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单词 SchroderBernsteinTheoremProof
释义

Schröder Bernstein Theorem: Proof


Let A and B be two nonempty sets; and let there be, in addition, two one-one functions f:AB and g:BA. We propose to show that A and B are equinumerous i.e., they are in one to one correspondence.

Consider the notation:

g-1(x),ifxg(B)
f-1(g-1(x)),ifg-1(x)f(A)
g-1(f-1(g-1(x))),iff-1(g-1(x))g(B)
..

Define, for each xA, the order of it, denoted by (x), to be the number of such preimageMathworldPlanetmath(s) which exist. In a similer way, we’d be able to define the order of an element yB, i.e., by considering the sequence f-1(y),g-1(f-1(y)),.

Now define, for each xA,

ϕ(x):=f(x),(x)=
=f(x),(x)=2n, for some nω
=b,(x)=2n+1,i.e.,bB:g(b)=x

Notice that if the order is infiniteMathworldPlanetmath, ϕ(x)=f(x) is also infinite. Because, otherwise x would have to have a finite order. On the other hand, if yB and (y) is infinite, then f-1(y) exists and has an infinite order; call the latter one x. This means, ϕ maps the infinite order elements of A bijectively onto the infinite order elements of B.

Next, if (x)=2n, then the order of f(x) is sheer 2n+1. Similer to the above para, if for yB, the order is 2n+1, as the order is non-zero, f-1(y) exists and it must have order 2n. To formally show it you need a tedious inductive reasoning!

Last, if (x)=2n+1, then the order is 1, and so, g-1(x) exists and the order of g-1(x) is sheer 2n(looking upon g-1(x) as an element of B). Conversely, similer to above, if there is an element y of order 2n in B, take x=g(y) and the order of x is indeed 2n+1. All that you need to convince a sceptic is a long, tedious, involved inductionMathworldPlanetmath!!

What we learn from what precedes is that the one-one function ϕ maps the infinite order elements onto infinite order elements, odd order onto odd order, and even order onto even order; a simple set theory reveals that ϕ is a one-one map from A onto B. This completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

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更新时间:2025/5/4 4:37:06