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Question: Suppose $f:X\to Y$ and $g:Y\to X$ two arbitrary functions. Prove that $\{A,B\}$ is a partition for X and $\{C,D\}$ is a partition for Y such that :
$$\begin{align} f(A) = C\end{align}$$ $$\begin{align} g(D) = B\end{align}$$ Definitions that I know:
I have to show that $\bigcup_{i\in I} \{A,B\}_{i\in I} = X$ and $\bigcup_{j\in J} \{C,D\}_{j\in J} = Y$ and also if $i\neq j$ then $\{A,B\}_{i\in I}\cap \{A,B\}_{j\in J} = \varnothing$.
I know that by Axiom schema of specification and Axiom of pairing there is set $$\begin{align} D=\{A,B\} = \{x\in C : x=A \,\text{ or }\, x=B\}\end{align}$$ I have know idea how to start, just clear for what's going on? I want to show that there are exists such that partitions.

Gary
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tent123
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  • Are you trying to prove that there exist such partitions? You don't mention that. You just say "Prove..." without saying what A,B,C,D are. – coffeemath Apr 22 '22 at 19:41
  • @coffeemath yes, I m trying to prove that there are exist such that partitions With the properties that I said .... to specify A, B , C, D My teacher said nothing, He asked me this question. – tent123 Apr 22 '22 at 19:56
  • also by Axiom of pairing there is set like {A,B} and , actually the question is like that. – tent123 Apr 22 '22 at 20:01
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    You may be overcomplicating things. A,B are simply two disjoint sets whose union is X. Similarly C,D are just two disjoint sets whose union is Y. The question is to show such sets exist having your two properties. By for example f(A) is meant the set of values f(a) as a ranges over A. – coffeemath Apr 22 '22 at 20:12
  • @coffeemath yep, thank you. I'm really beginner at the topic, to prove that such that sets exists, where should I really start? – tent123 Apr 22 '22 at 20:22
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    @coffeemath actually it is true but I want to show that there are exist such that partition – tent123 Apr 23 '22 at 18:36
  • no any Tips or guidance to start?? – tent123 Apr 24 '22 at 21:48
  • tent-- Are you still trying to prove there are such partition sets A,B,C,D? If so I have a class of counterexamples to the existence of A,B,C,D which I can put here. Let me know. – coffeemath Apr 25 '22 at 16:18
  • tent123: I have written up the details of a counterexample in my answer below. [I had deleted a smaller example but found this one in which $X,Y$ are infinite. – coffeemath Apr 28 '22 at 03:30

1 Answers1

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First note that by definition a partition of $X$ is any choice of two nonempty sets $A,B$ which are disjoint, and whose union is $X.$ Simalarly for the partition of $Y$ into two sets $C,D.$ The question is whether, for any sets $X,Y$ and any functions $f:X \to Y,\ g:Y \to X$ there are partitions of $X$ into $A,B$ and of $Y$ into $C,D$ such that $$f(A)=C,\ \ f(D)=B.$$

The point of this answer is to show the answer to the question posed is no. For this we let $X$ denote the set of nonzero integers, and $Y$ the set of positive integers. We define $f:X \to Y$ by $f(x)=|x|,$ the absolute value of the integer $x.$ And we define $g:Y \to X$ by $g(y)=y,$ which we may do since any positive integer is also a nonzero integer and so is in set $X.$ Now $A$ is a collection of nonzero integers. First we note that it cannot be the case that, for each nonzero integer $x,$ our set $A$ contains either $x$ or $-x$ (or both). This is because if it did then every $y$ in $Y$ would be $g(x),$ i.e. $C=g(X)$ would be the entire set $Y.$ Tht is not possible because the complement of $C$ in $Y$ is by definition $D,$ so $D$ would be empty, while we need $D$ nonempty in order for $C,D$ to be a partition of $Y.$

So we know there is a specific pair $-e,e$ of integers neither of which lies in $A.$ It follows we have $-e,e \in B$ since $A,B$ partitions $X.$ Also the positive integer $k=|e|$ is not in $C,$ so it must lie in $D.$ But then we do not have $g(D)=B$ because $B$ has in it the negative integer $-k$ which is not $g(y)$ for any $y$ in $D.$ Thus we have arrived at a contradiction and the desired sets $A,B,C,D$ do not exists in our example.

coffeemath
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