# Countability proof.

• Nov 29th 2010, 08:01 AM
mathsohard
Countability proof.
I need help on proving

(1) prove that if B http://www.mathhelpforum.com/math-he...ab35a80a37.png A and A is countable, then B is countable

(2) prove that if B http://www.mathhelpforum.com/math-he...ab35a80a37.png A, A is infinite, and B is finite, then A\B is infinite

I don't even know where to start... THx
• Nov 29th 2010, 08:07 AM
mathsohard
I have a theorem which says
Suppose A is a set. The following statements are equivalent:
1. A is countable
2. Either A = empty or there is a function f:z+ -> A that is onto.
3. There is a function f:A -> z+ that is one-to-one

I am supposed to use those however I am stuck....
• Nov 29th 2010, 08:22 AM
FernandoRevilla
If $\displaystyle A$ is denumerable then $\displaystyle A=\{a_1,a_2,\ldots\}$ . If $\displaystyle B=\emptyset$ then, $\displaystyle B$ is finite . If $\displaystyle B\neq \emptyset$ let $\displaystyle n_1$ be the least positive integer such that $\displaystyle a_{n_1}\in B$, let $\displaystyle n_2$ be the least positive integer such that $\displaystyle n_2>n_1$ and $\displaystyle a_{n_2}\in B$ ...

Could you continue?

Regards.

Fernando Revilla
• Nov 29th 2010, 08:33 AM
emakarov
You can use the equivalent condition #3 to show (1). Namely, suppose $\displaystyle B\subseteq A$ and there exists a one-to-one function $\displaystyle f:A\to\mathbb{Z}^+$. Consider the restriction of f on B, i.e., the function whose domain is B and that acts just like f on any $\displaystyle x\in B$. Is this restriction still one-to-one?

For (2), note that $\displaystyle A = B\cup(A\setminus B)$. What happens when $\displaystyle A\setminus B$ is finite?
• Nov 29th 2010, 08:40 AM
mathsohard
I get the (2) part , however could you explain more on (1) part please???
• Nov 29th 2010, 09:14 AM
emakarov
Quote:

could you explain more on (1) part please???
Whom are you asking? Fernando suggests using condition #2 to prove (1), i.e., that B is countable. Going with #2, let's assume that there is an onto function $\displaystyle f:\mathbb{Z}^+\to A$ and that B is nonempty, i.e., there exists some fixed $\displaystyle x_0\in B$. I think it is easier to construct another function $\displaystyle g:\mathbb{Z}^+\to B$ as follows:

$\displaystyle g(n)= \begin{cases} x_0 & f(n)\notin B\\ f(n) & \text{otherwise} \end{cases}$

You have to check that indeed $\displaystyle g:\mathbb{Z}^+\to B$ and that g is onto.

If you are going ti use #3 to show (1), then assume that there exists an $\displaystyle f:A\to\mathbb{Z}^+$. Consider $\displaystyle g(x)$ such that $\displaystyle g(x)=f(x)$ for $\displaystyle x\in B$ and $\displaystyle g(x)$ is undefined otherwise. You need to show that $\displaystyle g:B\to\mathbb{Z}^+$ and that g is one-to-one.