Let f: Ax AA and let CCA. We say that C is closed under f to mean that Vx, y E [f(x, y) E C]. For BCA, we call CCA a closure of B under f to mean that C is the smallest (with respect to containment) subset of A that is closed and contains B. Prove that for every B C A, a closure of B exists.

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter1: Fundamentals
Section1.2: Mappings
Problem 27E: 27. Let , where and are nonempty. Prove that has the property that for every subset of if and...
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Let f: Ax AA and let CCA. We say that
C is closed under f to mean that Vx, y E
[f(x, y) E C]. For BCA, we call CCA a
closure of B under f to mean that C is the
smallest (with respect to containment)
subset of A that is closed and contains B.
Prove that for every BCA, a closure of B
exists.
Transcribed Image Text:Let f: Ax AA and let CCA. We say that C is closed under f to mean that Vx, y E [f(x, y) E C]. For BCA, we call CCA a closure of B under f to mean that C is the smallest (with respect to containment) subset of A that is closed and contains B. Prove that for every BCA, a closure of B exists.
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