Originally Posted by

**issacnewton** Hi

I am proving that any subset of a countable set is also countable.

I will show my work here. Let set A be countable. We have as one of the cases, where A is an empty set , but by definition its finite and hence countable.

So I am not going to consider the empty subset of A as its trivial. So there are

two cases

Case 1) A is a finite set.

Let $\displaystyle A=\{a_1,a_2,\cdots,a_n\}$ and let

$\displaystyle B\subseteq A$

then $\displaystyle B=\{b_1,b_2,\cdots, b_m\}$

where $\displaystyle m\leqslant n$ and

$\displaystyle b_i \in A \;\; \forall \; i$

Then its clear that there exists a bijection from B to {1,2,..m} for some m in N

Hence B is finite and countable

Case 2) A is inifnite

Let $\displaystyle B\subseteq A$

Since A is infinite , $\displaystyle A \thicksim N$ so there exists a bijection from A to N. Hence there exists a bijection from N to A.

$\displaystyle f:N\rightarrow A$

Since $\displaystyle B\subseteq A$ ,

$\displaystyle \forall \;\;b\in B \;\;\exists \; n_1 \in N \backepsilon$

$\displaystyle f(n_1)=b$

Now we define a function

$\displaystyle g:N\rightarrow B$

from above arguments, its clear that g is onto. I have to prove that B is also countable, so it could mean that B is either finite or infinite or empty. So how

do I proceed now ? I have checked some proofs of this theorem online and I had

trouble understanding it , so I decided to post the question.

thanks