From Jech's Set Theory, we have the following theorems on cardinal exponentiation (a Corollary on page 49):
Theorem. For all $\alpha,\beta$, the value of $\aleph_{\alpha}^{\aleph_{\beta}}$ is always either:
- $2^{\aleph_{\beta}}$; or
- $\aleph_{\alpha}$; or
- $\aleph_{\gamma}^{\mathrm{cf}\;\aleph_{\gamma}}$ for some $\gamma\leq\alpha$ where $\aleph_{\gamma}$ is such that $\mathrm{cf}\;\aleph_{\gamma}\leq\aleph_{\beta}\lt\aleph_{\gamma}$.
Here, $\mathrm{cf}\;\aleph_{\gamma}$ is the cofinality of $\aleph_{\gamma}$: the cofinality of a cardinal $\kappa$ (or of any limit ordinal) is the least limit ordinal $\delta$ such that there is an increasing $\delta$-sequence $\langle \alpha_{\zeta}\mid \zeta\lt\delta\rangle$ with $\lim\limits_{\zeta\to\delta} = \kappa$. The cofinality is always a cardinal, so it makes sense to understand the operations above as cardinal operations.
Corollary. If the Generalized Continuum Hypothesis holds, then
$$\aleph_{\alpha}^{\aleph_{\beta}} = \left\{\begin{array}{lcl}
\aleph_{\alpha} &\quad & \mbox{if $\aleph_{\beta}\lt\mathrm{cf}\;\aleph_{\alpha}$;}\\
\aleph_{\alpha+1} &&\mbox{if $\mathrm{cf}\;\aleph_{\alpha}\leq\aleph_{\beta}\leq\aleph_{\alpha}$;}\\
\aleph_{\beta+1} &&\mbox{if $\aleph_{\alpha}\leq\aleph_{\beta}$.}
\end{array}\right.$$
So, under GCH, for all cardinals $\kappa$ with cofinality greater than $\aleph_0$ have $\kappa^{\aleph_0} = \kappa$, and for cardinals $\kappa$ with cofinality $\aleph_0$ (e.g., $\aleph_0$, $\aleph_{\omega}$), we have $\kappa^{\aleph_0} = 2^{\kappa}$. (In particular, it is not the case the cardinality of $A^{\mathbb{N}}$ is necessarily less than the cardinality of $\mathcal{P}(A)$).
Then again, GCH is usually considered "boring" by set theorists, from what I understand.
(Note that $|\Bbb C|=|\Bbb R|$.)
If we assume the axiom of choice, then indeed the answer is that the set of countable subsets of $\Bbb R$ has the same cardinality as $\Bbb R$ itself.
The proof is simple. Using the axiom of choice, choose an enumeration of each such subset (i.e. a bijection between the countable set and $\Bbb N$). Next note the following cardinal arithmetic:
$$|\Bbb{R^N}|=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=|\Bbb R|.$$
Therefore the set of sequences with real coefficients have the same cardinality as the real numbers. Now send each countable set to its enumeration, and we're done.
Without the axiom of choice, however, it is consistent that there are strictly more countable subsets. For example in Solovay's model where every set is Lebesgue measurable this is true.
Best Answer
The proof is very intuitive (as you probably are feeling). But it can be written elaborately as follows, if you wish.
Your claim: For any finite set F, there exists an infinite subset I.
Try to prove: Let $F$ be a finite set defined as $F = \{f_1, f_2, \ldots , f_n\}$, where $n = 1, 2, \ldots$
Let $I$ be an infinite set defined as $I = \{i_1, i_2, \ldots, i_n, \ldots\}$, where n = 1, 2, ...
If I is a subset of F, then every element in I is also an element in F. If F contains finitely many elements, then only finitely many elements of I could belong to F.
However, I is infinite by definition, so clearly not all elements of I are contained in F.
Therefore, I is not a subset of F. This implies the claim is false.
Hence, for any finite set F, there does not exist an infinite subset I.
There is actually a proof you can probably find which does the same thing, just it takes a different angle: Prove that every subset of a finite set is finite. You can probably look this up somewhere!
I don't believe there are infinite planets in the universe. There are a large number, but it is not infinite. I don't believe anything in the universe is infinite, so there shouldn't be anything to reconcile here. Inverted World is sci-fi, so it's not even a theory. Just a nice tale!