1.3Comp071110

# 1.3Comp071110 - 1 1.3. Complementation Theorems. Recall the...

This preview shows pages 1–3. Sign up to view the full content.

1 1.3. Complementation Theorems. Recall the Complementation Theorem from section 1.1. It first appeared in print in [Fr92], Theorem 3, p. 82, (in a slightly different form), where we presented some precursors of BRT. The Complementation Theorem is closely related to the standard Contraction Mapping Theorem. We discuss the connection below. The Complementation Theorem is also closely related to a well known theorem, due to von Neumann in [VM44], and subsequent developments in graph theory. We discuss this at the end of this section. COMPLEMENTATION THEOREM. For all f SD there exists A INF such that fA = N\A. COMPLEMENTATION THEOREM (with uniqueness). For all f SD there exists a unique A N with fA = N\A. Moreover, A INF. Before giving the proof of the Complementation Theorem (with uniqueness), we discuss some alternative formulations. The Complementation Theorem (without uniqueness) is written above as a statement of EBRT in A,fA on (SD,INF). Strictly speaking, we cannot express the uniqueness within BRT. DEFINITION 1.3.1. A . B is the disjoint union of A and B, and is defined as A B if A,B are disjoint; undefined otherwise. E.g., A . B = C if and only if A B = C A B = . Note that there are other equivalent ways of writing fA = N\A. E.g., we can write fA = N\A. A = N\fA. A . fA = N. The first evaluates the action of f on A. The second asserts that A is a fixed point (of the operator that sends each B to N\fB.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 The third asserts that N is partitioned into A and fA. Proof: Let f SD. Note that for all A N, n fA if and only if n f(A [0,n)). We inductively define a set A N as follows. Suppose n 0 and we have defined membership in A for all 0 i < n. We then define n A if and only if n f(A [0,n)). Since f SD, we have for all n, n A n fA as required. Now suppose fB = N\B. Let m be least such that A,B differ. Then m B m fB m f(B [0,m)), and m A m f(A [0,m)). Since A [0,m) = B [0,m), we have m A m B. This is a contradiction. Hence A = B. If A is finite then fA is finite and N\A is infinite. Hence A is infinite. QED It will be convenient to use the following terminology. Let f:X k X. DEFINITION 1.3.2. Let f be a multivariate function with domain X (see Definitions 1.1.8 - 1.10). A complementation of f is a set A X such that fA = X\A. Thus we can restate the Complementation theorem (with uniqueness) as follows. COMPLEMENTATION THEOREM (with uniqueness). Every f SD has a unique complementation. Note that we have proved the Complementation Theorem (with uniqueness) within the base theory RCA 0 of Reverse Mathematics. See [Si99]. The Complementation Theorem is obviously a particularly
This is the end of the preview. Sign up to access the rest of the document.

## 1.3Comp071110 - 1 1.3. Complementation Theorems. Recall the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online