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The set of all strictly increasing sequences ($a_n$) of natural numbers has cardinality $\mathbb{N}$, $\mathcal{P}(\mathbb{N})$ or $\mathcal{P}( \mathcal{P}(\mathbb{N}))$?

I answered $\mathcal{P}(\mathcal{P}(\mathbb{N}))$ because as the set can be described by $X=\{\{\ldots, a_n, \ldots \},\{\ldots,b_n,\ldots\}, \ldots\}$ as it is not known if there are any limited sequences (so, assuming all have infinite elements), the cardinality of $X$ could not be the first option $\mathbb{N}$ because considering any function $\phi : \mathbb{N} \rightarrow X$ there would always be elements not listed from $X$. So I thought it would possible to say that $|X|\ge | \mathcal{P}(\mathbb{N})|$ and thinking that trying to build any bijection $\varphi: \mathcal{P}(\mathbb{N}) \rightarrow X$ would also not be possible as there would be elements from $X$ not listed too. So than I got $|X| = |\mathcal{P}( \mathcal{P}(\mathbb{N}))|$, but I would like to know if I answered correct and, if not, why?

hmakholm left over Monica
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Marcos
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    The set of strictly increasing sequences of natural numbers can be seen to be in bijection with the set of sequences of positive natural numbers $(a_1,a_2,a_3,\dots)\mapsto (a_2-a_1,a_3-a_2,a_4-a_3,\dots)$. What do you know about the cardinality of $\Bbb N^{\Bbb N}$? (Remember something special about $2^{\Bbb N}$) – JMoravitz Jan 22 '18 at 17:19
  • What on earth did you hope to gain by pointlessly switching out and in of math mode dozens of times? – hmakholm left over Monica Jan 22 '18 at 17:24
  • Related: https://math.stackexchange.com/questions/2439340/proving-the-set-of-the-strictly-increasing-sequences-of-natural-numbers-is-not-e/2439666, https://math.stackexchange.com/questions/461054/how-many-monotonically-increasing-sequences-of-natural-numbers – Carmeister Jan 22 '18 at 19:46
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    @JMoravitz, that's not a bijection as there's no way to recover $a_1$, although it could likely be adjusted into a bijection. – Wildcard Jan 23 '18 at 01:09
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    Good point @Wildcard. Adjust it to $(a_1,a_2,a_3,\dots)\mapsto (a_1,a_2-a_1,a_3-a_2,\dots)$ then, or if you want to include zeroes as a possibility, $\mapsto (a_1,a_2-a_1-1,a_3-a_2-1,\dots)$ – JMoravitz Jan 23 '18 at 02:20

3 Answers3

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Each strictly increasing sequence of natural numbers corresponds to a particular infinite subset of $\mathbb N$. So there can't be more such sequences than there are subsets.

On the other hand, for every (finite or infinite) subset there is an infinite subset (which again corresponds to a strictly increasing sequence) -- for example, just multiply every number in the original subset by $2$ and then add in all of the odd numbers. So there can't be more subsets than there are sequences.

By Cantor-Bernstein, then, the set of strictly increasing sequences has the same cardinality as $\mathcal P(\mathbb N)$.

hmakholm left over Monica
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Let $X$ be the set of strictly increasing sequences of natural numbers.

Define $F:X\to\mathcal P(\Bbb N)$ as $F(\{a_n\})=\{a_n:n\in\Bbb N\}$.

This function is clearly injective, and its image is precisely the set $Y=\{A\subset\Bbb N:A\text{ is infinite}\}$.

Since the complementary of $Y$ in $\mathcal P(\Bbb N)$ is the set of finite subsets of $\Bbb N$, which is countable, then $|X|=|\mathcal P(\Bbb N)|$.

ajotatxe
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Let $X$ be the set of strictly increasing sequences of natural numbers. Then there is an injective map from $X$ to $P(\mathbb{N})$ where $(a_{n})\to\{a_{n}\mid n\in\mathbb{N}\}$. Hence $$ |X|\leq|P(\mathbb{N})|\tag{0}\label{0} $$ There is also a surjective map from $X$ to $\mathbb{N}\times (\mathbb{N}\setminus 0)^{\mathbb{N}}$ where $a_{n}\to (a_{0}, \Delta a_n )$. Here $\Delta a_n=a_{n+1}-a_{n}$. Hence $$ |X|\geq|\mathbb{N}\times (\mathbb{N}\setminus 0)^{\mathbb{N}}|=|\mathbb{N}^{\mathbb{N}}|=|2^{\mathbb{N}}|=|P(\mathbb{N})|\tag{1}\label{1} $$ since $$ |2^{\mathbb{N}}| ≤ |\mathbb{N}^{\mathbb{N}}| ≤ |(2^\mathbb{N})^\mathbb{N}| = |2^{\mathbb{N}\times \mathbb{N}}| = |2^{\mathbb{N}}|. $$ By Cantor-Bernstein namely, $\eqref{0}$ and $\eqref{1}$, $$ |X|=|P(\mathbb{N})|. $$

Sri-Amirthan Theivendran
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  • For this problem it might be convenient to define $\mathbb{N}$ as the set of all positive integers (rather than nonnegative). Then define $\theta$ from the set of strictly increasing sequences of natural numbers to the set of all sequences of natural numbers, by $\theta (a_1, a_2, a_3, \ldots) = (a_1, a_2 - a_1, a_3 - a_2, \ldots)$. It is then pretty straightforward to show directly that $\theta$ is both injective and surjective. This reduces the problem to showing that $\lvert \mathbb{N}^{\mathbb{N}}\rvert = \lvert P(\mathbb{N})\rvert$, as your chain of inequalities shows nicely. – mathmandan Jan 22 '18 at 20:00