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

Note: Questions 1-5 form a coherent sequence and questions 6-10 form a separate coherent sequence. You can jump between the sequences.

  1. Let be a complete metric space. A function is called a contraction if there exists a such that for all , . Show that if is a contraction, then for any , the sequence converges. Show further that it converges exponentially quickly (i.e. the distance between the th term and the limit point is bounded above by for some )

  2. (Banach contraction mapping theorem) Show that if is a complete metric space and is a contraction, then has a unique fixed point.

  3. If we only require that for all , then we say is a weak contraction. Find a complete metric space and a weak contraction with no fixed points.

  4. A function is convex if , for all and . A function is strongly convex if you can subtract a positive parabaloid from it and it is still convex. (i.e. is strongly convex if is convex for some .) Let be a strongly convex smooth function from to , and suppose that the magnitude of the second derivative is bounded. Show that there exists an such that the function given by is a contraction. Conclude that gradient descent with a sufficiently small constant step size converges exponentially quickly on a strongly convex smooth function.

  5. A finite stationary Markov chain is a finite set of states, along with probabilistic rule for transitioning between the states, where represents the space of probability distributions on . Note that the transition rule has no memory, and depends only on the previous state. If for any pair of states , the probability of passing from to in one step is positive, then the Markov chain is ergodic. Given an ergodic finite stationary Markov chain, use the Banach contraction mapping theorem to show that there is a unique distribution over states which is fixed under application of transition rule. Show that, starting from any state , the limit distribution exists and is equal to the stationary distribution.

  6. A function from a partially ordered set to another partially ordered set is called monotonic if implies that . Given a partially ordered set with finitely many elements, and a monotonic function from to itself, show that if or , then is a fixed point of for all .

  7. A complete lattice is a partially ordered set in which each subset of elements has a least upper bound and greatest lower bound. Under the same hypotheses as the previous exercise, extend the notion of for natural numbers to for ordinals , and show that is a fixed point of for all with or and all ( means there is an injection from to , and means there is no such injection).

  8. (Knaster-Tarski fixed point theorem) Show that the set of fixed points of a monotonic function on a complete lattice themselves form a complete lattice. (Note that since the empty set is always a subset, a complete lattice must be nonempty.)

  9. Show that for any set , forms a complete lattice, and that any injective function from to defines a monotonic function from to . Given injections and , construct a subset of and a subset of of such that and .

  10. (Cantor–Schröder–Bernstein theorem) Given sets and , show that if and , then . ( means there is an injection from to , and means there is a bijection)

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Tomorrow's AI Alignment Forum Sequences post will be "Approval-directed agents: overview" by Paul Christiano in the sequence Iterated Amplification.

The next post in this sequence will be released on Saturday 24th November, and will be 'Fixed Point Discussion'.

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on shortens all distances but is strictly monotonic.

#6: (the "show that if" condition follows from the property, the question is likely misstated)

The iteration is so long that it must visit an element twice. We can't have a cycle in the order so the repetition must be immediate.

Thanks, I actually wanted to get rid of the earlier condition that for all , and I did that.

Ex 6:

If at any point , then we're done. So assume that we get a strict increase each time up to . Since there are only elements in the entire poset, and is monotone, has to equal .

Ex 7:

For a limit ordinal , define as the least upper bound of for all . If , then the set for is a set of size that maps into a set of size by taking the value of the element. Since there are no injections between these sets, there must be two ordinals such that. Since is monotone, that implies that for every ordinal , and thus is a fixed point. Since this proves the exercise.

Ex 8:

Starting from , we can create a fixed point via iteration by taking , and iterating times as demonstrated in Ex 7. Call this fixed point . Suppose there was a fixed point such that and . Then at some point , but , which breaks the monotonicity of unless . So generated this way is always the smallest fixed point greater than .

Say we have fixed points . Then let be the least upper bound of , and generate a fixed point from . So will be greater than each element of since is monotone, and is the smallest such fixed point as shown in the above paragraph. So the poset of fixed points is semi-complete with upper bounds.

Now take our fixed points again. Now let be the greatest lower bound of , and generate a fixed point . Since and is monotonic, , and so is a lower bound of . It has to be the greatest such bound because itself is already the greatest such bound in our poset, and is monotonic.

Thus the lattice of fixed points has all least upper bounds and all greatest lower bounds, and is thus complete!


We wish to show that the terms of form a Cauchy sequence, which suffices to demonstrate they converge in a complete space. Take , and WLOG . Then we know from the definition of contraction that . This converges to 0 as m increases, so the sequence is Cauchy.

It's easy to see that this makes the rate of convergence between terms of the Cauchy sequence exponentially quick. Intuitively that seems like it ought to make the sequence converge to its limit with the same speed, but I don't think that can be made rigorous without more steps.


Take a sequence . This converges to some . Suppose was not a fixed point. Then choose an . A sequence which converges to a limit has, for every , some such that . Then we know that but , contradicting the contraction condition. So there is at least one fixed point, .

Suppose there are two fixed points, , for distinct and . If so, , which again contradicts the contraction condition. So there is at most one fixed point.


Take as the space , with the usual metric. Define . This is a weak contraction (toward infinity) and has no fixed points within this space.

Answer to question 1.

Let for arbitrary . Call . Then by induction () (power series simplification)

Therefore ie is a cauchy sequence. However is said to be complete, which by definition means any cauchy sequence is convergent. So and So converges exponentially quickly

Answer to question 2.

From part 1, as is continuous, So is a fixed point. Suppose and are both fixed points of a contraction map. Then and so therefore so . Thus has a unique fixed point.

Answer to question 3.

is a metric space. Its the real line with normal distance. Let . Then is a contraction map because is differentiable and has the property . However no fixed point exists as . This works because the sequence generated from repeated applications of will tend to infinity, despite successive terms becoming ever closer.

For Q2, I believe you aren't done:

You have established that there is at most one fixed point, but not that a fixed point exists.


Assume WLOG Then by monotonicity, we have If this chain were all strictly greater, than we would have istinct elements. Thus there must be some uch that By induction, or all


Assume nd construct a chain similarly to (6), indexed by elements of If all inequalities were strict, we would have an injection from o L.


Let F be the set of fixed points. Any subset S of F must have a least upper bound n L. If x is a fixed point, done. Otherwise, consider which must be a fixed point by (7). For any q in S, we have Thus s an upper bound of S in F. To see that it is the least upper bound, assume we have some other upper bound b of S in F. Then

To get the lower bound, note that we can flip the inequalities in L and still have a complete lattice.


P(A) clearly forms a lattice where the upper bound of any set of subsets is their union, and the lower bound is the intersection.

To see that injections are monotonic, assume nd s an injection. For any function, If nd that implies or some which is impossible since s injective. Thus s (strictly) monotonic.

Now s an injection Let e the set of all points not in the image of and let ote that since no element of s in the image of Then On one hand, every element of A not contained in s in y construction, so On the other, clearly so QED.


We form two bijections using the sets from (9), one between A' and B', the other between A - A' and B - B'.

Any injection is a bijection between its domain and image. Since nd s an injection, s a bijection where we can assign each element o the uch that Similarly, s a bijection between nd Combining them, we get a bijection on the full sets.