0

I came across this, $f(A)\setminus f(F) \neq f(A\setminus F)$. I'm not sure that I completely understand why they cannot be equal. So could someone give an example for which $f(A)\setminus f(F) \neq f(A\setminus F)$?

I think an example of this might help me to better understand.

Nosrati
  • 29,995
user6259845
  • 913
  • Let $A =\mathbb R $ $F=[0,2\pi]$ and $f (x)=sin x $. $f (A)=f (F)=[-1,1] $ so $f (A)\setminus f (F)=\emptyset $. $A\setminus F=(-\infty,0)\cup (2\pi,\infty) $ and $f (A\setminus F)=[-1,1] $. – fleablood Feb 17 '17 at 05:05
  • Here's silly analogy which may/may not work. Image your had a wardrobe with two suits. Suit one has one jacket and one pair of trousers. Suit two has one jacket and two pairs of trousers; one's a spare. If someone trashes your second jacket you must throw away both pair of trousers as you can't wear them in a suit any more. You are left with one suit. But if someone trashes the spare trousers, you can still wear suit with the original trousers. You have two suits. Trashing the input is fine if you have a spare input. But trashing the output destroys the output and that is significant. – fleablood Feb 17 '17 at 06:11

2 Answers2

1

Hint: $f(x)=|x|\,$, $A=\mathbb{R}\,$, $F=[0,\infty)\,$.

I'm not sure that I completely understand why they cannot be equal

They can be equal, but in general they don't have to be equal.

dxiv
  • 76,497
  • I guess I'm still confused on why they don't have to be equal isn't $f(A)∖f(F)$ the same thing as saying $f(A∖F)$ since they both exclude F? – user6259845 Feb 17 '17 at 05:08
  • @user6259845 Work out the example above and it should become more clear. They don't mean the same thing. The former excludes the values $f(F)$ from the range, the latter excludes the part $F$ from the domain. – dxiv Feb 17 '17 at 05:12
  • For this inequality to work the function must have some $x\ne y $ with $f (x)=f (y) $. Set minus ing the domain may allow you to throw away the y but keep the x so that the f (y)=f (x) is still there even though y has been thrown away. But in the range throwing away $f (x)=f (y) $ throws the value away entirely. – fleablood Feb 17 '17 at 05:27
  • The latter does exclude F. But if there are other values that not in F that get mapped to the same values any way the mapped values are not excluded after all. But in the former we explicitly are excluding the mapped value. Ooh! Note that f (A)/f (F) $\subset$ f (A/F)! That does have to be true! – fleablood Feb 17 '17 at 05:45
  • @fleablood $f(A) \setminus f(F) \subseteq f(A \setminus F)$ Indeed, that's always true. – dxiv Feb 17 '17 at 05:48
  • 1
    I bring it up as I think it illustrates what's going on and will make it clearer to the poster. ... well, I did when I typed it. – fleablood Feb 17 '17 at 05:54
1

Let $x\ne y $ but $f (x)=f (y) $. Then $\{x\}\ne\{y\} $ but $\{f (x)\} =\{f (y)\} $.

So "setminusing" $\{y\} $ from $\{x\} $ (or vice versa) will do nothing to the resulting image, because although we lose the $y $ we keep the $x$ and the resulting image is the same.

But as $f (x) $ does equal $f (y) $, "setminusing" $\{f(y)\} $ from $\{f(x)\} $ will eliminate everything and nothing will be left.

In other words: $f (\{x\}\setminus \{y\})=f (\{x\})=\{f (x)\} $. But $f (\{x\})\setminus f(\{y\})=f (\{x\})\setminus f (\{x\})=\emptyset $.

That's an extreme example but I hope that explains why.

Any non-injective function where A is one "preimage" of a set and B is another will do. Let $f=\sin $ and A is all the reals while $B $ is just a $[-\pi/2,\pi/2] $ interval. Or $f (x)=x^2$ and A are the negative numbers and B are the positive, etc.

fleablood
  • 124,253