How many infinite subsets does $\mathbb N$ have?

Question

I attempted to write a derivation of the answer, but was told my mathematics was wrong; please correct me.

The cardinality of $\mathbb N$ is $\aleph_0$.
From this set, we can generate another infinite subset by excluding $1$ element.
There are $\aleph_0$ such possible subsets that can be generated like this.
We can generate an infinite subset by excluding $2$ elements from $\mathbb N$.
There are $\aleph_0 \choose 2$ possible subsets that can be generated like this.

In general, for any $i$ from $0$ to $\aleph_0$ we can generate $\aleph_0 \choose i$ such possible subsets by excluding $i$. To find the total number of possible subsets, we simply sum all the combinations.
$$\sum_{i = 0}^{n} {n \choose i} = 2^n$$

Based on the above: $$\sum_{i = 0}^{\aleph_0} {\aleph_0 \choose i} = 2^{\aleph_0}$$

$2^{\aleph_0} = \aleph_1$
$\therefore$  the number of infinite subsets of $\mathbb N$ is $\aleph_1$.

I realise that I excluded the number of infinite subsets who have infinite complements.

To account for this, merely combine any $k$ $i$ used in the selection above, and exclude all multiples of the products of $i_1*i_2*i_3*...*i_k$.
We have $\aleph_0$ such sets of $i$ with numbers increasing from $0$ to $\aleph_o$.

I didn't consider this when I first wrote it out, and only realised it after. I haven't yet updated my proof to include it. However, this wasn't the problem with my proof; I was told I did "bad mathematics".

Answer 1

There are $\aleph_0 \choose 2$ possible subsets that can be generated like this.

No, the term $x\choose y$ is well defined only for finite numbers, not for cardinals. The number of subsets of $\mathbb{N}$ without $n$ points is $\aleph_0$. Introducing a new symbol is not only not needed, but also misleading, since you've concluded incorrect things from that.

Based on the above: $$\sum_{i = 0}^{\aleph_0} {\aleph_0 \choose i} = 2^{\aleph_0}$$

No, that is not correct. You've applied a formula that is valid for finite numbers only, not for arbitrary cardinals.

By my previous remark the number of subsets of $\mathbb{N}$ without $n$ points is $\aleph_0$. And so the proper left side is $\sum_{i=0}^\infty\aleph_0$, but

$$\sum_{i=0}^\infty \aleph_0=\aleph_0$$

which doesn't lead us to the correct answer. Because the collection of all complements of finite subsets is just not big enough, it doesn't cover all infinite subsets even closely.

$2^{\aleph_0} = \aleph_1$

No, that is continuum hypothesis. $2^{\aleph_0}$ is $2^{\aleph_0}$.

the number of infinite subsets of $\mathbb N$ is $\aleph_1$.

Actually it is $2^{\aleph_0}$ by the previous remark. And your final result is only coincidentally correct.

One proper solution is as follows:

1. The number of all finite subsets of $\mathbb{N}$ is $\aleph_0$.
2. The number of all subsets of $\mathbb{N}$ is $2^{\aleph_0}$.
3. The number of all infinite subsets is $2^{\aleph_0}$, because that is the only solution to the equation $$\aleph_0+x=2^{\aleph_0}$$ by the fact that for infinite cardinals $\aleph_0\leq \kappa_1\leq \kappa_2$ we have $\kappa_1+\kappa_2=\kappa_2$.

I was introduced to $\aleph_0$ and $\aleph_1$ by a friend who gave me the definition $\aleph_{n+1}=2^{\aleph_n}$.

Not good. The proper definition of $\aleph_{n+1}$ is: the smallest cardinal greater than $\aleph_n$. Or in other words: the successor cardinal. Which is a valid definition at least under the Axiom of Choice. It is then the generalized continuum hypothesis that $\aleph_{n+1}=2^{\aleph_n}$, which is a statement independent of ZFC. And so we can work in a model where it does or does not hold. Be careful.

Answer 2

$\Bbb N$ has $2^{\aleph_0}$ subsets. The number of subsets of size $n$ for any $n$ finite is countable. The union of a countable number of countable sets is countable, so $\Bbb N$ has a countable number of finite sets...

Answer 3

Some of the math you did (plugging cardinalities into the combination function for instance) isn't valid. The function $(n,k)\mapsto {n\choose k}$ takes nonnegative integers as arguments; the expression ${\aleph_0 \choose n}$ is meaningless (if such a thing has been defined and is useful, it's probably the case that you haven't learned about it in your class).

Let $X$ be the set of infinite subsets of $\mathbb{N}$. To prove the claim, note that $\vert \mathscr{P}(\mathbb N)\vert =\mathfrak{c}$. Further, there are $\aleph_0$ finite subsets of $\mathbb N$ (since the set of all finite subsets of $\mathbb N$ is a countable union of countable sets). Therefore, $$\aleph_0+\vert X\vert = \mathfrak c$$ Since $\aleph_0+\alpha=\alpha$ for any infinite cardinal $\alpha$, we have $\vert X \vert = \mathfrak c$.