I was looking to generate an injection from $\mathbb{N^N}$ to $\mathbb{R}$, and this is what i created:
$$
\begin{align*}
g: \mathbb{N^N} \hookrightarrow \mathbb{R} \\
f \mapsto p_1^{f(0)}p_2^{f(1)}p_3^{f(2)} \dots
\end{align*}
$$
Where $p_i$ is the $i$th prime number.
It looks like it creates an injection due to the fundamental theorem of arithmetic, but as it maps to only natural numbers, I can't see why its wrong, because clearly $|\mathbb{N^N|} > |\mathbb{N}|$?
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chin123
- 11
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3If $f(n) > 0$ for all $n$, what natural number is $g(f)$? – Alex Kruckman Feb 15 '19 at 05:54
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1The infinite product does not converge. – Kavi Rama Murthy Feb 15 '19 at 05:55
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If the sequence is infinite, the result is not a not a natural number as it is infinitely large. – fleablood Feb 15 '19 at 05:56
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1Oh, it should be infinity right? I completely missed that. Thank you! – chin123 Feb 15 '19 at 05:58
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It's an injection from all finite sequences but ... that's the rub... the cardinality of finite things are countable but the cardinality of infinite things are not. More or less-- that's an overgeneralization but it's mostly right. – fleablood Feb 15 '19 at 06:01