# Proof that the set is countable - my idea The set: S := <mo fence="false" stretchy="false

Proof that the set is countable - my idea
The set: is finite or $\mathbb{N}\mathrm{\setminus }T$ is finite}
Since all finite subsets of $\mathbb{N}$ are countable, can i just prove that since $\mathbb{N}\mathrm{\setminus }T$ would simply be $\left\{t;t\in \mathbb{N}$ $t\notin T\right\}$... All though i do have the question how could $\mathbb{N}\mathrm{\setminus }T$ be a finite set?
You can still ask an expert for help

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Darrell Valencia
Explanation:
Let ${X}_{n}$ be the set of all subsets of $\mathbb{N}$ of cardinality n. Then $|{X}_{n}|\le |{\mathbb{N}}^{n}|=|\mathbb{N}|.$. Let F be the set of all finite subsets of $\mathbb{N}.$. Then $F=\bigcup _{n\ge 0}{X}_{n}.$. Can you now finish?
###### Did you like this example?
Jovany Clayton
Step 1
Let

$=\left\{\mathbb{N}\setminus T:T\in F\right\}$
Thus there is a bijection $\varphi :I\to F$ where one has
$\varphi \left(A\right)=\mathbb{N}\setminus A$
Step 2
Now suppose we are given a bijection $\rho :F\to \mathbb{N}$
Consider the map $\beta :S\to \mathbb{N}$ defined via

note that this is a bijection.