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How can I approach this?

I have to find the cardinality of the set of the functions from $\mathbb{Z} \to \mathbb{Z}$ and I have no idea on how to solve it.

Can someone hint me here?

The approach itself is what confuses me... how do I try to map this to something else since its a set of functions?

Brian M. Scott
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JOX
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    You can have functions that map functions to functions, for example the derivative can be thought of as a function that maps certain kinds of functions to functions. As an example the function $p: x \mapsto x^2$ gets mapped to the function $q: x \mapsto 2x$. Do you have any other detail on what about it is confusing? (or was that the main thing) –  Mar 27 '15 at 04:30
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    How you approach this will depend on what you already know. Do you know that $|\wp(\Bbb Z)|=|\Bbb R|$, for instance? – Brian M. Scott Mar 27 '15 at 04:34
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    @BrianM.Scott, I just proved something similar about $ |R| \leq |℘(N)|$ but I'm still unable to work with this specific proof because I don't know how to interpret "set of functions from Z to Z". – JOX Mar 27 '15 at 04:40
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    A function from $\Bbb Z$ to $\Bbb Z$ is just that: a function that assigns to each integer an integer. An example is the function $f(n)=n+1$. Another is $f(n)=n^2$. Another is that $f(n)$ is the $n$-th digit in the decimal expansion of $\pi-3$ if $n$ is positive, the $(-n)$-th digit in the decimal expansion of $\sqrt2-1$ if $n$ is negative, and $0$ if $n=0$. You can think of such a function as a double-ended infinite sequence: the one that I just mentioned is $$\langle\ldots,1,2,4,1,4,0,1,4,1,6,9,\ldots\rangle;,$$ for instance. – Brian M. Scott Mar 27 '15 at 04:49

3 Answers3

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One approach is to hypothesize some cardinal $\kappa$ for which you can prove $\kappa \leq \left|\mathbb{Z}^{\mathbb{Z}}\right|$ and $\left|\mathbb{Z}^{\mathbb{Z}}\right| \leq \kappa$.

Here's a hint: $$\left|\mathbb{Z}^{\mathbb{Z}}\right| \leq \left|(2^{\mathbb{Z}})^{\mathbb{Z}}\right| = 2^{|\mathbb{Z} \times \mathbb{Z}|}$$

Amit Kumar Gupta
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The given set has cardinality $\omega^\omega$. Clearly, this is greater than or equal to $2^\omega$. Conversely, it's less than or equal to $(2^\omega)^\omega=2^{\omega^2}=2^\omega$. Thus, by CSB, the cardinality is equal to $2^\omega=\mathbb R$.

Nishant
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$ A:=\{ f : \mathbb{Z}\rightarrow \mathbb{ Z} \} ,\ B:=\{ f: \mathbb{ Z}\rightarrow \{ 0,1 \} \} $ so that $$ |A|\geq |B|= 2^{\aleph_0 }= \mathfrak{c}$$ where $ \mathfrak{c}=|\mathbb{ R}|$ and $ \aleph_0=|\mathbb{N}|$ (cf. Schroder-Bernstein theorem)

Since $|\mathbb{Z}|=|\mathbb{N}|$, then $|A|=|A'|$ where $A':=\{f : \mathbb{N}\rightarrow \mathbb{N}\}$. And $|B|=|B'|$ where $B':=\{f : \mathbb{ N}\rightarrow \{0 , 1\} \}$

Now we will match an element in $A'$ into an element in $B'$.

For $f\in A'$, then define $$ 1\underbrace{0\cdots 0}_{f(1)-\text{times}} 1\underbrace{0\cdots 0}_{f(2)-\text{times}} 1 \cdots $$

so that we have $F\in B'$ : $$F(1)=1,\ F(i)=0 \ (2\leq i \leq 1+ f(1)),\ \cdots $$

HK Lee
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    $\mathbb Z^{\mathbb Z}$ (your $A$) is not a superset of $2^{\mathbb Z} = {0,1}^{\mathbb Z}$ (your $B$) in the usual representation of functions as sets in ZF. Furthermore, why are you writing $\mathcal N$ for the cardinality of $\mathbb R$ instead of the usual $\mathfrak c$, and $\mathcal N_0$ instead of $\aleph_0$? – kahen Mar 27 '15 at 14:12
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    TeX tips: Do not use \bf Z. Instead use \mathbb Z to get the usual symbol for the integers. I also see that you forgot a \text in the \underbrace construction. – kahen Mar 27 '15 at 14:13