# Algebraic Topology Waterloo 1978 by P. Hoffman, V. Snaith

By P. Hoffman, V. Snaith

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Example text

Hint. The result is known for ﬁnite , so we proceed by induction. Choose 0 , 1 , . . so that the supremum{ i + 1 : i = 0, 1, . . } = and let Gi ⊆ N × X be universal for Σ 0i X . Put G(α, x) ⇐⇒ i ¬Gi (α)i , x and show that G is universal for Σ 0 X by verifying that each P in Σ 0 P(x) ⇐⇒ i X satisﬁes ¬Pi (x) 0 with each Pi in Σ i . In the exercises of the next section we will extend this result to show that each Σ 0 is Y-parametrized for each perfect product space Y. 7. Suppose R0 , R1 , · · · are all subsets of the same space X and we take R(x, s) ⇐⇒ Rs (x).

Now the set G ⊆ X × N deﬁned by G (x, α) ⇐⇒ G x, α(0) is also open since G is and the map (x, α) → x, α(0) is continuous, hence F is Π 11 . Thus every closed set is Π 11 and then, by deﬁnition, every Σ 11 set is Σ 12 , from which Σ 1n ⊆ Σ 1n+1 follows immediately by induction. The remaining inclusions in the diagram are trivial. 2] 1E. The projective sets 31 To prove the closure properties of the Lusin pointclasses we need maps that allow us to code inﬁnite sequences of irrational by single irrationals.

3. Let f : R → R be a continuous function on the line. Prove that the relations P(x, y) ⇐⇒ f (x) = y, Q(x) ⇐⇒ f (x) exists are both Π 03 . Hint. Let r0 , r1 , . . be an enumeration of all rational numbers and put R(x, y, s, k, m) ⇐⇒ rm = 0 & 1 1 . 6] 1C. 5. Uniformization. Clearly R is a closed relation. It is easy to verify that 1 =⇒ R(x, y, s, k, m) . k+1 The second assertion is proved similarly, starting with the relation P(x, y) ⇐⇒ (∀s)(∃k)(∀m) 0 < |rm | < S(x, s, k, m, n) ⇐⇒ rm = 0 & rn = 0 & 1 rm {f(x + rm ) − f(x)} − 1 rn {f(x + rn ) − f(x)} ≤ 1 s+1 .