Function $\mathbb Z\times \mathbb Z \to \mathscr P(\mathbb Z)$ defined by $f(a,b) = \{a,b\}$. Show f is not one-to-one. Show f is not onto.

I have encountered this problem in my studies and to be honest with you, I am just having some trouble grasping the notation. I have reread the section in my textbook and consulted a few of my peers, but I have yet to understand. I figured I would just give a shot to posting it here to see if I gain any insight. I apologize if this is not welcome, but I thought I would give it a try. Thank you much!

Consider the function $f \colon \mathbb Z\times \mathbb Z \to \mathscr P(\mathbb Z)$ defined by $f(a,b) = \{a,b\}$ As usual, $\mathscr P(\mathbb Z)$ denotes the power set of the integers $\mathbb Z$

a) Give a specific example to show $f$ is not one-to-one

b) Give a specific example to show $f$ is not onto

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Did you try that at all? How about adding to the post some details about what you tried? – Asaf Karagila Oct 24 '12 at 6:58

Hints:

1. How many elements can $f(a,b)$ have?
2. You are looking at ordered pairs $(a,b)$ on the left hand side.

As you added that you are confused by notation, I'll try to make things more clear:

• $\mathbb{Z}\times \mathbb{Z}$ is the set of ordered pairs of integers, e.g. $(1,2)$, $(0,0)$, $(-1,8)$ are in there (and $(1,2)\neq (2,1)$).
• $\mathscr{P}(\mathbb{Z})$ is the power set of $\mathbb{Z}$, i.e. the set of all subsets of $\mathbb{Z}$, e.g. $\emptyset$, $\{1\}$, $\{1,3\}$, $\{1,2,3,5\}$ are in there.
• The function now associates to every ordered pair the set containing the two numbers in the coordinates, e.g. $f(1,2)=\{1,2\}$, the set containing $1$ and $2$. $f(0,0)=\{0,0\}=\{0\}$ since in sets double entries can be left out.
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Hint:

The infinite sets may confuse you. You can try showing that $g:\{0,1\}\times\{0,1\}\to\mathcal{P}(\{0,1\})$ given by $g(a,b)=\{a,b\}$ is neither one-to-one nor onto. The set $\{0,1\}$ has exactly four subsets, so you just check each case. The counterexamples you get will be counterexamples you can use also with $f$ instead of $g$.

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