Diagonalization Fixed Point Exercises

Scott Garrabrant; SamEisenstat

Fixed Points

Diagonalization Fixed Point Exercises

by Scott Garrabrant, Sam Eisenstat

4 min read18th Nov 201813 comments

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Fixed Point Theorems

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Diagonalization Fixed Point Exercises

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This is the second of three sets of fixed point exercises. The first post in this sequence is here, giving context.

Recall Cantor’s diagonal argument for the uncountability of the real numbers. Apply the same technique to convince yourself than for any set $S$ , the cardinality of $S$ is less than the cardinality of the power set $P (S)$ (i.e. there is no surjection from $S$ to $P (S)$ ).
Suppose that a nonempty set $T$ has a function $f$ from $T$ to $T$ which lacks fixed points (i.e. $f (x) \neq x$ for all $x \in T$ ). Convince yourself that there is no surjection from S to $S \to T$ , for any nonempty $S$ . (We will write the set of functions from $A$ to $B$ either as $A \to B$ or $B^{A}$ ; these are the same.)
For nonempty $S$ and $T$ , suppose you are given $g : S \to T^{S}$ a surjective function from the set $S$ to the set of functions from $S$ to $T$ , and let $f$ be a function from $T$ to itself. The previous result implies that there exists an $x$ in $T$ such that $f (x) = x$ . Can you use your proof to describe $x$ in terms of $f$ and $g$ ?
Given sets $A$ and $B$ , let $C o m p (A, B)$ denote the space of total computable functions from $A$ to $B$ . We say that a function from $C$ to $C o m p (A, B)$ is computable if and only if the corresponding function $f^{'} : C \times A \to B$ (given by $f^{'} (c, a) = f (c) (a))$ is computable. Show that there is no surjective computable function from the set $S$ of all strings to $C o m p (S, {T, F})$ .
Show that the previous result implies that there is no computable function $h a l t (x, y)$ from $S \times S \to {T, F}$ which outputs $T$ if and only if the first input is a code for a Turing machine that halts when given the second input.
Given topological spaces $A$ and $B$ , let $C o n t (A, B)$ be the space with the set of continuous functions from $A$ to $B$ as its underlying set, and with topology such that $f : C \to C o n t (A, B)$ is continuous if and only if the corresponding function $f^{'} : C \times A \to B$ (given by $f^{'} (c, a) = f (c) (a)$ ) is continuous, assuming such a space exists. Convince yourself that there is no space $X$ which continuously surjects onto $C o n t (X, S)$ , where $S$ is the circle.
In your preferred programming language, write a quine, that is, a program whose output is a string equal to its own source code.
Write a program that defines a function $f$ taking a string as input, and produces its output by applying $f$ to its source code. For example, if $f$ reverses the given string, then the program should outputs its source code backwards.
Given two sets $A$ and $B$ of sentences, let $S y n (A, B)$ be the set of all functions from $A$ to $B$ defined by substituting the Gödel number of a sentence in $A$ into a fixed formula. Let $S_{0}$ be the set of all sentences in the language of arithmetic with one unbounded universal quantifier and arbitrarily many bounded quantifiers, and let $S_{1}$ be the set of all formulas with one free variables of that same quantifier complexity. By representing syntax using arithmetic, it is possible to give a function $f \in S y n (S_{1} \times S_{1}, S_{0})$ that substitutes its second argument into its first argument. Pick some coding of formulas as natural numbers, where we denote the number coding for a formula $φ$ as $┌ φ ┐$ . Using this, show that for any formula $ϕ \in S_{1}$ , there is a formula $ψ \in S_{0}$ such that $ϕ (┌ ψ ┐) \leftrightarrow ψ$ .
(Gödel's second incompleteness theorem) In the set $S_{1}$ , there is a formula $\neg B e w$ such that $\neg B e w (┌ ψ ┐)$ holds iff the sentence $ψ$ is not provable in Peano arithmetic. Using this, show that Peano arithmetic cannot prove its own consistency.
(Löb's theorem) More generally, the diagonal lemma states that for any formula $ϕ$ with a single free variable, there is a formula $ψ$ such that, provably, $ϕ (┌ ψ ┐) \leftrightarrow ψ$ . Now, suppose that Peano arithmetic proves that $B e w (ψ) \to ψ$ for some formula $ψ$ . Show that Peano arithmetic also proves $ψ$ itself. Some facts that you may need are that (a) when a sentence $ψ$ is provable, the sentence $B e w (ψ)$ is itself provable, (b) Peano arithmetic proves this fact, that is, Peano arithmetic proves $B e w (ψ) \to B e w (B e w (ψ))$ , for any sentence $ψ$ and (c) Peano arithmetic proves the fact that if $χ$ and $χ \to ζ$ are provable, then $ζ$ is provable.
(Tarski's theorem) Show that there does not exist a formula $ϕ$ with one free variable such that for each sentence $ψ$ , the statement $ϕ (┌ ψ ┐) \leftrightarrow ψ$ holds.
Looking back at all these exercises, think about the relationship between them.

Please use the spoilers feature - the symbol '>' followed by '!' followed by space -in your comments to hide all solutions, partial solutions, and other discussions of the math. The comments will be moderated strictly to hide spoilers!

I recommend putting all the object level points in spoilers and including metadata outside of the spoilers, like so: "I think I've solved problem #5, here's my solution <spoilers>" or "I'd like help with problem #3, here's what I understand <spoilers>" so that people can choose what to read.

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Diagonalization Fixed Point Exercises

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[-]Adele Lopez5y4

Ex 8: (in Python, using a reversal function)

def f(s):
return s[::-1]

dlmt = '"""'
code = """def f(s):
return s[::-1]

dlmt = '{}'
code = {}{}{}
code = code.format(dlmt, dlmt, code, dlmt)
print(f(code))"""
code = code.format(dlmt, dlmt, code, dlmt)
print(f(code))

[-]Rafael Harth5y4

Ex 4

Given a computable function $ϕ : S \mapsto Comp (S, {T, F})$ , define a function $f : S \mapsto {T, F}$ by the rule $f (s) = {\begin{matrix} T & ϕ (s) (s) = F F & ϕ (s) (s) = T \end{matrix}$ . Then $f$ is computable, however, $f \notin ϕ (S)$ because for $s \in S$ , we have that $ϕ (s) (s) = F ⟹ f (s) = T ⟹ ϕ (s) \neq f$ and $ϕ (s) (s) = T ⟹ f (s) = F ⟹ ϕ (s) \neq f$ .

Ex 5:

We show the contrapositive: given a function halt, we construct a surjective function $f$ from $S$ to $Comp (S, {T, F})$ as follows: enumerate all turing machines, such that each corresponds to a string. Given a $t \in S$ , if $t$ does not decode to a turing machine, set $f (t) \equiv F$ . If it does, let $[t]$ denote that turning machine. Let $f (t)$ with input $s \in S$ first run halt $(t, s)$ ; if halt returns $F$ , put out $F$ , otherwise $[t]$ will halt on input $s$ ; run $[t]$ on $s$ and put out the result.

Given a computable function $g : S \mapsto {T, F}$ , there is a string $t^{*} \in S$ such that $[t^{*}]$ implements $g$ (if the turing thesis is true). Then $g = f (t^{*})$ , so that $f$ is surjective.

Ex 6:

Let $h : R \mapsto S$ be a parametrization of the circle given by $h (r) \mapsto (cos (r), sin (r))$ . Given $p \in S$ and $r \in R$ , write $p + r$ to denote the point $h (p^{'} + r)$ , where $p^{'} \in h^{- 1} ({p})$ . First, note that, regardless of the topology on $X$ , it holds true that if $f : X \mapsto S$ is continuous, then so is $f_{r} : x \mapsto f (x) + r$ for any $r \in R$ , because given a basis element $(h (a), h (b))$ of the circle, we have $f_{r}^{- 1} (B) = f^{- 1} (h (a) - r, h (b) - r)$ which is open because $f$ is continuous.

Let $ϕ$ be a continuous function from $X$ to $Cont (X, S)$ . Then $ϕ^{'} : X \times X \mapsto S$ is continuous, and so is the diagonal function $h : x \mapsto ϕ (x, x)$ . Fix any $r \in (0, 2 π)$ , then $f : X \mapsto S$ given by $f : x \mapsto (ϕ^{'} (x, x) + r))$ is also continuous, but given any $x \in X$ , one has $ϕ (x) (x) = ϕ^{'} (x, x) \neq ϕ^{'} (x, x) + r = f (x)$ and thus $ϕ (x) \neq f$ . It follows that $ϕ$ is not surjective.

Ex 7:

I did it in java. There's probably easier ways to do this, especially in other languages, but it still works. It was incredibly fun to do. My basic idea was to have a loop iterate 2 times, the first time printing the program, the second time printing the printing command. Escaping the " characters is the biggest problem, the main idea here was to have a string q that equals " in the first iteration and " + q + " in the second. That second string (as part of the code in an expression where a string is printed) will print itself in the console output. Code:

package maths;public class Quine{public static void main(String[]args){for(int i=0;i<2;i++){String o=i==1?""+(char)34:"";String q=""+(char)34;q=i==1?q+"+q+"+q:q;String e=i==1?o+"+e);}}}":"System.out.print(o+";System.out.print(o+"package maths;public class Quine{public static void main(String[]args){for(int i=0;i<2;i++){String o=i==1?"+q+""+q+"+(char)34:"+q+""+q+";String q="+q+""+q+"+(char)34;q=i==1?q+"+q+"+q+"+q+"+q:q;String e=i==1?o+"+q+"+e);}}}"+q+":"+q+"System.out.print(o+"+q+";"+e);}}}

[-]Czynski5y3

For #6, I have what looks to me like a counterexample. Possibly I am using the wrong definition of continuous function? I am taking this one as my source.

Take as the topological space $X$ the real line under the Euclidean topology. Let $f^{'} (x, y)$ be the point in $S_{1}$ at $(x + y)$ radians rotation. This is surjective; all points in $S_{1}$ are mapped to infinitely many times. It is also continuous; For every $(x, y) \in X, X$ and neighborhood $N (f^{'} (x, y))$ there is a neighborhood $M (x, y)$ such that $\forall p \in M$ , $f (p) \in N$

The partial functions f(x) from $X \to S_{1}$ are also continuous by the same argument.

[-]Adele Lopez5y2

Currying doesn't preserve surjectivity. As a simple example, you can easily find a surjective function $2 \times 2 \to 2$ , but there are no surjective functions $2 \to 2^{2}$ .

[-]Czynski5y2

Ah, yes, that makes sense. I got distracted by the definition of $C o n t (A, B)$ 's topology

and applied it to surjectivity as well as continuity.

[-]Gurkenglas5y3

Let f be such a surjection. Construct a member of P(S) outside f's image by differing from each f(x) in whether it contains x.

A nonempty set has functions without a fixed point iff it has at least two elements. It suffices to show that there is no surjection from S to S -> 2, which is analogous to #1.

T has only one element. Use that one.

#7 Haskell

source = "main = putStrLn (\"source = \" ++ show source ++ \"\\n\" ++ source)"
main = putStrLn ("source = " ++ show source ++ "\n" ++ source)

Is #8 supposed to read "Write a program that takes a function f taking a string as input as input, and produces its output by applying f to its source code. For example, if f reverses the given string, then the program should outputs its source code backwards."?

If so, here.

source = "onme = putStrLn $ f $ \"source = \" ++ show source ++ \"\\n\" ++ source"
onme f = putStrLn $ f $ "source = " ++ show source ++ "\n" ++ source

[-]Rafael Harth5y3

Ex 1

Exercise 1: Let $f : S \mapsto P (S)$ and let $A = {x \in S | x \notin f (x)}$ . Suppose that $A \in f (S)$ , then let $x \in S$ be an element such that $f (x) = A$ . Then $x \notin f (x) = A ⟹ x \in A$ by definition, and $x \in A = f (x) ⟹ x \notin A$ . So $x \notin A ⟺ x \in A$ , a contradiction. Hence $A \notin f (S)$ , so that $f$ is not surjective.

Ex 2

Exercise 2: Since $T$ is nonempty, it contains at least one element $t_{0}$ . Let $h : T \mapsto T$ be a function without a fixed point, then $t_{0} \neq h (t_{0}) =: t_{1}$ , so that $t_{0}$ and $t_{1}$ are two different elements in $T$ (this is the only thing we shall use the function $h$ for).

Let $Ψ : S \mapsto T^{S}$ for $S$ nonempty. Suppose by contradiction that $Ψ$ is surjective. Define a map $f : S \mapsto P (S)$ by the rule $f (x) = {s \in S | Ψ (x) (s) = t_{0}}$ . Given any subset $U \subset S$ , let $g : S \mapsto T$ be given by $g : x \mapsto {\begin{matrix} t_{0} & x \in U t_{1} & x \notin U \end{matrix}$ Since $Ψ$ is surjective, we find a $s^{*} \in S$ such that $Ψ (s^{*}) = g$ . Then $f (s^{*}) = U$ . This proves that $f$ is surjective, which contradicts the result from (a).

Ex 3

Exercise 3: By (2) we know that $T = {x}$ , and so $f = {(x, x)}$ and $g = {(s, h) | s \in S}$ where $h = {(s, x) | s \in S}$ . That means $x = g (s) (s^{'}) = f^{n} (g (s) (s^{'}))$ for any $s, s^{'} \in S$ . and $n \in N$ .

[-]James Payor5y2

Q7 (Python):

Y = lambda s: eval(s)(s)
Y('lambda s: print("Y = lambda s: eval(s)(s)\\nY({s!r})")')

Q8 (Python):

Not sure about the interpretation of this one. Here's a way to have it work for any fixed (python function) f:

f = 'lambda s: "\\n".join(s.splitlines()[::-1])'

go = 'lambda s: print(eval(f)(eval(s)(s)))'

eval(go)('lambda src: f"f = {f!r}\\ngo = {go!r}\\neval(go)({src!r})"')

[-]Czynski5y2

I am confused by the introductory statement for #9. Is this an accurate rephrasing?

By representing syntax using arithmetic, it is possible to define a function $f$ as follows:

Define $f (s \in S_{1}, t \in S_{1})$ with image in $S_{0}$ , such that:
$f$ substitutes the Goedel-number of $t$ into $s$ (creating $s^{'}$ ) and then substitutes the Goedel-number of $s^{'}$ into some fixed formula to get a result in $S_{0}$ .

[-]Vladimir Mikulik5y2

I’m confused about Q9.

Given the way $S y n (A, B)$ is defined, it’s unclear to me how we ensure type correctness of $S y n (S_{1} \times S_{1}, S_{0})$ . In what sense is $S_{1} \times S_{1}$ a set of sentences (rather than a set of pairs of sentences)? What does an element of that set look like?

[-]Scott Garrabrant5y3

Yeah, it is just functions that take in two sentences and put both their Godel numbers into a fixed formula (with 2 inputs).

[-]Czynski5y1

Minor correction for #7: You probably want to say "nonempty quine" or "nontrivial quine". The trivial quine works in many languages.

[-]Czynski5y1

My nontrivial answer for Q7, in Python:

with open("foo.py", "r") as foo:
print foo.read()

And for Q8:

def f(string):
return ''.join([chr(ord(c)+1) for c in string])

with open("foo.py", "r") as foo:
print f(foo.read())

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