My Math Forum Given f is one to one, prove $f(A \backslash B)=f(A)\backslash f(B)$

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 January 20th, 2018, 01:41 PM #1 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 Given f is one to one, prove $f(A \backslash B)=f(A)\backslash f(B)$ Given $f:X\rightarrow Y$, $A, B \subset X$, and $f$ is one to one, prove that $f(A \backslash B)=f(A)\backslash f(B)$. I have a start, but I'm not really sure where to go from here:
 January 20th, 2018, 02:05 PM #2 Senior Member   Joined: Aug 2012 Posts: 2,050 Thanks: 587 Right, that's the left to right direction. Now you just need the right to left direction.
January 20th, 2018, 02:26 PM   #3
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 Originally Posted by Maschke Right, that's the left to right direction. Now you just need the right to left direction.
This isn't quite true. He has proved that $f(A \setminus B) \subset f(A) \cap f(B^c)$ but there is an argument missing which should explain why $f(A) \cap f(B^c) \subset f(A) \setminus f(B)$. Of course, he hasn't yet used that $f$ is one-to-one.

Next, you should prove that for any subset, $A \subset X$, we have $f(A)^c = f(A^c)$. In other words, the operation of taking complements commutes with any one-to-one function. The proof of this is essentially one line which consists of stating the definition of a one-to-one function.

With this in hand, $f(B^c) = f(B)^c$ so it follows that
$f(A) \cap f(B^c) = f(A) \setminus f(B)$
which was the missing piece in the forward direction of the proof.

The converse statement is almost identical. Try it and let us know if you get stuck.

Last edited by skipjack; January 20th, 2018 at 03:57 PM.

 January 20th, 2018, 05:25 PM #4 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 Unfortunately, the part you just mentioned "consists of one line" is where I've been stuck this entire time! I'm not quite sure how to formalize it.
January 20th, 2018, 07:33 PM   #5
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 Originally Posted by John Travolski Unfortunately, the part you just mentioned "consists of one line" is where I've been stuck this entire time! I'm not quite sure how to formalize it.
Well notice that for ANY function, we have $f(A)^c \subset f(A^c)$. Do you see why? Thus, you only need to figure out why the opposite inclusion holds for one-to-one functions (i.e. $f(A^c) \subset f(A)^c$).

However, this follows since if $y \in f(A^c) \cap f(A)$, then there exists $x_1 \in A$ and $x_2 \in A^c$ such that $f(x_1) = f(x_2) = y$. This contradicts the assumption that $f$ is one-to-one and thus $f(A^c) \cap f(A) = \emptyset$ or equivalently, $f(A^c) \subset f(A)^c$.

 January 21st, 2018, 09:15 AM #6 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 Wait, isn't it only true that $f(A)^C\subset f(A^C)$ if $f$ is onto (surjective)? That's not known; all we know is that it's one-to-one.
January 21st, 2018, 12:05 PM   #7
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 Originally Posted by John Travolski Wait, isn't it only true that $f(A)^C\subset f(A^C)$ if $f$ is onto (surjective)? That's not known; all we know is that it's one-to-one.
Set complementation is relative and $f(A)^c$ is implied to mean with respect to its image. Put another way, every map is surjective onto its image.

Last edited by skipjack; January 24th, 2018 at 10:44 AM.

January 22nd, 2018, 05:09 PM   #8
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Quote:
 Originally Posted by SDK Set complementation is relative and $f(A)^c$ is implied to mean with respect to its image. Put another way, every map is surjective onto its image.
Okay, so then explain where I'm going wrong with this example:

Define $f:X\rightarrow Y$ where $X=\lbrace 1, 2 \rbrace$, $Y=\lbrace 1, 2, 3 \rbrace$, in such a way that $f(\lbrace 1 \rbrace) = \lbrace 1 \rbrace$ and $f(\lbrace 2 \rbrace) = \lbrace 2 \rbrace$. Now let $A=\lbrace 1 \rbrace\rightarrow f(A)=\lbrace 1 \rbrace$ So that means that $A^C=\lbrace 2 \rbrace \rightarrow f(A^C)= \lbrace 2 \rbrace$. Yet $(f(A))^C= \lbrace 2, 3 \rbrace$.

In that example, $f$ is one-to-one yet $f(A^C)$ does not equal $(f(A))^C$. What am I not understanding here?

Last edited by skipjack; January 24th, 2018 at 10:45 AM.

 January 24th, 2018, 05:03 AM #9 Senior Member   Joined: Sep 2016 From: USA Posts: 476 Thanks: 263 Math Focus: Dynamical systems, analytic function theory, numerics In your example, image$(f) = \{1,2\}$ and indeed $f(A^c) = \{2\} = f(A)^c$ with respect to the image of $f$. You can add anything you like to the codomain of $f$, but this doesn't make it relevant. Without some further context (and motivation) in mind, set complementation is always implied to mean with respect to the image. Last edited by skipjack; January 24th, 2018 at 10:46 AM.

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