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Consider the function $\theta:\mathcal{P}(\mathbb{Z}) \to \mathcal{P}(\mathbb{Z})$ defined as $\theta(X) = \bar{X}$. Is $\theta$ injective? Is it surjective? Bijective? Explain.

I know how to prove this for more typical functions like $f: x \mapsto x^2$, but how would I prove injectivity and surjectivity for this particular function?

Here are my sketches:

Injectivity (Contrapositive):

Suppose $X,Y \in \mathcal{P}(\mathbb{Z})$ and $\theta(X) = \theta(Y)$.

$\ldots$

Therefore $X = Y$ (If injective)

Surjectivity:

Suppose $B \in \mathcal{P}(\mathbb{Z})$

$\ldots$

There exists $A \in \mathcal{P}(\mathbb{Z}$) for which $\theta(A) = B$ (If surjective)

I really only need an explanation for either injectivity or surjectivity, then I'll try to figure out the other one myself.

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  • $\begingroup$ Let me get this straight: $\mathcal P(\Bbb Z)$ is the power set of $\Bbb Z$, and $\bar X$ is the complement of $X$ in $\Bbb Z$, $\Bbb Z \setminus X$, right? $\endgroup$ – Robert Lewis Dec 10 '14 at 4:06
  • $\begingroup$ Yep, that is correct. Although my textbook didn't specify what $\bar{X}$ was, I don't know what else it could be. $\endgroup$ – St Vincent Dec 10 '14 at 4:09
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$\textbf{Injectivity-}$ If $X^c=Y^c$ then $(X^c)^c=(Y^c)^c$ implying $X=Y$

$\textbf{Surjective-}$ Let $B \in \mathcal{P}(\mathbb{Z})$ then $\theta(B^c)=B$. Done

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