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There are more numbers in $\mathbb{R}$ than in $\mathbb{N}$. There are as many vectors in $\mathbb{R}^n, n \in \mathbb{N}$ as numbers in $\mathbb{R}$.

How many real functions are there? If I denote $\{f,f: \mathbb{R} \rightarrow \mathbb{R}\}=\mathbb{R}^\mathbb{R}$ I get another quite.. very big, infinite, uncountable set, right?

Is $|\mathbb{R}^\mathbb{R}| > |\mathbb{R}|$?.. another uncountable infinite beyond continuity?

What about $\left|(\mathbb{R}^\mathbb{R})^{(\mathbb{R}^\mathbb{R})}\right|$ then? How far can we build huge sets in this direction?
Is the number of various infinites itself countable? $\left\{|\mathbb{N}|,|\mathbb{R}|,|\mathbb{R}^\mathbb{R}|,\left|(\mathbb{R}^\mathbb{R})^{(\mathbb{R}^\mathbb{R})}\right|,\dots\right\}$ would be, right?

Or maybe $|\mathbb{R}^\mathbb{R}| = |\mathbb{R}|$ and we fall back on our feet?

iago-lito
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  • Very similar question: https://math.stackexchange.com/questions/477/cardinality-of-set-of-real-continuous-functions –  Oct 20 '17 at 11:37
  • @Jazzachi Yeah, but this is about continuous functions only, right? ;) What about all functions (even ugly ones)? I think that $\mathcal{C}(\mathbb{R}) \subset \mathbb{R}^\mathbb{R}$ so $|\mathcal{C}(\mathbb{R})| < |\mathbb{R}^\mathbb{R}|$, right? – iago-lito Oct 20 '17 at 11:39
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    These questions drove Cantor crazy, be careful! Anyway generalized continuum hypothesis should guarantee that the set of cardinals is countable. On the other hand Cardinal numbers are also written as $$\aleph_0,;\aleph_1,;\ldots,;\aleph_n,;\ldots$$ where $\aleph_{n+1}=2^{\aleph_n}$ is the cardinality of the subsets set of a set having cardinality $\aleph_n$ – Raffaele Oct 20 '17 at 12:43
  • @Raffaele And this is a very nice way of writing down the concern I had for this infinite series of infinites. Cheers for that :) And don't hesitate to add this pointer to paw88789's answer below ;) – iago-lito Oct 20 '17 at 12:57

1 Answers1

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The set $S$ of functions $\mathbb{R}\to\mathbb{R}$ has the same cardinality as the power set of $\mathbb{R}$.

Step 1: Associate to each function in $S$ its graph in $\mathbb{R}^2$. This graph can be considered as a subset of $\mathbb{R}^2$. And no two functions in $S$ have the same graph. So the cardinality of $S$ is not greater than the cardinality of the power set of $\mathbb{R}^2$ which has the same cardinality as the power set of $\mathbb{R}$. So the cardinality of $S$ is not greater than the cardinality of the power set of $\mathbb{R}$.

Step 2: Let $A$ be the set of subsets of $\mathbb{R}$ that include the element $0$. So $A$ has the same cardinality as the power set of $\mathbb{R}$. For each element $K$ of $A$, we can find a function in $S$ whose range is $K$ (namely take the function from $\mathbb{R}\to\mathbb{R}$ that sends each element of $K$ to itself and each nonelement of $K$ to $0$). This gives a mapping from $S$ to $A$ that is onto. So the cardinality of $S$ is at least as great as the cardinality of $A$ which is equal to the cardinality of the power set of $\mathbb{R}$.

paw88789
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  • Great! This answers it. And so there are as many infinites than continuous powersets that you can make powersets of, which sounds infinite countable. Cheers :) PS: with your permission, I'll take time to edit your post and rewrite it using $\forall, \exists, \wp, $ etc.. It has taken me a while to get them right from the text, even though it's crystal-clear ;) – iago-lito Oct 20 '17 at 12:53
  • @iago-lito If you do your edits, I would prefer that you do them underneath, leaving the original post as it was, with the alternative version underneath. Thanks. – paw88789 Oct 20 '17 at 15:25
  • Okay. Anyway, maybe it was just me having a hard time.. now it's clear :) – iago-lito Oct 20 '17 at 15:28