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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
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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Julian Kuelshammer Oct 24 '12 at 6:59
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Thanks for being kind and helpfully informing me. To be totally honest with you, I have having trouble understanding the notation behind this. I have been rereading this section in my textbook and am still trying to understand. I thought I would just try this out to see if I got any insight faster. Next time, I will tag this as homework as to not spam your query. Thanks again! –  Drew Hagen Oct 24 '12 at 7:04
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2 Answers

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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