Question

Show that ifand are sets with the same cardinality, then $\mathbf{\left|}\mathbf{A}\mathbf{\right|}\mathbf{\le }\mathbf{\left|}\mathbf{B}\mathbf{\right|}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\left|}\mathbf{B}\mathbf{\right|}\mathbf{\le }\mathbf{\left|}\mathbf{A}\mathbf{\right|}$ .

Short answer

A and B are sets with the same cardinality and only if
$\left|A\right|\le \left|B\right|and\left|B\right|\le \left|A\right|$

Step-by-step solution

Definition

The function is one-to-one if and only if f = (a) = f (b) implies that a = b for all a and b in the domain.

f is a one –to-one correspondence if and only if f is one-to-one and onto.

X is a subset of Y if every element of X is also an element of Y. Notation: $X\subseteq Y$

Definition (1):There is a one-to-one correspondence from and if and only if$\left|A\right|=\left|B\right|$

Definition (2): There is a one-to-one function from to if and only if$\left|A\right|\le \left|B\right|$

Proof for the first case

Given, A and B are sets with the same cardinality.

To Proof:$\left|A\right|\le \left|B\right|and\left|B\right|\le \left|A\right|$

IN FIRST CASE

A and B are sets with the same cardinality.

By definition (1), we know that there exists a one-to-one correspondence f from A to B .

By the definition of a one-to-one correspondence, is the also a one-to-one function from to .

By definition (2), we the obtain$\left|A\right|\le \left|B\right|$.

Proof for the second case

IN SECOND CASE

andare sets with the same cardinality is equivalent with andare sets with the same cardinality.

|B| = |A|

By definition (1), we know that there exists a one-to-one correspondence from B to A.

By the definition of a one-to-one correspondence, is the also a one-to-one function from B to A.

By definition (2), we the obtain$\left|B\right|\le \left|A\right|$.

Therefore,

and are sets with the same cardinality and only if $\left|A\right|\le \left|B\right|and\left|B\right|\le \left|A\right|$