Question

proof problemmz

Answer 1

wait how do I replace the $ @wio

Answer 2

Ummm, by hand?

Answer 3

Let \( f: X \rightarrow Y\) and \( E \subseteq Y\) Prove that \( X \backslash \bar f^{-1}(E) \subseteq \bar f^{-1}(Y \backslash E)\)

Answer 4

k .. hold on getting mai attempz

Answer 5

By complement definition, \\


\( X \backslash \bar f^{-1}(E) = [x: x \in X, x \notin f^{-1}(E)]\) \\

There are X elements that don't belong in \(f^{-1}(E)]\) and \(f(x) \notin E\). If \(Y\) elements belong in \(f(x)\), then they must belong to \(f^{-1}(x)\). Therefore, \(x \in f^{-1}(Y \setminus E)\)\\

Answer 6

...

Answer 7

hold on

Answer 8

http://jsfiddle.net/wio_dude/9QLqf/
Try using this

Answer 9

Consider an element \[x \epsilon X/f^-1(E)\], then x is an element in some set \[N \subseteq X\] since x \[x \notin f^-1(E)\]. This implies that the image under f of an element x must be an element in the compliment of E in Y. Hence, \[x \in f^-1(Y/E)\]

Answer 10

so the element x under the image f must be an element in the compliment of E in Y

Answer 11

Whait a minute, why is it \(\bar f\)? What's the bar for? is \(\bar f^{-1}\) different from \(f^{-1}\)?

Answer 12

blame my book

Answer 13

By definition right? Since Y/E = { y in Y : y not in E}. Yes, what is f bar?

Answer 14

Blame your book? What if you are leaving out something important?

Answer 15

What if \(f^{-1}:Y\to X\) while \(\bar{f}^{-1}:\mathcal P(Y)\to \mathcal P(X)\).

Answer 16

Because that would explain a few things.

Answer 17

Definition 5.3.8 states that we let \(f : X \rightarrow Y\). For each set \(B \in \mathcal P \left({Y}\right),\) define the function \(\bar f^{-1}: \mathcal P \left({Y}\right) \rightarrow \mathcal P \left({X}\right) \) by \(\bar f^{-1}(B) : [ x \in X :f(x) \in B]\)

Answer 18

http://jsfiddle.net/wio_dude/9QLqf/
Try using this

Answer 19

got it

Answer 20

negation would be

we let \(f : X \rightarrow Y\). For each set \(B \in \mathcal P \left({Y}\right),\) define the function \(\bar f^{-1}: \mathcal P \left({Y}\right) \rightarrow \mathcal P \left({X}\right) \) by \(\bar f^{-1}(B) : [ x \notin X :f(x) \notin B]\)

Answer 21

Wait, so \(\bar{f}^{-1}\) is the compliment?

Answer 22

You are confusing me.

Answer 23

argh hhhsdafdslksdjfl;ajflsd;kdslkafjsdl;kfjsdlk I don't know what the heck is going on...

Answer 24

lol

Answer 25

maybe I'll just submit this coo coo ness and get revision feedback...and do it again.. there are unlimited revisions.

Answer 26

but first revision gets some points back.. second afterwards... nada

Answer 27

I want the truth.

Answer 28

s***

Answer 29

Exercise 5.3.11\\

Let \(f: X \rightarrow Y\) and \(E \subseteq Y\) Prove that \(X \backslash \bar f^{-1}(E) \subseteq \bar f^{-1}(Y \backslash E)\)\\

Definition 5.3.8 states that we let \(f : X \rightarrow Y\). For each set \(B \in \mathcal P \left({Y}\right),\) define the function \(\bar f^{-1}: \mathcal P \left({Y}\right) \rightarrow \mathcal P \left({X}\right) \) by \(\bar f^{-1}(B) : [ x \in X :f(x) \in B]\)\\


By complement definition, \\


\(X \backslash \bar f^{-1}(E) = [x: x \in X, x \notin f^{-1}(E)]\) \\

There are X elements that don't belong in \(f^{-1}(E)]\) and \(f(x) \notin E\).\\

If \(Y\) elements belong in \(f(x)\), then they must belong to \(f^{-1}(x)\). Therefore, \(x \in f^{-1}(Y \setminus E)\)\\

Answer 30

that's what I have. my original question was how do I reword this to make it sound better and I saw what rbauer4 and it's kind of what I want to type just couldn't find the right words.

Answer 31

.__________________. @satellite73 helppppppppp :/

Answer 32

Ok hmmmm

Answer 33

what do you mean by X elements? X is already the domain, how can it be a number?

Answer 34

compliment definition

Answer 35

so the x doesn't belong to the inverse of E and f(x) = E

Answer 36

no, \(X\) is a set, not a number.

Answer 37

the set x doesn't belong to E?

Answer 38

No, it doesn't.

Answer 39

Consider if \(Y = \{0,1\}\) and \(E=\{0\} \subseteq Y\)

Answer 40

so it should be written as set x

Answer 41

Then suppose \(X = \{true,false\}\)

Answer 42

Okay, let's just suppose \(B = \bar{f}^{-1}(E)\).

Answer 43

We want to show \(X\setminus B\subseteq \bar{f}^{-1}(Y\setminus E)\)

Answer 44

yeah but there's compliment defintiions happening twice.

Answer 45

so B is gone and the E is out too
X subset bar f^-1(y)

Answer 46

\[
\forall x\quad x\in \bar{f}^{-1}(Y\setminus E) \implies x\in X\setminus B
\]

Answer 47

mhm

Answer 48

\[
\forall x\quad f(x)\in Y\setminus E \implies x\in X\setminus B
\]

Answer 49

f(x) in Y implies that x belongs to X

Answer 50

We can expand the second set difference

Answer 51

\[
\forall x\quad f(x)\in Y\land f(x)\notin E \implies x\in X\land x\notin B
\]

Answer 52

We know \(f(x)\in Y\) and \(x\in X\) are true already.... \[
\forall x \quad f(x)\notin E \implies x\notin B
\]

Answer 53

Hmmmm \[
\forall x \quad f(x)\notin E \implies x\notin \bar{f}^{-1}(E)
\]

Answer 54

Contrapositive \[
\forall x\quad x\in \bar{f}^{-1}(E)\implies f(x)\in E
\]

Answer 55

\[
\forall x\quad f(x)\in E\implies f(x) \in E
\]

Answer 56

not going there... ... I really like the one for the y the one rbauer posted

Answer 57

Sure, be prepared to be told to rewrite.

Answer 58

even if I do get it right I still need to revise anyway... happened to my previous assignment.

Answer 59

Just pretend to prove it, lol.

Answer 60

da mn ok so what happens if for all x x belonging to e the inverse of e rather.. then the f(x) belongs to e?

Answer 61

AWWWWW! That means f(x) is in E.

Answer 62

are we proving that something isn't a subset to something... ?

Answer 63

element chase ftw

Answer 64

Let \( f: X \rightarrow Y\) and \( E \subseteq Y\)
Prove that \( X \setminus \bar f^{-1}(E) \subseteq \bar f^{-1}(Y \setminus E)\)
\[
\begin{array}{lcc}

& X \setminus \bar {f}^{-1}(E) &\subseteq& \bar f^{-1}(Y \setminus E) \\

\forall x & x\in \bar f^{-1}(Y \setminus E)&\implies & x\in X \setminus \bar {f}^{-1}(E) \\

\forall x & f(x)\in Y \setminus E&\implies & x\in X \setminus \bar {f}^{-1}(E) \\
\forall x & f(x)\in Y \land f(x)\notin E&\implies & x\in X \land x\notin \bar {f}^{-1}(E) \\
\forall x & f(x)\notin E&\implies & x\notin \bar {f}^{-1}(E) \\
\forall x & x\in \bar {f}^{-1}(E) &\implies & f(x)\in E \\
\forall x & f(x)\in E &\implies & f(x)\in E \\

\end{array}
\]
This to me is a golden proof.