Explore BrainMass
Share

Explore BrainMass

    Examining Proof about Cardinality

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Let A and B be sets. Show that A x B and B x A have equal cardinality by constructing an explicit bijection between the two sets. Then use the following proposition to prove that multiplication is commutative. (Let n, m be natural numbers. Then nxm=mxn)

    Proposition: Cardinal arithmetic

    a) Let X be a finite set, and let x be an object which is not an element of X. Then X U (union) {x} is finite and #(X U {x})= #(X)+1

    b) Let X and Y be finite sets. Then X U Y is finite and # (X U Y) is less than or equal to #(X) + #(Y). If in addition X and Y are disjoint (i.e., X intersection Y = the empty set), then #(X U Y)= #(X) + #(Y)

    c) Let X be a finite set, and let Y be a subset of X. Then Y is finite, and #(Y) is less than or equal to #(X), If in addition Y does not equal X (i.e. Y is a proper subset of X), then we have #(Y) is less than #(X)

    d) If X is a finite set, and f:X-->Y is a function, then f(X) is a finite set with #(f(X)) less than or equal to #(X). If in addition f is one to one, then #(f(X)) = #(X)

    e) Let X and Y be finite sets. Then cartesian product X x Y is finite and # (X x Y) = #(X) x #(Y)

    f) Let X and Y be finite sets. Then the set Y^X is finite and #(Y^X)= #(Y)^#(X)

    © BrainMass Inc. brainmass.com October 10, 2019, 4:22 am ad1c9bdddf
    https://brainmass.com/math/discrete-math/examining-proof-about-cardinality-457749

    Solution Preview

    Consider the mapping f:AxB-->BxA which sends (a,b) to (b,a). This mapping is injective. Indeed, let f(a,b)=f(a',b'). We have f(a,b)=(b,a), and f(a',b')=(b',a'). Therefore (b,a)=(b',a'). This implies that b=b' and a=a'. Thus (a,b)=(a',b'). ...

    Solution Summary

    The proof about cardinality is examined in the following posting.

    $2.19