There is a function $f:\Bbb{N}\times\Bbb{N_+}\to \Bbb{Q}$ , $f(n,m):=\frac{n}{m}$.

Let $F:\wp(\Bbb{Q})\to\wp(\Bbb{N\times N_+})$ be a function such that $F(A):=f^{-1}[A].$

(here $\wp(X):=\{B\mid B\subseteq X\}$)

  1. Is function $F$ surjective?
  2. Is function $F$ injective?
  3. Find $F[\{\{-1\},\{1\}\}]$ and $F[\{\{-1,1\}\}]$

I had this exercise on my exam and didn't understand what this function actually does. Can anyone give some hints or say how it should be done properly

  • $\begingroup$ To make sure: is $\varphi(X)$ a notation for the powerset of $X$? $\endgroup$ – drhab Feb 9 '17 at 13:38
  • $\begingroup$ @drhab Yes, originally it was some other letter but I couldn't find that one $\endgroup$ – UfmdFkiF Feb 9 '17 at 13:41
  • $\begingroup$ @TheMeff in general powerset is notated using "\mathcal P" or "\wp" or some other kind of "p". $\endgroup$ – Masacroso Feb 9 '17 at 13:44
  • $\begingroup$ HINT: any set-valued function induced from a function is bijective (and have inverse). $\endgroup$ – Masacroso Feb 9 '17 at 13:46

Hint. The function $F$ takes a set of rational numbers to the set of all pairs of integers with nonnegative denominator that represent those rational numbers as fractions in the obvious way.

That's "what this function actually does".

Can you finish? Perhaps start with a few examples - e.g. (3).

  1. Certianly $\{(1,1)\}$ is a subset of $\mathbb{N}\times \mathbb{N}_+$. So if $F$ was surjective then this set must be in image of $F$. That means, for some subset $A$ of $\mathbb{Q}$, you must have that $f^{-1}[A]=\{(1,1)\}$. Well since $f(1,1)=1$, this means that $A$ must contain $1$ as an element. But then you see that there's a little problem here (look at the definition of $f$).

  1. Is $F$ injective? Well for $F$ to be injective you need for every subsets $A,B$ of $\mathbb{Q}$, if $F(A)=F(B)$ then $A=B$. That means $f^{-1}(A)=f^{-1}(B) \implies A=B$. But there is a little problem with that happening. Suppose you have a subset $A$ of $\mathbb{Q}$, which does not contain $-1$ (or pick any negative rational that's not contained in $A$. If $A$ contains all negative rationals you can instead pull out a negative rational). If you add $-1$ to $A$, does that change $f^{-1}(A)$?

  1. Hopefully you've understood how $F$ works reasonably well at this point. Now you can simply apply definition. $A=\{\{-1\},\{1\}\}$, then $F(A)=F(\{-1\})\cup F(\{1\})=f^{-1}(-1)\cup f^{-1}(1)$, and you can do the same thing with the second one. It shouldn't be too hard to identify these sets

Definition 0. Given a function $f : A \rightarrow B$, there's another function $\mathcal{P}(f) : \mathcal{P}(B) \rightarrow \mathcal{P}(A)$ defined by $$\mathcal{P}(f)(S) = f^{-1}(S).$$

Proposition 0. Given a function $f : A \rightarrow B$, we have

  • $f$ is injective iff $\mathcal{P}(f)$ is surjective
  • $f$ is surjective iff $\mathcal{P}(f)$ is injective

The proof is left as an exercise.

Ergo, since the function $f : \mathbb{N} \times \mathbb{N}_{>0} \rightarrow \mathbb{Q}$ you describe is surjective but not injective, hence $\mathcal{P}(f)$ is injective but not surjective.

  • 1
    $\begingroup$ $F$ is not injective (there are negative rationals so $f$ is not surjective). (But this makes it clear that if it were to $\mathbb Q_+$ then $F$ would be injective) $\endgroup$ – spaceisdarkgreen Feb 10 '17 at 7:09

I saw this through your duplication today and being that you've already been chastised for bad form I will answer a few of your questions from the duplicate since they indicate where you're having trouble:

You ask "why does $f^{-1}$ even exist since f isn't injective?" If you're confused about this, then there's no hope for the rest. The key is to notice that in the definition of $F$, $f^{-1}$ is applied to $A$ which is a subset of $\mathbb Q,$ not an element of $\mathbb Q.$ Thus, $f^{-1}$ does not refer to the nonexistent inverse function $\mathbb Q\to \mathbb N \times \mathbb N_+$ but rather the always existent set inverse function $ f^{-1}:\mathcal{P}(\mathbb Q)\to \mathcal{P}(\mathbb N\times \mathbb N_+)$ that takes a subset of $\mathbb Q$ and returns its preimage under the function $f.$

For instance let's consider a subset of $\mathbb Q.$ How about $\{1/3\}.$ Then $f^{-1}(\{1/3\})$ is the set of all pairs in $\mathbb N\times\mathbb N_+$ that map to $1/3$ under $f.$ This would include $(1,3),$ but also $(2,6)$ and $(3,9),$ etc. So, we have $$ F(\{1/3\}) = \{(1,3),(2,6),(3,9),(4,12),\ldots\}.$$ Note that the function takes a subset of $\mathbb Q$ (i.e. an element of $\mathcal P(\mathbb Q)$) and maps it to a subset of $\mathbb N\times \mathbb N_+$ (an element of $\mathcal P(\mathbb N\times\mathbb N_+)).$

Let's do another one: $$ F(\{0,1/2\}) =\{(0,1),(1,2),(0,2),(2,4),(0,3),(3,6),\ldots\}.$$ Does that make sense?

Now reread the short, hinty answer you receieved from Ethan above. Hopefully it makes more sense now. (Although, when he said "non-negative" and "integers", I think he was confused and thought $\mathbb N$ was $\mathbb Z$. I'm assuming that $\mathbb N = \{0,1,2,\ldots\}$ and $\mathbb N_+ = \{1,2,3,\ldots\}.$)

Moving onto the first part of the problem, for $F$ to be surjective, every possible subset of $\mathbb N\times \mathbb N_+$ must be the image under $F$ of some subset of the rationals. As a previous answerer indicated, the subset $\{(1,1)\}$ is a counterexample. Can you see why it can't be the image of any subset of $\mathbb Q$? Notice in both my examples above the image was an infinite set.

For injectivity you need each subset of $\mathbb Q$ to map to a unique subset of $\mathbb N\times\mathbb N_+.$ As the previous answer said, the trick is to consider sets involving negative rational numbers.

For point three, note they are applying $F$ as a set function (just like $f^{-1}$ above except it maps a subset of the domain to the subset of the range that is its image). $F[]$ takes $\mathcal P(\mathcal P(\mathbb Q)) \to \mathcal P(\mathcal P(\mathbb N\times \mathbb N_+)).$ In the first case you are asked to apply it to $\{\{-1\},\{1\}\}.$ So there are two domain points in this subset, $\{-1\}$ and $\{1\}.$ We know that $$F(\{-1\}) = \emptyset$$ since $f$ doesn't touch the negative rationals. We also know that $$F(\{1\} )= \{(1,1),(2,2),(3,3),\ldots\}.$$ So these are our two image points and we have $$F[\{\{-1\},\{1\}\}] = \{\emptyset, \{(1,1),(2,2),(3,3),\ldots\}\}.$$

This is probably a little confusing, so let's compare to a simple function $g:\mathbb R\to \mathbb R.$ Like let's say $g(x) = x^2.$ Then say we are to consider it as a set function and are asked for $g[\{2,10,-12,-2\}].$ The answer is $$g[\{2,10,-12,-2\}] = \{4,100,144\},$$ right? What we did above is the same thing only the function $F$ had a sets of sets for its domain and range.

Now for the other one, $F[\{\{-1,1\}\}].$ Notice now there is only one element in the subset of the domain that we need to find the image of. Our answer will be a set consisting of a single subset of $\mathbb N\times \mathbb N_+.$ I'll leave it to you.

  • $\begingroup$ But why $\{(1,1)\}$ can't be an image? $\endgroup$ – UfmdFkiF Feb 10 '17 at 6:58
  • $\begingroup$ @TheMeff Can you find a subset of rational numbers that maps to it? Read the other answer's point 1. The 'little problem here' is that having $1$ anywhere in the preimage generates an infinite set for the image. In fact, the image $ F(\mathcal P(\mathbb Q))$ only consists of infinite sets. (Other than the empty set.) Look at my examples. Do you see why pretty much any set in the image is going to be infinite like those? $\endgroup$ – spaceisdarkgreen Feb 10 '17 at 7:03
  • $\begingroup$ Oh, now I see. And in $F(\{0,1/2\}) =\{(0,1),(1/2),(0,2),(2,4),(0,3),(3,6),\ldots\}.$ why there is (1/2)? shouldn't it be (1,2)? $\endgroup$ – UfmdFkiF Feb 10 '17 at 7:10
  • $\begingroup$ @TheMeff Yes, that's a typo $\endgroup$ – spaceisdarkgreen Feb 10 '17 at 7:13
  • $\begingroup$ To be sure, $F$ isn't injective because e.g. adding {-1} to the set does not change the value? $\endgroup$ – UfmdFkiF Feb 10 '17 at 7:19

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