [t_\mathsf {rec}]$, and that ${\mathsf {dom}} \langle t_\mathsf {rec} \rangle \subseteq {\mathbb {N}}$. Therefore it suffices to show by induction on n that

$$ \forall n \in {\mathbb {N}}\, \forall g,g' \in X. n \in {\mathsf {dom}} \langle g \rangle \cap {\mathsf {dom}} \langle g' \rangle \rightarrow g(n) = g'(n) $$

If $n=0$ then by (5), $g(0) = g'(0) = t_i$. Assume that $n>0$. Since $n \in {\mathsf {dom}} \langle g \rangle \cap {\mathsf {dom}} \langle g' \rangle$ and since $\mathsf {seqFin}[g] \wedge \mathsf {seqFin}[g']$, $n-1 \in {\mathsf {dom}} \langle g \rangle \cap {\mathsf {dom}} \langle g' \rangle$. By the induction hypothesis, $g(n-1) = g'(n-1)$, hence by (5), $g(n) = g'(n) = t_s$, and overall $t_\mathsf {rec}$ is a function.

We prove now by induction on n that $\forall n\in {\mathbb {N}}. n\in {\mathsf {dom}} \langle t_\mathsf {rec} \rangle$ (and hence ${\mathsf {dom}} \langle t_\mathsf {rec} \rangle ={\mathbb {N}}$). Let $g_0=\{\langle 0,t_i \rangle \}$. Since $\varphi \{\emptyset /f,g_0/g\}$, it holds that $g_0\in X$ and hence $0 \in {\mathsf {dom}} \langle t_\mathsf {rec} \rangle$. Assume that $n \in {\mathsf {dom}} \langle t_\mathsf {rec} \rangle$. Then there exists $g \in X$ s.t. $n \in {\mathsf {dom}} \langle g \rangle$. If $n+1 \in {\mathsf {dom}} \langle g \rangle$ we are done. Otherwise, let $g' :=g \cup \{\langle n+1 , t_s\langle n+1 , g(n) , \bar{x} \rangle \rangle \}$. Obviously $\varphi \{g/f,g'/g\}$, hence $g' \in X$, and hence $n+1 \in {\mathsf {dom}} \langle g \rangle$ as desired.

Overall we proved (1) ($t_\mathsf {rec}$ is a function with domain ${\mathbb {N}}$). Proving (2) is done by (5) and a straightforward ${\mathsf {TC}}$-induction.

Proof

(part of the proof of Proposition 2). We sketch the proof for $\mathsf {cauchySeq}[a] \rightarrow \mathsf {converge}[a]$. So assume that $\mathsf {cauchySeq}[a]$. It is straightforward to show that a is bounded. We prove that $\mathsf {liminf}\langle a \rangle =\mathsf {limsup}\langle a \rangle$. It is straightforward to show that $\mathsf {liminf}\langle a \rangle \subseteq \mathsf {limsup}\langle a \rangle$ (even without any assumption on a). Assume for contradiction that $\exists q \in \mathsf {limsup}\langle a \rangle \setminus \mathsf {liminf}\langle a \rangle$. Since $\mathsf {dedCut}[\mathsf {limsup}\langle a \rangle ]$ and $\mathsf {dedCut}[\mathsf {liminf}\langle a \rangle ]$ it also holds that

$$\begin{aligned} \exists q,q',q'' \in {\mathbb {Q}}. q<q'<q'' \wedge q,q',q'' \in \mathsf {limsup}\langle a \rangle \setminus \mathsf {liminf}\langle a \rangle . \end{aligned}$$

Let $\varepsilon = q''-q'$. Since $\mathsf {cauchySeq}[a]$, there exists $N \in {\mathbb {N}}$ s.t.

$$\begin{aligned} \forall m,k \ge N. |a(m)-a(k)| < \varepsilon . \end{aligned}$$

(6)

Since $q' \notin \mathsf {liminf}\langle a \rangle$, there exists $k' \ge N$ s.t. $q' \notin a(k')$, and hence $a(k') \le q < q'$. Since $q'' \in \mathsf {limsup}\langle a \rangle$, there exists $k'' \ge N$ s.t. $q'' \in a(k'')$, and hence $q'' < a(k'')$. Putting it together, it holds that

$$\begin{aligned} a(k')< q'< q'' < a(k'') \end{aligned}$$

Hence $|a(k'')-a(k')| > q''-q' = \varepsilon$, that contradicts (6).

Proof

(proof of Proposition 3 ).

By Proposition 2-(1) it follows that $\mathsf {converge}[a,\mathsf {limit}\langle a \rangle ]$. Since the term $\mathsf {limit}\langle \cdot \rangle$ is a term of $\mathcal{{L}}_\mathsf {PZF}$), and since $a \in \mathsf {U}$, by the $\mathsf {U}$-closure scheme it holds that $\mathsf {limit}\langle a \rangle \in \mathsf {U}$. Hence not just $\mathsf {dedCut}[\mathsf {limit}\langle a \rangle ]$ (Proposition 1), but also $\mathsf {limit}\langle a \rangle \in {\mathbb {R}}$, and the claim follows.

1.1 A.1 Uniformly Convergence and Power Series

In this section we want to talk about sequences of (real) functions. We use the standard procedure of Currying. Define the following:

Definition 13

( ${\mathsf {PZF}^{\mathsf {U}}}$ )(sequence of functions).

• ${\mathsf {seqFunc}}[F] :={\mathsf {func}} [F] \wedge {\mathsf {dom}} [F]={\mathbb {N}}\times {\mathbb {R}}\wedge {\mathsf {rng}} \langle F \rangle \subseteq {\mathbb {R}}$.

• ${\mathsf {uniformlyConvergence}}[F,f] :={\mathsf {seqFunc}}[F] \wedge {\mathsf {func}} [f,{\mathbb {R}},{\mathbb {R}}] \wedge \forall \varepsilon \in {\mathbb {R}}^+ \, \exists N \in {\mathbb {N}}\, \forall x \in {\mathbb {R}}\, \forall n \ge N. |F(n,x)-f(x)| < \varepsilon$.

Proposition 6

(${\mathsf {PZF}^{\mathsf {U}}}$). Assume that ${\mathsf {uniformlyConvergence}}[F,f]$. Then:

• 1. $\forall x \in {\mathbb {R}}. \mathsf {converge}[\lambda n \in {\mathbb {N}}. F(n,x),f(x)]$.

• 2. $\forall x \in {\mathbb {R}}. \mathsf {limit}\langle \lambda n \in {\mathbb {N}}.F(n,x) \rangle = f(x)$.

Proof

• 1. Let $x \in {\mathbb {R}}$. From ${\mathsf {uniformlyConvergence}}[F,f]$ it follows that $\forall \varepsilon \in {\mathbb {R}}^+ \, \exists N \in {\mathbb {N}}\, \forall x \in {\mathbb {R}}\, \forall n \ge N. |F(n,x)-f(x)| < \varepsilon$. Hence $\forall \varepsilon \in {\mathbb {R}}^+ \, \exists N \in {\mathbb {N}}\, \forall n \ge N. |F(n,x)-f(x)| < \varepsilon$, and hence $\mathsf {converge}[\lambda n \in {\mathbb {N}}. F(n,x),f(x)]$ as desired.

• 2. It follows immediately from previous item and Proposition 2-(2).

Proposition 7

(${\mathsf {PZF}^{\mathsf {U}}}$). Assume that: ${\mathsf {uniformlyConvergence}}[F,f]$, ${\mathsf {Ufunc}}[F]$, and that $\forall n \in {\mathbb {N}}. {\mathsf {continuous}}[\lambda x\in {\mathbb {R}}.F(n,x)]$. Then ${\mathsf {continuous}}[f]$.

Proof

For every $x\in {\mathbb {R}}$, condition (3) of Definition 11 is proved in the usual way. It remains to prove that ${\mathsf {Ufunc}^{\mathbb {R}}}[f]$ holds. Since ${\mathsf {Ufunc}}[F]$, and since for every $x \in {\mathbb {R}}$, ${\mathbb {N}}\times \{x\} \in \mathsf {U}$, it follows that $F \upharpoonright {\mathbb {N}}\times \{x\} \in \mathsf {U}$ and hence for every $x\in {\mathbb {R}}$,

$$\begin{aligned} \lambda n \in {\mathbb {N}}.F(n,x) = \lambda n \in {\mathbb {N}}.(F \upharpoonright {\mathbb {N}}\times \{x\})(n,x) \in \mathsf {U}. \end{aligned}$$

(7)

Let $A \in {\mathsf {P}^{{\mathbb {R}}}}$. By Proposition 6-(2),

$$ f \upharpoonright A = \lambda x \in A.f(x) = \lambda x \in A. \mathsf {limit}\langle \lambda n \in {\mathbb {N}}.F(n,x) \rangle $$

and by Proposition 6-(1) for every $x \in A$, the sequence $\lambda n \in {\mathbb {N}}.F(n,x)$ converges. By (7), for every $x \in A$, $\lambda n \in {\mathbb {N}}.F(n,x) \in \mathsf {U}$, and since $\mathsf {limit}\langle \cdot \rangle$ maps convergent sequences from $\mathsf {U}$ to real numbers (in $\mathsf {U}$), we conclude that

$$ \lambda x \in A. \mathsf {limit}\langle \lambda n \in {\mathbb {N}}.F(n,x) \rangle \in \mathsf {U}$$

and the claim follows.

Remark 8

To ease notations, Definition 13 and hence Proposition 7 deal only with sequences of functions from ${\mathbb {R}}$ to ${\mathbb {R}}$. However, it can easily be extended to deal with sequences of functions from open segments to ${\mathbb {R}}$.

Definition 14

(${\mathsf {PZF}^{\mathsf {U}}}$)(Series). Define the following term that maps sequences $a \in \mathsf {U}$ such that $\mathsf {converge}[\lambda n\in {\mathbb {N}}.\sum _{k=0}^{n} a(k)]$ to their sum:

$$ \mathsf {series}\langle a \rangle :=\mathsf {limit}\langle \lambda n\in {\mathbb {N}}.\sum _{k=0}^{n} a(k) \rangle .$$

We denote $\mathsf {series}\langle a \rangle$ by the usual convention $\sum _{k=0}^{\infty } a(k)$.

Proposition 8

( ${\mathsf {PZF}^{\mathsf {U}}}$ )(Power Series).

Let $a \in \mathsf {U}$ be a sequence, and let $c \in {\mathbb {R}}^+$. If $\mathsf {converge}[\lambda n \in {\mathbb {N}}.|a(n)| \cdot c^n]$, then $\forall x\in (-c,c). \mathsf {converge}[\lambda n\in {\mathbb {N}}.\sum _{k=0}^{n} a(k) \cdot x^k]$, and $\lambda x \in (-c,c).\sum _{k=0}^{\infty } a(k) \cdot x^k$ is continuous (on $(-c,c)$).

Proof

(proof of Proposition 8 ).

The proof that $\forall x\in (-c,c). \mathsf {converge}[\lambda n\in {\mathbb {N}}.\sum _{k=0}^{n} a(k) \cdot x^k]$ is standard hence omitted. Define the functions $f :=\lambda x \in (-c,c).\sum _{k=0}^{\infty } a(k) \cdot x^k$, and $F = \lambda n \in {\mathbb {N}}, \lambda x \in (-c,c).\sum _{k=0}^{n} a(k) \cdot x^k$. We wish to apply Proposition 7 to F and f, and conclude that ${\mathsf {continuous}}[f]$.

The proof for ${\mathsf {uniformlyConvergence}}[F,f]$ is also standard and omitted. By Proposition 5 and straightforward induction, it holds that

$$ \forall n\in {\mathbb {N}}. {\mathsf {continuous}}[\lambda x \in (-c,c).\sum _{k=0}^{n} a(k) \cdot x^k]. $$

It remains to prove that ${\mathsf {Ufunc}}[F]$. Let $X \in \mathsf {U}$ s.t. $X \subseteq {\mathbb {N}}\times {\mathbb {R}}$. By the closure properties of $\mathsf {U}$, it is obvious that $\big (\lambda \langle n,r \rangle \in X. \sum _{k=0}^{n} a(k) \cdot x^k \big ) \in \mathsf {U}$, and the claim follows.

Corollary 2

(${\mathsf {PZF}^{\mathsf {U}}}$). Let $a \in \mathsf {U}$ be a sequence of rational numbers. If for every $c \in {\mathbb {R}}^+$, $\mathsf {converge}[\lambda n \in {\mathbb {N}}.|a(n)| \cdot c^n]$, then $\forall x\in {\mathbb {R}}. \mathsf {converge}[\lambda n\in {\mathbb {N}}.\sum _{k=0}^{n} a(k) \cdot x^k]$, and $\lambda x \in {\mathbb {R}}.\sum _{k=0}^{\infty } a(k) \cdot x^k$ is continuous (on ${\mathbb {R}}$).

Proof

It is straightforward from previous proposition.

1.2 A.2 Elementary Functions

Define the following sequence:

$$ a^{\sin } = \lambda n\in {\mathbb {N}}. \left\{ \begin{array}{@{}ll} \frac{1}{n!} &{} \text { if } n \equiv 1 \pmod 4 \\ -\frac{1}{n!} &{} \text { if } n \equiv 3 \pmod 4 \\ 0 &{} \text { else } \\ \end{array} \right. $$

It is straightforward to show that for every $c \in {\mathbb {R}}^+$, $\mathsf {converge}[\lambda n \in {\mathbb {N}}.|a^{\sin }(n)| \cdot c^n]$. Hence by Corollary 2 the following function is continuous:

$$ \sin = \lambda x \in {\mathbb {R}}. \sum _{k=0}^{\infty } a^{\sin }(k) \cdot x^k $$

Similarly we can define the other trigonometric functions, the inverse trigonometric functions, the hyperbolic/inverse hyperbolic functions, the exponentiation, and the logarithmic functions. Using Corollary 2 or Proposition 8, we can then prove that those functions are continuous. (Note that not all of these functions have domain ${\mathbb {R}}$.) Next, we define the elementary functions to be all the functions obtained by adding, subtracting, multiplying, dividing, and composing any of the previously mentioned functions. By Proposition 5, we conclude that every elementary function is a continuous function.