(This post was originally published on Nov 30th 2017, and is 1 of 4 posts brought forwarded today as part of the AI Alignment Forum launch sequence on fixed points.)
1 Introduction
Before the work of Turing, one could justifiably be skeptical of the idea
of a universal computable function.
After all, there is no computable function f:N×N→N such that for all computable g:N→N there is some index ig such that f(ig,n)=g(n) for all n.
If there were, we could pick g(n)=f(n,n)+1, and then
g(ig)=f(ig,ig)+1=g(ig)+1,
a contradiction.
Of course, universal Turing machines don't run into this obstacle; as Gödel
put it, "By a kind of miracle it is not necessary to distinguish orders, and the
diagonal procedure does not lead outside the defined notion." [1]
The miracle of Turing machines is that there is a partial computable function
f:N×N→N∪{⊥} such that for all partial computable g:N→N∪{⊥} there is an index i such that f(i,n)=g(n) for all n.
Here, we look at a different "miracle", that of reflective oracles [2,3].
As we will see in Theorem 1, given a reflective oracle O, there is a (stochastic) O-computable function f:N×N→{0,1} such that for any (stochastic) O-computable function g:N→{0,1}, there is some index i such that f(i,n) and g(n) have the same distribution for all n.
This existence theorem seems to skirt even closer to the contradiction
mentioned above.
We use this idea to answer "in spirit" the converse Lawvere problem posed in [4]. These methods also generalize to prove a similar analogue of the ubiquitous converse Lawvere problem from [5]. The original questions, stated in terms of topology, remain open, but I find that the model proposed here, using computability, is equally satisfying from the point of view of studying reflective agents. Those references can be consulted for more motivation on these problems from the perspective of reflective agency.
Section 3 proves the main lemma, and proves the converse Lawvere theorem for reflective oracles. In section 4, we use that to give a (circular) proof of Brouwer's fixed point theorem, as mentioned in [4]. In section 5, we prove the ubiquitous converse Lawvere theorem for reflective oracles.
2 Definitions
For any measurable space X, the set of probability measures on X is denoted ΔX.
A probabilistic oracle is a map N→[0,1].
A probabilistic oracle machine is a probabilistic Turing machine that can query a probabilistic oracle
O.
On a given query n∈N, it gets the answer 1 with probability O(n) and the answer 0 otherwise.
Different queries are independent events, so this completely determines
the behavior of such a machine.
We denote by ϕOi the stochastic partial O-computable function N→Δ(N∪{⊥}), where ⊥ represents nonhalting, computed by the probabilistic Turing machine with
index i.
The notation ϕOi(n)↓ indicates the event that ϕOi halts on input n, and ϕOi(n)↓=m is the event that ϕOi(n) halts and outputs m.
Finally, ϕOi(n)↑ is the event that ϕOi does not halt on input n.
We use ⟨⋅⟩ to represent a computable coding function, in order to biject arbitrary
countable sets to the naturals for the purpose of computability.
A reflective oracle is a probabilistic oracle O such that for all i,n∈N and p∈[0,1]Q,
P(ϕOi(n)↓=1)>p⟹O(⟨i,n,p⟩)=1P(¬ϕOi(n)↓=0)<p⟹O(⟨i,n,p⟩)=0.
For more on reflective oracles, see Fallenstein et al., 2015 [2].
A function f:N→[0,1] is O-computable if there is an index i such that for all n∈N, we have
P(ϕOi(n)↓∈{0,1})=1
and
P(ϕOi(n)↓=1)=f(n).
That is, ϕOi represents f by probabilistically outputting either 0 or 1.
For any m∈N, a function f:N→[0,1]m is O-computable if each coordinate fj:N→[0,1] for 1≤j≤m is O-computable.
A function f:N→[0,1]N is O-computable if the corresponding function N→[0,1] given by ⟨n,m⟩↦f(n)(m) is O-computable.
For any point p∈[0,1], a probabilistic oracle O is compatible with p if for all r∈[0,1]Q, we have O(⟨r⟩)=0 if p<r and O(⟨r⟩)=1 if p>r.
More generally, for any m∈N and any p∈[0,1]m, a probabilistic oracle O is compatible with p if for all j such that 1≤j≤m and all r∈[0,1]Q, we have O(⟨j,r⟩)=0 if pj<r and O(⟨j,r⟩)=1 if pj>r.
A function f:[0,1]m→[0,1]m is O-computable if for each coordinate fj:[0,1]m→[0,1], there is an index i such that whenever P is compatible with p, we have
P(ϕO,Pi(0)↓∈{0,1})=1
and
P(ϕO,Pi(0)↓=1)=fj(p).
A map f:N→[0,1]N is O-computably ubiquitous if for every O-computable map e:N→[0,1]N, there is some n∈N such that f(n)=e(n).
3 Converse Lawvere property
We first need a lemma that tells us that we can replace certain partial
O-computable functions with total ones when working relative to a reflective
oracle.
As discussed in the introduction, this contrasts strongly with the situation for computable functions.
All of our theorems will make essential use of this lemma.
Lemma 1 (totalizing): There is a computable map τ:N→N such that for all i,n∈N and any reflective oracle O, we have
P(ϕOτ(i)(n)↓∈{0,1})=1
and for b∈{0,1},
P(ϕOτ(i)(n)↓=b)≥P(ϕOi(n)↓=b).
Proof: We construct τ using a recursive procedure that ensures that P(ϕOτ(i)(n)↓=b) upper-bounds P(ϕOi(n)↓=b) using what may be thought of as repeated subdivision or binary search.
This is essentially the same as the function flip in [3], but we handle computability issues differently.
Let S⊆[0,1]×[0,1] be the set of pairs of dyadic rationals (p,q) such that p<q.
We recursively define a stochastic O-computable function t:N×N×S→Δ{0,1}; the intent is to have P(t(i,n,p,q)=1) equal
P(ϕOτ(i)(n)↓=1)−pq−p
if that quantity is in the interval [0,1], and take the closest possible value, either 0 or 1, otherwise.
Then, we will be able to define ϕOτ(i)(n) to be t(i,n,0,1).
Construct t so that a call t(i,n,p,q) first queries O(⟨i,n,r⟩), where r=p+q2, and also flips a fair coin C.
Then, it either outputs 0, 1, or the result of a recursive call, as follows:
t(i,n,p,q)=⎧⎪
⎪
⎪⎨⎪
⎪
⎪⎩0if O(⟨i,n,r⟩)=0 and C=0t(i,n,p,r)if O(⟨i,n,r⟩)=0 and C=11if O(⟨i,n,r⟩)=1 and C=0t(i,n,r,q)if O(⟨i,n,r⟩)=1 and C=1.
We can now choose τ so that ϕOτ(i)(n)=t(i,n,0,1).
This algorithm upper bounds the probabilities P(ϕOi(n)↓=0) and P(ϕOi(n)↓=1) by binary search.
Once the initial call t(i,n,0,1) has recursed to t(i,n,p,q), it has already halted outputting 1 with probability p, confirming that this is an upper bound since it received answer 1 from a call to O(⟨i,n,p⟩).
Similarly, it has output 0 with probability 1−q, confirming this bound since it received the answer 0 from a call to O(⟨i,n,q⟩).
Further, since each call to t halts without recursing with probability 12, t halts almost surely.
Thus, we get the totality property
P(ϕOτ(i)(n)↓∈{0,1})=1
and the bounds
P(ϕOτ(i)(n)↓=b)≥P(ϕOi(n)↓=b)
for b∈{0,1}.
□
Now we can prove our main theorem.
Theorem 1 (converse Lawvere for reflective oracles): Let O be a reflective oracle.
Then, there is an O-computable map f:N→[0,1]N such that for all O-computable g:N→[0,1], there is some index i such that g=f(i).
Proof: Using τ from the totalizing lemma, let
f(i)(n)=P(ϕOτ(i)(n)↓=1).
Given any O-computable g:N→[0,1], there is some i such that
P(ϕOi(n)↓=1)=g(n)P(ϕOi(n)↓=0)=1−g(n).
Then,
f(i)(n)=P(ϕOτ(i)(n)↓=1)≥P(ϕOi(n)↓=1)=g(n)
and similarly 1−f(i)(n)≥1−g(n), so f(i)=g.
□
This theorem gives a computable analogue to the problem posed in [4].
The analogy would be strengthened if we worked in a cartesian closed category
where the present notion of O-computability gave the morphisms, and where [0,1]N is an exponential object.
In addition, the totalizing lemma would have a nice analogue in this setting.
I expect that all this can be done using something like an effective topos
[6], but I leave this for future work.
4 Recovering Brouwer's fixed point theorem
As mentioned in [4], the intermediate value theorem would follow from the
existence of a space with the converse Lawvere property, that is, a space
X that surjects onto the mapping space [0,1]X.
Further, Brouwer's fixed point theorem on the n-ball, Bn, would follow from the existence of a topological space X with a surjection X→(Bn)X.
We can do something similar to conclude Brouwer's fixed point theorem from
the converse Lawvere theorem for reflective oracles.
Of course, this is circular; Kakutani's fixed point theorem, a generalization
of Brouwer's fixed point theorem, is used to prove the existence of reflective
oracles.
Still, it is interesting to see how this can be done.
We start with two technical lemmas telling us some basic facts about reflective-oracle-computable functions [0,1]m→[0,1]m.
Using these, it will be easy to derive Brouwer's fixed point theorem.
Lemma 2 (continuous implies relatively computable): Take m∈N and let h:[0,1]m→[0,1]m be continuous.
Then, there is a (deterministic) oracle O such that h is O-computable.
Proof: For each coordinate hj of h, each rectangle
R=[ℓ1,u1]×⋯×[ℓm,um]⊆[0,1]m
with rational endpoints, and each pair of rationals ℓ0,u0∈[0,1]Q with ℓ0<u0, let O(⟨j,ℓ0,u0,…,ℓm,um⟩) be 1 if hj(R)⊆[ℓ0,u0] and 0 otherwise.
To see that h is O-computable, we compute any hj(p) for any j with 1≤j≤m, and any p∈[0,1]m, making use of O and any oracle P compatible with p.
We proceed by a search process similar to the argument in the totalizing
lemma.
Start with R0=[0,1]m, ℓ00=0 and u00=1.
At each step s, perform an exhaustive search for a rectangle
Rs+1=[ℓs+11,us+11]×⋯×[ℓs+1m,us+1m]⊆Rs
and points ℓs+10,us+10 such that a query to P(⟨k,ℓs+1k⟩) returns 1 for all k, a query to any P(⟨k,us+1k⟩) returns 0, a query to O(⟨j,ℓs+10,us+10,Rs+1⟩) returns 1, and where either ℓs+10=ℓs0 and us+10=13ℓs0+23us0, or ℓs+10=23ℓs0+13us0 and us+10=us0.
In the first case, output 0 with probability 13 and continue with probability 23, and in the second case, output 1 with probability 13 and continue with probability 23.
By construction, p∈Rs and hj(Rs)⊆[ℓs0,us0] at each stage s.
Since hj is continuous, there is some neighbourhood of p on which its image is contained in either [ℓs0,13ℓs0+23us0] or [23ℓs0+13us0,us0].
There are two possibilities to consider.
If p∈intRs, then there is some rectangle R contained in such a neighbourhood of p, and with p∈intR.
This rectangle would be chosen if considered, so the algorithm will move
beyond step s.
The remaining possibility is that p is on the boundary of Rs; say, pk=ℓsk.
Since Rs was chosen previously though, we know that querying P(k,ℓs+1k) has returned 1 at least once, so P(k,ℓs+1k)≠0.
Thus, the algorithm will almost surely eventually accept Rs or another rectangle.
Putting this together, this algorithm almost surely halts, outputting either
0 or 1.
By stage s, it has halted outputting 1 with probability ℓs0 and outputting 0 with probability 1−us0, so overall it outputs 1 with probability hj(p).
Thus, h is O-computable.
□
Lemma 3 (composition): Let O be a reflective oracle and let g:N→[0,1]m and h:[0,1]m→[0,1]m be O-computable.
Then, h∘g:N→[0,1]m is O-computable.
Proof: For each j with 1≤j≤m, take ij∈N such that ϕOij witnesses the computability of gj.
Then, O(⟨ij,n,r⟩)=0 if gj(n)<r and O(⟨ij,n,r⟩)=1 if gj(n)>r, so O lets us simulate a probabilistic oracle compatible with g(n).
Hence, by the O-computability of h, for each k with 1≤k≤m, we have a probabilistic O-machine that always halts, outputting either 0 or 1, and that on input n outputs 1 with probability hk∘g(n).
□
Theorem 2 (Brouwer's fixed point theorem): Take m∈N and h:[0,1]m→[0,1]m.
Then, h has a fixed point.
Proof: By Lemma 2, we have an oracle O such that h is O-computable.
By relativizing the construction of a reflective oracle [2,3] to O, we get a reflective oracle ˜O above O.
Notice that h is ˜O-computable.
Letting f be the ˜O-computable function
f(i)(n)=P(ϕ˜Oτ(i)(n)↓=1)
constructed in the converse Lawvere theorem for reflective oracles, define
fm:N→([0,1]m)N by
fm(⟨i1,…,im⟩)(n)=(f(i1)(n),…,f(im)(n)).
The rest will now follow the proof of Lawvere's fixed point theorem [7].
Define g:N→[0,1]m by g(n)=h(fm(n)(n)); this is ˜O-computable by Lemma 3.
Now, by converse Lawvere theorem, for each coordinate 1≤j≤m of g, there is some ij∈N such that gj=f(ij).
Letting i=⟨i1,…,im⟩, we have
gj(i)=hj(fm(i)(i))=hj(g(i)),
so g(i)=h(g(i)), and so g(i) is a fixed point of h.
□
5 Ubiquitous converse Lawvere property
Theorem 3 (ubiquitous converse Lawvere): Let O be a reflective oracle.
Then, there is an O-computable, O-computably ubiquitous map f:N→[0,1]N.
Proof: This follows by a combination of the recursion theorem and the totalizing
lemma.
We use the same map f from Theorem 1.
Let e:N→[0,1]N be any O-computable map.
There is a computable map s:N→N such that for all i,n∈N, we have
P(ϕOs(i)(n)↓∈{0,1})=1,
and
e(i)(n)=P(ϕOs(i)(n)↓=1).
By the recursion theorem, there is some i such that ϕOs(i)=ϕOi.
Then,
e(i)(n)=P(ϕOs(i)(n)↓=1)=P(ϕOi(n)↓=1)=P(ϕOτ(i)
(This post was originally published on Nov 30th 2017, and is 1 of 4 posts brought forwarded today as part of the AI Alignment Forum launch sequence on fixed points.)
1 Introduction
Before the work of Turing, one could justifiably be skeptical of the idea of a universal computable function. After all, there is no computable function f:N×N→N such that for all computable g:N→N there is some index ig such that f(ig,n)=g(n) for all n. If there were, we could pick g(n)=f(n,n)+1, and then g(ig)=f(ig,ig)+1=g(ig)+1, a contradiction. Of course, universal Turing machines don't run into this obstacle; as Gödel put it, "By a kind of miracle it is not necessary to distinguish orders, and the diagonal procedure does not lead outside the defined notion." [1]
The miracle of Turing machines is that there is a partial computable function f:N×N→N∪{⊥} such that for all partial computable g:N→N∪{⊥} there is an index i such that f(i,n)=g(n) for all n. Here, we look at a different "miracle", that of reflective oracles [2,3]. As we will see in Theorem 1, given a reflective oracle O, there is a (stochastic) O-computable function f:N×N→{0,1} such that for any (stochastic) O-computable function g:N→{0,1}, there is some index i such that f(i,n) and g(n) have the same distribution for all n. This existence theorem seems to skirt even closer to the contradiction mentioned above.
We use this idea to answer "in spirit" the converse Lawvere problem posed in [4]. These methods also generalize to prove a similar analogue of the ubiquitous converse Lawvere problem from [5]. The original questions, stated in terms of topology, remain open, but I find that the model proposed here, using computability, is equally satisfying from the point of view of studying reflective agents. Those references can be consulted for more motivation on these problems from the perspective of reflective agency.
Section 3 proves the main lemma, and proves the converse Lawvere theorem for reflective oracles. In section 4, we use that to give a (circular) proof of Brouwer's fixed point theorem, as mentioned in [4]. In section 5, we prove the ubiquitous converse Lawvere theorem for reflective oracles.
2 Definitions
For any measurable space X, the set of probability measures on X is denoted ΔX.
A probabilistic oracle is a map N→[0,1]. A probabilistic oracle machine is a probabilistic Turing machine that can query a probabilistic oracle O. On a given query n∈N, it gets the answer 1 with probability O(n) and the answer 0 otherwise. Different queries are independent events, so this completely determines the behavior of such a machine.
We denote by ϕOi the stochastic partial O-computable function N→Δ(N∪{⊥}), where ⊥ represents nonhalting, computed by the probabilistic Turing machine with index i. The notation ϕOi(n)↓ indicates the event that ϕOi halts on input n, and ϕOi(n)↓=m is the event that ϕOi(n) halts and outputs m. Finally, ϕOi(n)↑ is the event that ϕOi does not halt on input n.
We use ⟨⋅⟩ to represent a computable coding function, in order to biject arbitrary countable sets to the naturals for the purpose of computability.
A reflective oracle is a probabilistic oracle O such that for all i,n∈N and p∈[0,1]Q, P(ϕOi(n)↓=1)>p⟹O(⟨i,n,p⟩)=1P(¬ϕOi(n)↓=0)<p⟹O(⟨i,n,p⟩)=0. For more on reflective oracles, see Fallenstein et al., 2015 [2].
A function f:N→[0,1] is O-computable if there is an index i such that for all n∈N, we have P(ϕOi(n)↓∈{0,1})=1 and P(ϕOi(n)↓=1)=f(n). That is, ϕOi represents f by probabilistically outputting either 0 or 1.
For any m∈N, a function f:N→[0,1]m is O-computable if each coordinate fj:N→[0,1] for 1≤j≤m is O-computable.
A function f:N→[0,1]N is O-computable if the corresponding function N→[0,1] given by ⟨n,m⟩↦f(n)(m) is O-computable.
For any point p∈[0,1], a probabilistic oracle O is compatible with p if for all r∈[0,1]Q, we have O(⟨r⟩)=0 if p<r and O(⟨r⟩)=1 if p>r. More generally, for any m∈N and any p∈[0,1]m, a probabilistic oracle O is compatible with p if for all j such that 1≤j≤m and all r∈[0,1]Q, we have O(⟨j,r⟩)=0 if pj<r and O(⟨j,r⟩)=1 if pj>r.
A function f:[0,1]m→[0,1]m is O-computable if for each coordinate fj:[0,1]m→[0,1], there is an index i such that whenever P is compatible with p, we have P(ϕO,Pi(0)↓∈{0,1})=1 and P(ϕO,Pi(0)↓=1)=fj(p).
A map f:N→[0,1]N is O-computably ubiquitous if for every O-computable map e:N→[0,1]N, there is some n∈N such that f(n)=e(n).
3 Converse Lawvere property
We first need a lemma that tells us that we can replace certain partial O-computable functions with total ones when working relative to a reflective oracle. As discussed in the introduction, this contrasts strongly with the situation for computable functions. All of our theorems will make essential use of this lemma.
Lemma 1 (totalizing): There is a computable map τ:N→N such that for all i,n∈N and any reflective oracle O, we have P(ϕOτ(i)(n)↓∈{0,1})=1 and for b∈{0,1}, P(ϕOτ(i)(n)↓=b)≥P(ϕOi(n)↓=b).
Proof: We construct τ using a recursive procedure that ensures that P(ϕOτ(i)(n)↓=b) upper-bounds P(ϕOi(n)↓=b) using what may be thought of as repeated subdivision or binary search. This is essentially the same as the function flip in [3], but we handle computability issues differently. Let S⊆[0,1]×[0,1] be the set of pairs of dyadic rationals (p,q) such that p<q. We recursively define a stochastic O-computable function t:N×N×S→Δ{0,1}; the intent is to have P(t(i,n,p,q)=1) equal P(ϕOτ(i)(n)↓=1)−pq−p if that quantity is in the interval [0,1], and take the closest possible value, either 0 or 1, otherwise. Then, we will be able to define ϕOτ(i)(n) to be t(i,n,0,1).
Construct t so that a call t(i,n,p,q) first queries O(⟨i,n,r⟩), where r=p+q2, and also flips a fair coin C. Then, it either outputs 0, 1, or the result of a recursive call, as follows: t(i,n,p,q)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩0if O(⟨i,n,r⟩)=0 and C=0t(i,n,p,r)if O(⟨i,n,r⟩)=0 and C=11if O(⟨i,n,r⟩)=1 and C=0t(i,n,r,q)if O(⟨i,n,r⟩)=1 and C=1. We can now choose τ so that ϕOτ(i)(n)=t(i,n,0,1).
This algorithm upper bounds the probabilities P(ϕOi(n)↓=0) and P(ϕOi(n)↓=1) by binary search. Once the initial call t(i,n,0,1) has recursed to t(i,n,p,q), it has already halted outputting 1 with probability p, confirming that this is an upper bound since it received answer 1 from a call to O(⟨i,n,p⟩). Similarly, it has output 0 with probability 1−q, confirming this bound since it received the answer 0 from a call to O(⟨i,n,q⟩). Further, since each call to t halts without recursing with probability 12, t halts almost surely. Thus, we get the totality property P(ϕOτ(i)(n)↓∈{0,1})=1 and the bounds P(ϕOτ(i)(n)↓=b)≥P(ϕOi(n)↓=b) for b∈{0,1}. □
Now we can prove our main theorem.
Theorem 1 (converse Lawvere for reflective oracles): Let O be a reflective oracle. Then, there is an O-computable map f:N→[0,1]N such that for all O-computable g:N→[0,1], there is some index i such that g=f(i).
Proof: Using τ from the totalizing lemma, let f(i)(n)=P(ϕOτ(i)(n)↓=1). Given any O-computable g:N→[0,1], there is some i such that P(ϕOi(n)↓=1)=g(n)P(ϕOi(n)↓=0)=1−g(n). Then, f(i)(n)=P(ϕOτ(i)(n)↓=1)≥P(ϕOi(n)↓=1)=g(n) and similarly 1−f(i)(n)≥1−g(n), so f(i)=g. □
This theorem gives a computable analogue to the problem posed in [4]. The analogy would be strengthened if we worked in a cartesian closed category where the present notion of O-computability gave the morphisms, and where [0,1]N is an exponential object. In addition, the totalizing lemma would have a nice analogue in this setting. I expect that all this can be done using something like an effective topos [6], but I leave this for future work.
4 Recovering Brouwer's fixed point theorem
As mentioned in [4], the intermediate value theorem would follow from the existence of a space with the converse Lawvere property, that is, a space X that surjects onto the mapping space [0,1]X. Further, Brouwer's fixed point theorem on the n-ball, Bn, would follow from the existence of a topological space X with a surjection X→(Bn)X. We can do something similar to conclude Brouwer's fixed point theorem from the converse Lawvere theorem for reflective oracles. Of course, this is circular; Kakutani's fixed point theorem, a generalization of Brouwer's fixed point theorem, is used to prove the existence of reflective oracles. Still, it is interesting to see how this can be done.
We start with two technical lemmas telling us some basic facts about reflective-oracle-computable functions [0,1]m→[0,1]m. Using these, it will be easy to derive Brouwer's fixed point theorem.
Lemma 2 (continuous implies relatively computable): Take m∈N and let h:[0,1]m→[0,1]m be continuous. Then, there is a (deterministic) oracle O such that h is O-computable.
Proof: For each coordinate hj of h, each rectangle R=[ℓ1,u1]×⋯×[ℓm,um]⊆[0,1]m with rational endpoints, and each pair of rationals ℓ0,u0∈[0,1]Q with ℓ0<u0, let O(⟨j,ℓ0,u0,…,ℓm,um⟩) be 1 if hj(R)⊆[ℓ0,u0] and 0 otherwise. To see that h is O-computable, we compute any hj(p) for any j with 1≤j≤m, and any p∈[0,1]m, making use of O and any oracle P compatible with p.
We proceed by a search process similar to the argument in the totalizing lemma. Start with R0=[0,1]m, ℓ00=0 and u00=1. At each step s, perform an exhaustive search for a rectangle Rs+1=[ℓs+11,us+11]×⋯×[ℓs+1m,us+1m]⊆Rs and points ℓs+10,us+10 such that a query to P(⟨k,ℓs+1k⟩) returns 1 for all k, a query to any P(⟨k,us+1k⟩) returns 0, a query to O(⟨j,ℓs+10,us+10,Rs+1⟩) returns 1, and where either ℓs+10=ℓs0 and us+10=13ℓs0+23us0, or ℓs+10=23ℓs0+13us0 and us+10=us0. In the first case, output 0 with probability 13 and continue with probability 23, and in the second case, output 1 with probability 13 and continue with probability 23.
By construction, p∈Rs and hj(Rs)⊆[ℓs0,us0] at each stage s. Since hj is continuous, there is some neighbourhood of p on which its image is contained in either [ℓs0,13ℓs0+23us0] or [23ℓs0+13us0,us0]. There are two possibilities to consider. If p∈intRs, then there is some rectangle R contained in such a neighbourhood of p, and with p∈intR. This rectangle would be chosen if considered, so the algorithm will move beyond step s.
The remaining possibility is that p is on the boundary of Rs; say, pk=ℓsk. Since Rs was chosen previously though, we know that querying P(k,ℓs+1k) has returned 1 at least once, so P(k,ℓs+1k)≠0. Thus, the algorithm will almost surely eventually accept Rs or another rectangle.
Putting this together, this algorithm almost surely halts, outputting either 0 or 1. By stage s, it has halted outputting 1 with probability ℓs0 and outputting 0 with probability 1−us0, so overall it outputs 1 with probability hj(p). Thus, h is O-computable. □
Lemma 3 (composition): Let O be a reflective oracle and let g:N→[0,1]m and h:[0,1]m→[0,1]m be O-computable. Then, h∘g:N→[0,1]m is O-computable.
Proof: For each j with 1≤j≤m, take ij∈N such that ϕOij witnesses the computability of gj. Then, O(⟨ij,n,r⟩)=0 if gj(n)<r and O(⟨ij,n,r⟩)=1 if gj(n)>r, so O lets us simulate a probabilistic oracle compatible with g(n). Hence, by the O-computability of h, for each k with 1≤k≤m, we have a probabilistic O-machine that always halts, outputting either 0 or 1, and that on input n outputs 1 with probability hk∘g(n). □
Theorem 2 (Brouwer's fixed point theorem): Take m∈N and h:[0,1]m→[0,1]m. Then, h has a fixed point.
Proof: By Lemma 2, we have an oracle O such that h is O-computable. By relativizing the construction of a reflective oracle [2,3] to O, we get a reflective oracle ˜O above O. Notice that h is ˜O-computable. Letting f be the ˜O-computable function f(i)(n)=P(ϕ˜Oτ(i)(n)↓=1) constructed in the converse Lawvere theorem for reflective oracles, define fm:N→([0,1]m)N by fm(⟨i1,…,im⟩)(n)=(f(i1)(n),…,f(im)(n)). The rest will now follow the proof of Lawvere's fixed point theorem [7].
Define g:N→[0,1]m by g(n)=h(fm(n)(n)); this is ˜O-computable by Lemma 3. Now, by converse Lawvere theorem, for each coordinate 1≤j≤m of g, there is some ij∈N such that gj=f(ij). Letting i=⟨i1,…,im⟩, we have gj(i)=hj(fm(i)(i))=hj(g(i)), so g(i)=h(g(i)), and so g(i) is a fixed point of h. □
5 Ubiquitous converse Lawvere property
Theorem 3 (ubiquitous converse Lawvere): Let O be a reflective oracle. Then, there is an O-computable, O-computably ubiquitous map f:N→[0,1]N.
Proof: This follows by a combination of the recursion theorem and the totalizing lemma. We use the same map f from Theorem 1. Let e:N→[0,1]N be any O-computable map. There is a computable map s:N→N such that for all i,n∈N, we have P(ϕOs(i)(n)↓∈{0,1})=1, and e(i)(n)=P(ϕOs(i)(n)↓=1). By the recursion theorem, there is some i such that ϕOs(i)=ϕOi. Then, e(i)(n)=P(ϕOs(i)(n)↓=1)=P(ϕOi(n)↓=1)=P(ϕOτ(i)