Question

Show that ifand are sets and $\mathbf{A}\mathbf{\subset }\mathbf{B}\mathbf{\text{then}}\mathbf{\left|}\mathit{A}\mathbf{\right|}\mathbf{<}\mathbf{\left|}\mathit{B}\mathbf{\right|}$

Short answer

f is a one-to-one function from $A,\left|A\right|\le \left|\mathit{B}\right|$.

Step-by-step solution

Definition

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

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

Definition : If there is a one-to-one function from to , then$A,\left|A\right|\le \left|\mathit{B}\right|$

Proof for the given statement

$a\in B$Given: A and B are sets with$\mathrm{A}\subset \mathrm{B}$

.

To proof:$A,\left|A\right|\le \left|\mathit{B}\right|$

By the definition of a subset: If $a\in A$then .

We can then define the function as:

$f:A\to B,f\left(a\right)=a$

We need to check that the function is one-to-one. Let f (a) = f (b).

By the definition of, we then obtain a = b .

Thus f is one-to-one.

Since f is a one-to-one function from $A,\left|A\right|\le \left|\mathit{B}\right|$(see definition ).