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1. Non-decreasing functions

Let A and B be partially-ordered sets. Recall that a function f:A→B is non-decreasing if x≤y implies f(x)≤f(y) for all x, y in A.

  1. Prove or disprove: The set S2 of all non-decreasing functions from ℕ to {0,1} is countable.

  2. Prove or disprove: The set S of all non-decreasing functions from ℕ to ℕ is countable.

(Assume the usual order on ℕ and {0,1}.)

1.1. Solution

  1. Proof: We'll show how to assign a unique element of ℕ to each function in S2. Given a non-decreasing function f from ℕ to {0,1}, let c(f) = 0 if f(x) = 0 for all x in ℕ. Otherwise, there is some least value mf such that f(mf) = 1; let c(f) = mf+1. If f and g are distinct non-decreasing functions, then there is some x such that f(x) != g(x). Suppose that f(x) = 1 and g(x) = 0. Then either g is the constant zero function, and c(f) = mf + 1 > 0 = c(g), or mg > x (because if g(y) = 1 for some y < x, g(x) would have to be 1 as well) and mf ≤ x; again this implies mf > mg. So we have shown that c is an injection from S2 to ℕ, implying that S2 is countable.1

  2. Disproof: There are a lot of ways to show S is uncountable. Here's a particularly slick way, by constructing an explicit bijection between S and ℕ, the set of all functions (monotone or not) from ℕ to ℕ. Given f in ℕ, let mf be the function given by mf(x) = ∑z≤x f(z). Then mf is non-decreasing (if x ≤ y, then mf(y) = mf(x) + ∑x<z≤y f(z) ≥ mf(x)). The map f↦mf is a bijection because it's invertible: given mf, we can compute f by letting f(x) = mf(x) - mf(x-1) (with f(0) = mf(0) as a special case). Since ℕ is uncountable, this implies S is also uncountable.

2. Lex and colex orders

Let A and B be totally ordered sets. Here are three partial orders on A×B:

Lexicographic order

(a,b) ≤L (a',b') iff a < a' or a = a' and b ≤ b'.

Colexicographic order

(a,b) ≤C (a',b') iff b < b' or b = b' and a ≤ a'.

Product order

(a,b) ≤× (a',b') iff a ≤ a' and b ≤ b'.

Prove or disprove: For any totally ordered sets A and B, the product order (≤×) = (≤L) ∩ (≤C).

2.1. Solution

Proof: Suppose (a,b) ≤× (a',b') in the product order. Then a' ≤ a and b ≤ b'. If a < a', then (a,b) ≤L (a',b'); if a = a', then b ≤ b' implies again (a,b) ≤L (a',b'). By symmetry, the same argument works for ≤C. If follows that (a,b) ≤× (a',b') implies (a,b) ≤L (a',b') and (a,b) ≤C (a',b'), or that (≤×) ⊆ (≤L) ∩ (≤C).

For the other direction, suppose (a,b) ≤L (a',b') and (a,b) ≤C (a',b'). From (a,b) ≤L (a',b'), we have a ≤ a'. From (a,b) ≤C (a',b'), we have b ≤ b'. It follows that (a,b) ≤× (a',b'). In set-theoretic terms, we thus have (≤×) ⊇ (≤L) ∩ (≤C).

Since we've proved subset in both directions, the two sets are equal.

(Note: This doesn't work for the product of three or more sets. Consider (0,1,0) and (1,0,1).)

3. Addition and negation

Suppose we have an addition operation on some set S (i.e., a function +:S×S→S, written in infix form between its arguments), and we know that addition satisfies the axioms

where x, y, z, are any elements of S.

Define the relation ~ on S×S by the rule (x,y) ~ (x',y') iff x+y' = x'+y.

  1. Show that ~ is an equivalence relation.
  2. Define the operation (x,y) + (x',y') = (x+x', y+y'). Show that if (x,y) ~ (z,z), then (x,y) + (x',y') ~ (x',y').
  3. Define -(x,y) = (y,x). Show that (x,y) + (x',y') + -(x',y') ~ (x,y).

3.1. Solution

  1. Reflexive: (x,y) ~ (x,y), since x+y = x+y. Symmetric: Suppose (x,y) ~ (x',y'). Then x+y' = x'+y. Using symmetry of equality gives x'+y = x+y', or (x',y') ~ (x,y). Transitive: Let (r,s) ~ (t,u) and (t,u) ~ (v,w). Then r+u = t+s and t+w = v+u. Add these two equations together to get r+u+t+w = t+s+v+u. Rewrite as r+w+(t+u) = v+s+(t+u) and cancel out the (t+u) terms to get r+w = v+s. Use the definition of ~ to get (r,s) ~ (v,w).
  2. Let (x,y) ~ (z,z); then x+z = z+y, which we can rewrite as x+z = y+z, which implies x = y by cancellation. Expand (x,y) + (x',y') as (x+x', y+y'). Observe that (x+x', y+y') ~ (x',y') if and only if x+x'+y' = x'+y+y', which we can rewrite as (x'+y')+x = (x'+y')+y. This follows from x=y.
  3. Compute (x',y') + -(x',y') = (x',y') + (y',x') = (x'+y',x'+y') = (z,z) where z = x'+y'. But then (x,y) + (x',y') + -(x',y') = (x,y) + (z,z) ~ (x,y) by the previous case.

Some examples of this construction in action: If S = ℕ and + is the usual addition operation, then (ℕ×ℕ)/~ represents ℤ, with + and - having their usual meanings. If S = ℤ - {0} and + is multiplication, then (S×S)/~ represents ℚ-{0} with + acting as multiplication and - acting as inverse.

  1. It's actually a bijection. (1)

2014-06-17 11:57