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Prove, that set $\{f \in \mathbb{N^N} \: | \:f \: $is strictly increasing $\}$ has the same cardinality as $\mathbb R$.

My attempts:

The beginning of this task was quite easy, but then I got stuck on constructing an injection between a set of function (let's call it $X$) and $\mathbb R$. I started with proving that $|\mathbb R| \geq |X|$:

  • $|\mathbb R| \geq |X|$ because if $\forall _f , f\in \mathbb{N^N}$, and $|\mathbb{N^N}|=|\mathbb R$|, then $X \subset\mathbb R$.

Then I tried to prove that $|\mathbb R| \leq |X|$, but I don't know how to do it. I tried to define a function $g(x)=x^3$, but the result is a number, not a function. Or maybe it is a correct solution?

If not, can you explain to me how can I construct an injective function from $\{f \in \mathbb{N^N} \: | \:f \: $is strictly increasing $\}$ to $\mathbb R$? Is it even possible?

Andrés E. Caicedo
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zu_m
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  • Hint: construct an injection ${ 0, 1 }^{\mathbb{N}} \to X$. Intuitively, this construction can be thought of as finding infinitely many properties of a function $f \in X$ such that every combination of these properties is represented by some function $f \in X$. – Adayah Aug 26 '17 at 22:11
  • You already have "an injective function from ${f \in \mathbb{N^N} : | :f : $is strictly increasing $}$ to $\mathbb R$", by your own admission, from the bijection between $\Bbb N^{\Bbb N}$ to $\Bbb R$, composed with inclusion. What your want it's an injective function the other way. This means take a number in $\Bbb R$ (or even easier: in $[0,1)$), and construct an increasing element of $\Bbb N^{\Bbb N}$ from that number in an injective way. – Arthur Aug 26 '17 at 22:12

3 Answers3

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The $\mathscr{P}(\mathbb{N})$, set of all subsets of $\mathbb{N}$, has the same cardinality as $\mathbb{R}$. Let $Y$ be the collection of all infinite subsets of $\mathbb{N}$. Since the collection of all finite subsets of $\mathbb{N}$ is countable, $Y$ has the same cardinality as $\mathscr{P}(\mathbb{N})$.

Let $X$ be the collection of all strictly increasing sequences from $\mathbb{N}$ to $\mathbb{N}$. A bijection of $X$ to $Y$ is given by $f \in X$ is mapped to $\mathrm{rang}(f) \in Y$.

William
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Hint: Try to encode each $f$ as a sequence of $0$s and $1$s.

user357980
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  • That would give us an injection $X \to { 0, 1 }^{\mathbb{N}}$ which is opposite to what we need. – Adayah Aug 26 '17 at 22:12
  • No, it is a bijection. – user357980 Aug 26 '17 at 22:29
  • Encoding each $f$ as a sequence of $0$s and $1$s sounds like assigning to each $f$ a distinct infinite binary sequence, but nothing in the answer seems to imply this assignment should be surjective. At least it wasn't clear to me. – Adayah Aug 27 '17 at 07:33
  • Since $f$ is strictly increasing, you can write $f$ exactly as a sequence of zeros and ones like as in this example: If $f = 1, 3, 4, 7, \dots,$ then the sequence is $1, 0, 1, 1, 0, 0, 1, \dots$. – user357980 Aug 28 '17 at 03:55
  • So it is not surjective, because the sequence $0, 0, 0, 0, 0, \ldots$ is not encoded. Anyway, I only meant to point out that the wording of the answer doesn't indicate that we want a surjective mapping. – Adayah Aug 28 '17 at 06:32
  • Good points. Its image is the set of zeros and ones that do not end in all zeros. Fortunately, the set of binary sequences that end in zeros and ones is countable (you can identify it with ${\emptyset} \cup \bigcup_{n=1}^{\infty}{0,1}^n$) and it is an exercise to prove that if $A$ is uncountable and $B$ is a countable subset of $A$, then there is a bijection between $A$ and $A\setminus B$. – user357980 Aug 30 '17 at 03:14
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An infinite subset $A$ of $\mathbb{N}$ can be coded by a strictly increasing function: Define $f(0) = \min(A)$, and recursively $f(n+1) = \min(A \setminus f[\{0,\ldots,n])$

Henno Brandsma
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