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With $f:X \rightarrow Y; A, B \subseteq X; C, D \subseteq Y$, I'm given these identities:

1. $f(A\cap B) \subseteq f(A) \cap f(B)$

2. $f(A\setminus B) \subseteq f(A) \setminus f(B)$

3. $f^{-1}(C\cap D) \subseteq f^{-1}(C) \cap f^{-1}(D)$

4. $A \subseteq f^{-1}(f(A))$

5.$f^{-1}(f^{-1}(C)) = C \cap f(X) \subseteq C$

Here's what I need to do:
a)Find examples of functions and non-empty sets A, B for which 1. and/or 2. aren't equal (i.e. they are $\subset$, but not $=$)
I believe this is the case for any non-injective function for which $a \neq b$ but $f(a) = f(b)$ , like $f(x) = |x|$, with $a = 1$ and $b=-1$. I'd like to hear some more examples of functions like this.

b)Here I need to find the same thing like a) but for identities 4 and 5, however I'm not sure what even to look for.

c) Prove 1 or 2 and 3 or 4.
Here's my not so successful attempt at proving 1:

$f(A \cap B) = \{y \in Y : y = f(x) \wedge x \in A \cap B\} = \{y \in Y : y = f(x) \wedge (x \in A \wedge x \in B)\} = \{y \in Y : (y = f(x) \wedge x \in A) \wedge (y = f(x) \wedge x \in B)\} = \{y \in Y : y = f(x) \wedge x \in A\} \cap \{y \in Y : y = f(x) \wedge x \in B\} = f(A) \cap f(B)$

And I made an analogous "proof" for 2, however, obviously it's not right as I get that $f(A \cap B) = f(A) \cap f(B)$ instead of $\subseteq$. And for the second part of c) (i.e. 3 and 4), I'm not even sure how to start.

Any help at solving a) b) and c) is highly appreciated.

Bane Bojanić
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3 Answers3

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1): $A\cap B\subseteq A\Rightarrow f\left(A\cap B\right)\subseteq f\left(A\right)$ and $A\cap B\subseteq B\Rightarrow f\left(A\cap B\right)\subseteq f\left(B\right)$ . Consequence: $$f\left(A\cap B\right)\subseteq f\left(A\right)\cap f\left(B\right)$$

2) is not true: let $f$ be a constant function and let $A\backslash B\neq\emptyset\wedge B\neq\emptyset$. Then $f\left(A\backslash B\right)\neq\emptyset$ and $f\left(A\right)\backslash f\left(B\right)=\emptyset$

3): $$x\in f^{-1}\left(C\cap D\right)\iff f\left(x\right)\in C\cap D\iff$$$$ f\left(x\right)\in C\wedge f\left(x\right)\in D\iff x\in f^{-1}\left(C\right)\cap f^{-1}\left(D\right)$$

4): $$x\in A\Rightarrow f\left(x\right)\in f\left(A\right)$$ combined with: $$f\left(x\right)\in f\left(A\right)\iff x\in f^{-1}\left(f\left(A\right)\right)$$

Adjusted 5): To be proved is probably $f\left(f^{-1}\left(C\right)\right)=C\cap f\left(X\right)$

(the notation $f^{-1}\left(f^{-1}\left(C\right)\right)$ makes no sense).

$$y\in f\left(f^{-1}\left(C\right)\right)\iff y=f\left(x\right)\text{ for some }x\in f^{-1}\left(C\right)\iff y\in C\cap f\left(X\right)$$

drhab
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  • thanks a lot! Do you know any examples of a set and a function for which in 4) we cannot use "="? And when it comes to $f^{−1}(f^{−1}(C))=C∩f(X) ⊆ C$ why do you say it makes no sense? Perhaps it's just untrue, which is possible, because the assingment does say "prove 3 or 5". – Bane Bojanić Nov 01 '14 at 13:05
  • To your last question: $f^{-1}(C)$ is a subset of $X$ (not of $Y$). Only subsets of $Y$ have preimages under $f$ so there is no preimage of $f^{-1}(C)$, i.e. $f^{-1}(f^{-1}(C))$ makes no sense. – drhab Nov 01 '14 at 13:09
  • On your first question: If $f$ is a constant function and $\emptyset\neq A$ then $f^{-1}(f(A)=X$. Any set $A$ with $\emptyset\neq A\neq X$ will do then – drhab Nov 01 '14 at 13:13
  • How come? Say that we have A = {1, 2}, f(x) = 5. In that case f⁻¹(f{A}) = R, then clearly A is a subset of R. – Bane Bojanić Nov 01 '14 at 13:44
  • Indeed, a subset. But normally a proper subset. So $\subseteq$ and also $\subset$ are okay here but $A=f^{-1}(f(A))$ is not okay and you asked for "a function wich in 4) we cannot use "$=$"". Btw, why do you speaking of $R$ here? The domain of $f$ in your question is $X$, not $R$. – drhab Nov 01 '14 at 14:36
  • Oh, I made a mistake, for some reason I forgot that the question was about "=" and not a subset in general. Big thanks, you helped me a lot. – Bane Bojanić Nov 01 '14 at 15:35
  • drhab, I have another question concerning 5. If f⁻¹ is Y' -> X, then why is not f⁻¹(f⁻¹)) X' -> Y'? (X' and Y' are subsets of X and Y) – Bane Bojanić Nov 01 '14 at 20:08
  • I believe f^-1 f^-1 here is meant to denote "inverse function of an inverse function". Or in other words, I need an example of a function for which the image of its inverse function of its inverse function is not completely equal to the original function. – Bane Bojanić Nov 01 '14 at 20:17
  • On your last comment: believe it or not, but that is definitely not true. You can only speak of an inverse function $f^{-1}$ if the original function $f$ is bijective, and nothing is said about that here. There are more things to mention that deny your belief. – drhab Nov 01 '14 at 21:40
  • On your former comment. Here $f^{-1}(D)$ is not the notation of the image of $D$ under function $f^{-1}$ but it is the notation of the preimage of $D$ under function $f$. You are confusing these concepts here. In this context you better forget about any function $f^{-1}$. – drhab Nov 01 '14 at 21:48
  • Generally speaking, is there a possibility that this was just an example of bad notation and was meant to be "inverse function of an inverse function" (let's just call it g for the sake of brevity)? In that case, is there an example of a function for which that function g ('s image) wouldn't be the same as f(X)? Once again, big thanks for your patience and helpfulness. I realize I might have asked a bad question now, but this came in my homework and I don't feel like losing points. – Bane Bojanić Nov 01 '14 at 22:20
  • No on your question. Btw, if $f:X\rightarrow Y$ then (if an inverse exists, i.e. if $f$ is bijective) $f^{-1}:Y\rightarrow X$ so that the function $f^{-1}\circ f^{-1}$ can only exist if $Y=X$. Next to that a correct notation for 'inverse of inverse' would not be $f^{-1}\circ f^{-1}$ but $\left(f^{-1}\right)^{-1}$. This with $\left(f^{-1}\right)^{-1}=f$. The 'inverse of inverse of $f$ is $f$ itself. I am convinced that the notation $f^{-1}\left(f^{-1}\left(C\right)\right)$ in 5) was wrong and was meant to be $f\left(f^{-1}\left(C\right)\right)$, so that it indeed equals $C\cap f\left(X\right)$. – drhab Nov 02 '14 at 08:28
  • I asked the professor about this, and it turns out that indeed it was a mistake and was fixed to $f(f^{-1}(C))$. Thanks once again. – Bane Bojanić Nov 04 '14 at 16:48
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for a), consider the function $f(x) = x^2 $ and put $A = [-1,0) $ and $B = (0,1] $, then

$$ A \cap B = \varnothing \implies f( A \cap B ) = \varnothing $$

but, $f(A) = f(B) = (0,1] $, hence

$$ f(A) \cap f(B) = (0,1] $$

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To show $X\subseteq Y$, it is often easiest to show that $x\in X\implies x\in Y$. For example, if $x\in f(A\cap B)$, then $x=f(y)$ for some $y\in A\cap B$. Then $y \in A$, so $x\in f(A)$, and similarly $x\in f(B)$. Hence, $x\in f(A)\cap f(B)$.

Sarastro
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