Let A and B be nonempty sets and f:A→B be a 1-1 function. Then f(X ∩ Y) = f(X) ∩ f(Y) for all non empty subsets X and Y of A.
I believe this statement is true?
Asked
Active
Viewed 2,607 times
0
Terry Schmidt
- 287
-
Yes, it is true. – Boris Novikov Nov 20 '13 at 19:14
-
This looks like a duplicate of ii) of http://math.stackexchange.com/q/435900/11994. – MarnixKlooster ReinstateMonica Nov 20 '13 at 19:43
2 Answers
1
If $f:A\to B$ is not one-one, we can find $x,y$ for which $x\neq y$ yet $z=f(x)=f(y)$. Consider $X=\{x\}$ and $Y=\{y\}$. The converse is also true: if $f(X)\cap f(Y)=f(X\cap Y)$ for any $X,Y$ then $f$ is one-one.
Pedro
- 122,002
1
It is alway true that $f\left(X\cap Y\right)\subset f\left(X\right)\cap f\left(Y\right)$. If $z\in f\left(X\right)\cap f\left(Y\right)$ then $z=f\left(x\right)=f\left(y\right)$ for some $x\in X$ and some $y\in Y$. If $f$ is injective then this leads to $x=y\in X\cap Y$, hence $z=f\left(x\right)=f\left(y\right)\in f\left(X\cap Y\right)$. So then also $f\left(X\right)\cap f\left(Y\right)\subset f\left(X\cap Y\right)$ and consequently $f\left(X\right)\cap f\left(Y\right)=f\left(X\cap Y\right)$.
The sets do not have to be non-empty to have this property.
drhab
- 151,093