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First off, this post is a question; I hope to get constructive feedback on this proof. I would not consider anything below correct, it just makes sense to me.

I am trying to learn Real Analysis and I came across the following two conjectures:

Let $f:A\to B$ and let $C,D\subset B$ then we have the following:

$\textit{ i})$ $f^{-1}(C\cap D)=f^{-1}(C)\cap f^{-1}(D)$

$\textit{ ii})$ $f^{-1}\left(D^c\right)=\left(f^{-1}(D)\right)^c$

The author suggested doing an element-wise proof assuming that $\textit{ x }$ was element of the left side of the equation, and then showing that it was also an element of the right side of the equation, likewise he showed the converse was true. So, since both sides of the equations are subsets of each other they must be equal.

I have written a couple proofs the same way that the author suggested, but I decided to try something different this time. Instead of writing an element-wise proof I tried to write direct proofs from the definitions of an inverse image, complement, union and intersection, along with some double negations. This is what I got:

The proof of $\textit{ i}$) :

$f^{-1}(C\cup D)=\{x\in A: f(x)\in (C\cup D)\}$(by the definition of an inverse image)

$ \space\space\space\space \space\space \space\space \space\space \space\space \space\space \space\space \space\space\space =\{x\in A: f(x)\in C \text{ or } f(x)\in D\}$ (by the definition of a union)

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space =\{x\in A: f(x)\in C\}\cup \{x\in A: f(x)\in D\}$ (by definition of a union)

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space =f^{-1}(C)\cup f^{-1}(D)$ (again by the definition of an inverse image)

And the proof of $\textit{ ii}$) :

$\space f^{-1}\left(D^c\right)= f^{-1}(\{x\in B: x\notin D\} )$(definition of the complement)

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space=f^{-1}(\{x: x\in B \text{ and } x\notin D\} )$ ( equivalent representation of the set)

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space =f^{-1}\left(B\cap D^c\right)$ (definition of the union)

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space =\left\{x\in A: f(x)\in \left(B\cap D^c\right)\right\} $ (definition of the inverse image)

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space =\left\{x\in A: f(x)\notin \left(B\cap D^c\right)^c\right\}$ (double negation of the set's property)

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space =\left\{x\in A: f(x)\notin \left(B^c\cup D\right)\right\}$ (DeMorgan's Law)

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space =\{x\in A: f(x)\notin (\emptyset \cup D)\} $ ($\textit{ B }$ is being regarded as the universal set)

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space =\{x\in A: f(x)\notin D\} $ (the union with the empty set is $\textit{ D }$ )

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space =\{x\in A: f(x)\in D\}^c $ (double negation of the set)

$\space \space \space \space \space \space \space \space \space \space \space \space \space \space = \left(f^{-1}(D)\right)^c $ (definition of the inverse image )

Is this a sufficient proof? Also, I do not have a formal education in math so I would greatly appreciate constructive criticism. Thanks in advance for your time.

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In (ii), you can skip the first few steps and directly write $\{x \in A : f(x) \in D^c\} = \{x \in A : f(x) \notin D\}$. Other than that your proofs are okay. –  hot_queen Jul 2 '13 at 8:25
    
@Ashutosh thanks, i guess I did add an awful lot of unnecessary steps. –  JimmyJackson Jul 2 '13 at 8:32
    
@user36342: Are you using Rudin's book? I'd also recommend using Bruckner's book as a beginner. –  hot_queen Jul 2 '13 at 8:39
    
@Ashutosh I am using an open source text by Jiˇrí Lebl. I will check out Bruckner's book. Thanks. And if you don't mind can I ask you a couple other quick questions to get my self on the right path? –  JimmyJackson Jul 2 '13 at 8:44
    
You should probably move to mathstack. There are a lot of people who can answer your questions there. –  hot_queen Jul 2 '13 at 8:48
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migrated from mathoverflow.net Jul 2 '13 at 9:29

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1 Answer

up vote 2 down vote accepted

The proof of $(i)$ is fine. The proof of $(ii)$ is not wrong, but it’s unnecessarily long: about half a dozen steps are just spinning your wheels. All you need is this:

$$\begin{align*} f^{-1}[B\setminus D]&=\{x\in A:f(x)\in B\setminus D\}\\ &\overset{(1)}=\{x\in A:f(x)\notin D\}\\ &=A\setminus\{x\in A:f(x)\in D\}\\ &=A\setminus f^{-1}[D]\;. \end{align*}$$

I’ve replaced your complement notation with set differences, since complementation must be relative to some set, and here you have complements both relative to $A$ and relative to $B$; I find it safer to specify where the complement is being taken.

$(1)$ is true because $f[A]\subseteq B$, so that specifying that an $f(x)$ is in $B$ says nothing. If you really wanted, you could expand that step like this,

$$\begin{align*} \{x\in A:f(x)\in B\setminus D\}&=\{x\in A:f(x)\in B\text{ and }f(x)\notin D\}\\ &=\{x\in A:f(x)\in B\}\cap\{x\in A:f(x)\notin D\}\\ &=A\cap\{x\in A:f(x)\notin D\}\\ &=\{x\in A:f(x)\notin D\}\;, \end{align*}$$

but I’d not bother.

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"I find it safer to specify where the complement is being taken." Yup. That's where Bourbaki's $\complement$ notation is also nice, $\complement_B D$. –  Daniel Fischer Jul 2 '13 at 10:39
    
The third line should read "$=A \setminus \{x\in A : f(x) \in D\}$", rather than $x\in D$. –  Theo Johnson-Freyd Jul 2 '13 at 17:47
    
@Theo: It should indeed; thanks. –  Brian M. Scott Jul 2 '13 at 19:02
    
@Daniel: Well, it’s certainly preferable to $D^c$ and the like, but it strikes me as pointless proliferation of notations. –  Brian M. Scott Jul 2 '13 at 19:16
    
Well, $\complement_A X \setminus \complement_A Y$ is IMO a little easier to parse than $(A\setminus X)\setminus(A\setminus Y)$. But not a big deal. I suppose one likes what one learned when one was young. –  Daniel Fischer Jul 2 '13 at 19:22
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