Image of difference of sets

Aug 2018
44
0
Denmark
Hey,

I want to show that \(\displaystyle f(A\backslash B) \subset f(A) \backslash f(B)\)

I have done the following:

Assume that \(\displaystyle y \in f(A\backslash B)\).
This means that \(\displaystyle \exists x\in A\backslash B: f(x)=y\) (by definition).
This means that \(\displaystyle x\in A \land x\notin B\).
This means that \(\displaystyle f(x) \in f(A) \land f(x) \notin B\).
This means that \(\displaystyle f(x) \in f(A) \backslash f(B)\)

Therefore \(\displaystyle f(A\backslash B) \subset f(A) \backslash f(B)\).

Is the above correct?
 

Plato

MHF Helper
Aug 2006
22,472
8,642
Hey,

I want to show that \(\displaystyle f(A\backslash B) \subset f(A) \backslash f(B)\)

I have done the following:

Assume that \(\displaystyle y \in f(A\backslash B)\).
This means that \(\displaystyle \exists x\in A\backslash B: f(x)=y\) (by definition).
This means that \(\displaystyle x\in A \land x\notin B\).
This means that \(\displaystyle f(x) \in f(A) \land f(x) \notin B\).
This means that \(\displaystyle f(x) \in f(A) \backslash f(B)\)

Therefore \(\displaystyle f(A\backslash B) \subset f(A) \backslash f(B)\).

Is the above correct?
Yes it is completely correct. Good work.

One comment: I would have preferred using \subseteq $\subseteq$
 
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Dec 2012
1,145
502
Athens, OH, USA
The statement is not true for arbitrary functions. For example, let A be the set of positive reals, B the set of negative reals and f the squaring function. Your conclusion says $A\subseteq\emptyset$ ??

One hypothesis to make the statement true is that f is one to one. in the 4th line of your proof, you need some reason why $f(x)\not\in f(B)$ is true; I hope you understand why f one to one guarantees this to be true.
 
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