# Thread: Prove: If f(A-B) = f(A)-f(B), then f is injective.

1. ## Prove: If f(A-B) = f(A)-f(B), then f is injective.

Book: Mathematical Analysis by Apostol

Theorem 2.2: $(a,b) = (c,d) \iff a = c \text{ and } b = d$.

$A - B = \{ x: x\in A, \text{ but } x \notin B\}$

Let $f: X \rightarrow Y$. Prove that if $f(A-B) = f(A) - f(B) \text {for all } A,B \subseteq X \text{, then } f \text{ is injective.}$

My attempt: Let $y \in f(A-B).$ Then $y \in f(A)-f(B)$, by assumption. Then, $\exists x \in A - B$ such that $y = f(x)$. Similarly, $\exists x_1 \in A$ such that $y = f(x_1)$.
Since $y \in f(A)-f(B)$, $y=f(x_1) \in f(A)$, but $y=f(x_1) \notin f(B)$. Then $x_1 \notin B$. Since $x \in A-B$, $x \in A$, but $x \notin B$.
So we have that $x,x_1 \in A$ and $x,x_1 \notin B$. Since $y = f(x)$ and $y = f(x_1)$, $f(x) = y = f(x_1)$.
This is equivalent to saying that for $(x,y) \in f$ and $(x_1, y) \in f$, we have that $(x, y) = (x_1, y)$.
By theorem 2.2, it must be that $x=x_1$. So $f(x) = f(x_1)$ implies that $x=x_1$. Therefore, f is injective.

Have I made any errors in my argument?

2. ## Re: Prove: If f(A-B) = f(A)-f(B), then f is injective.

Originally Posted by amthomasjr
Book: Mathematical Analysis by Apostol
Let $f: X \rightarrow Y$. Prove that if $f(A-B) = f(A) - f(B) \text {for all } A,B \subseteq X \text{, then } f \text{ is injective.}$
I simply do not follow any of that. You are told to $f: X\to Y$ is injective if $f(A\setminus B)=f(A)\setminus f(B)$ for all $A~\&~B$ subsets of $X$.
Suppose $f$ is not injective. Then $\exists \{s,t\}\subset X,~s\ne t\text{ BUT }f(s)=f(t)$.

What can you say about $\Large {f(\{s\}\setminus \{t\})\text{ and }f(\{s\})\setminus f(\{t\})~?}$

3. ## Re: Prove: If f(A-B) = f(A)-f(B), then f is injective.

Originally Posted by Plato
What can you say about $\Large {f(\{s\}\setminus \{t\})\text{ and }f(\{s\})\setminus f(\{t\})~?}$
$f(\{s\}) \setminus f(\{t\}) = \emptyset$, since $f(s) = f(t)$. But $f(\{s\}\setminus \{t\})$ cannot be empty since $s \ne t$

4. ## Re: Prove: If f(A-B) = f(A)-f(B), then f is injective.

Originally Posted by amthomasjr
$f(\{s\}) \setminus f(\{t\}) = \emptyset$, since $f(s) = f(t)$. But $f(\{s\}\setminus \{t\})$ cannot be empty since $s \ne t$
Is that a proof by contradiction?

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### f(a-b) f(a)-f(b)

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