Question

Show that there is no infinite set A such that$\left|\mathrm{A}\right|<\left|{\mathrm{Z}}^{+}\right|={\mathrm{N}}_0$

Short answer

A is not infinite.

Step-by-step solution

Step 1:

DEFINITIONS

A set is countable if it is finite or countably infinite.

A set is finite if it contains a limited number of elements (thus it is possible to list every single element in the set).

A set is countably infinite if the set contains an unlimited number of elements and if there is a one-to-one correspondence with the positive integers.

The function f is one-to-one if and only if $f\left(a\right)=f\left(b\right)$implies that 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(2): There is a one-to-one function from A to B if and only if $\left|A\right|=\left|B\right|$.

Step 2:

Given: $\left|A\right|<\left|Z^{+}\right|$

To proof: A is not infinite.

PROOF:

$\left|A\right|<\left|Z^{+}\right|$means $\left|A\right|<\left|Z^{+}\right|$and A and$Z^{+}$d o not have the same cardinality. Thus$\left|A\right|=\left|Z^{+}\right|$

is not true, which means that there does not exists a one-to-one correspondence from A to $Z^{+}$, and thus A cannot be countably infinite.

$\left|A\right|\le \left|Z^{+}\right|$

By definition (2), we then know that there is a one-to-one function from A to $Z^{+}$.

Then A has the same cardinality as a subset of $Z^{+}$.

In exercise , we will proof that the subset of a countable set is also countable. Thus A has the same cardinality as a countable set (which is the subset of the countable set $Z^{+}$) and thus A is countable as well.

We have then proved that if $\left|A\right|\le \left|Z^{+}\right|$, then A is countable.

Since A is not countably infinite and A is countable, A then has to be finite (not infinite).