Prove that if A is denumerable and B is a finite subset of A, then A \ B is denumerable.

Printable View

- Dec 6th 2005, 10:49 AMmathkiddenumerability
Prove that if A is denumerable and B is a finite subset of A, then A \ B is denumerable.

- Dec 6th 2005, 08:37 PMCaptainBlackQuote:

Originally Posted by**mathkid**

countably infinite.

Let $\displaystyle a_1,\ a_2,\ ...,\ a_n,\ ...$ be an enumeration of $\displaystyle A$.

We can rearrange this enumeration so that:**<you may need to prove that this can be done>**

$\displaystyle B\ =\ \{a_1,\ a_2,\ ...,\ a_{N_B}\}$

($\displaystyle N_B$ is the number of elements of $\displaystyle B$)

Then:

$\displaystyle A \backslash B\ =\ \{a_{N_B+1},\ a_{N_B+2},\ ...\}$.

So $\displaystyle c_1,\ c_2,\ ...,\ c_N,\ ...$ where $\displaystyle c_n\ =\ a_{N_B+n}$ is an enumeration of $\displaystyle A \backslash B$

RonL - Dec 9th 2005, 11:17 AMrgep
I like to use the definition of "countable" as "can be injected into the natural numbers". It is then immediate that a subset of a countable set is countable. "Denumerable" is countable and not finite, where "finite" means every injection of the set to itself is also a surjection. All that remains is to prove that the union of two finite sets is finite; use the lemma that a finite set injects into {1..n} for some natural number n.