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Edit: by now I've figured out i<=>ii! Just not iii<=>i

For an assignment I need to prove the following: i<=>ii<=>iii. My approach was to prove i<=>ii and i<=>ii. I'm mainly getting stuck on the 'iii<=>i' parts here.

(i) f is injective

(ii) For all A, B ⊆ X: f ( A ∩ B ) = f ( A ) ∩ f ( B ).

(iii) For all A, B ⊆ X: f ( A \ B ) = f ( A ) \ f ( B ).

I saw a proof like this before, but I did not really understand all of the answers, and as the question was posted years ago, I thought it'd be better to post my own question.

Marc
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2 Answers2

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Assume $f$ is injective and take an arbitrary $y$ in $f(A\backslash B)$. Since $f$ is injective, there is a unique $x$ such that $f(x)=y\in f(A\backslash B)$. For this $x$ we have that $x\in A\backslash B$ so $x\in A$ and $x\not\in B$, so $f(x)\in f(A)$ and $f(x)\not\in f(B)$. Since $y$ was arbitrary, $f(A\backslash B)\subset f(A)\backslash f(B)$. (i)

Now take a $y$ in $f(A)\backslash f(B)$, so $y\in f(A)$ and $y\not\in f(B)$. Again, since $f$ is injective, there is a unique $x$ such that $f(x)=y$. So for this $x$ we have $x\in A$ and $x\not\in B$, so $x\in A\backslash B$, so $f(x)=y\in f(A\backslash B)$. So $f(A)\backslash f(B)\subset f(A\backslash B)$. (ii)

Thus from (i) and (ii) we conclude that if $f$ is injective, $f(A)\backslash f(B)=f(A\backslash B)$. This establishes the forward implication($\Rightarrow$).

Now assume that $f(A)\backslash f(B)=f(A\backslash B)$ for all $A,B \in X$. Take an $f(x)=y\in f(A\backslash B)=f(A)\backslash f(B)$. Thus if $x\in A\backslash B$, then $f(x)\in A$ and $f(x)\not\in B$. (iii)

Now assume that $f$ is not injective. Then there exist $x_1,x_2\in X$ such that $f(x_1)=f(x_2)$. Choose disjoint sets $A,B$ such that $x_1\in A$ and $x_2\in B$. Then we have from (iii) that for $x_1\in A$ so $x_1\not\in B$ (because $A$ and $B$ are disjoint) that $y=f(x_1)\in f(A\backslash B)=f(A)\backslash f(B)$. So $y\not\in f(B)$. This is contradiction since we assumed that $f(x_1)=f(x_2)$ and since $x_2\in B$, $f(x_2)\in f(B)$, thus $f(x_1)=f(x_2)\in B$. Because assuming $f$ is not injective leads to a contradiction, $f$ must be injective.

So if $f(A)\backslash f(B)=f(A\backslash B)$, then $f$ is injective. This establishes the backward implication($\Leftarrow$).

Václav Mordvinov
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  • I don't quite understand this step: ''Since f is injective, there is a unique x such that f(x)=y∈f(A∖B). For this x we have that x∈A\B''

    I know that if f is injective that if f(a)∈f(A) --> a ∈ A, but how does this follow from its definition?

     – Marc Nov 16 '17 at 19:37
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    The $x$ is unique, that is what follows from the defintion of injectivity. For a given $y$, there is a unique $x$ such that $f(x)=y$. Because $y\in f(A\backslash B)$, we conclude furthermore that $x\in A\backslash B$. – Václav Mordvinov Nov 16 '17 at 20:46
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You can prove the above equivalence as follows: (i)$\Rightarrow$(ii)$\Rightarrow$(iii)$\Rightarrow$ (i)

(i)$\Rightarrow$(ii) Suppose $f$ is injective. Then to show that $f(A\cap B)=f(A)\cap f(B).$

let $x\in f(A\cap B)$. There exists some $y\in A\cap B$ such that $f(y)=x.$ $$\Rightarrow f(y)=x, y\in A\ \text{and}\ f(y)=x,\ y\in B$$ $$\Rightarrow x\in f(A)\ \text{and}\ x\in f(B)$$ $$\Rightarrow x\in f(A)\cap f(B)$$ $$\Rightarrow f(A\cap B)\subseteq f(A)\cap f(B)$$

conversely, suppose $z\in f(A)\cap f(B)$ Then, $$ z=f(u)\ \text{and}\ z=f(v)\ \text{for some } u\in A \text{and} \ v\in B$$ $$\Rightarrow f(u)=f(v)$$ $$\Rightarrow u=v \text{ (since $f$ is injective)}$$ $$\Rightarrow u=v\in A\cap B$$ $$\Rightarrow z\in f(A\cap B)$$ $$f(A)\cap f(B) \subseteq f(A\cap B)$$

(ii)$\Rightarrow $(iii)

suppose (ii) is true. Then $f(A\setminus B)=f(A\cap B^c)=f(A)\cap f(B^c)=f(A)\cap f(B)^c=f(A)\setminus f(B)$

(iii)$\Rightarrow$ (i)

We prove the contrapositive.

Suppose $f$ is not injective. Then for some $x\neq y$ we must have $f(x)=f(y)$. let $x\in A$ and $y\in X\setminus A$

Then $f(y)\in f(X\setminus A)=f(X)\setminus f(A)$ which implies that $f(y)\notin f(A)$ BUT we already have $f(y0=f(x)\in f(A)$- Contradiction. This implies that $f(A\setminus B)\neq f(A)\setminus f(B)$. Thus (iii)$\Rightarrow$(i) is established.

Dastan
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  • Isn't ''f(A)∩f(Bc)=f(A)∩f(B)c'' exactly what needs to be proven (and can thus not be used?) Because Bc is another way of writing A\B. – Marc Nov 16 '17 at 19:13
  • Yes, I agree.Actually here we need to show that $f(A^c)=f(A)^c$ – Dastan Nov 16 '17 at 19:31
  • Just to add to previous comments, the problem with (ii)$\Rightarrow$(iii) is that it implicitly uses $f(B^c)=f(B)^c$. What this actually means depends on what we take as the universal set in the definition of complement. One possible interpretation $f(X\setminus B)=Y\setminus f(B)$, which is definitely not true for functions which are not surjective. Another possible interpretation would be $f(A\setminus B)=f(A)\setminus f(B)$, which is actually the claim we are trying to prove. – Martin Sleziak Mar 24 '19 at 11:29