MHB Proof 4 Proof Questions w/ Fuzzy Sets, Functions & Bijections

[zack95]

Hi,
I have exam next week and Need to know the answer of thease questions that I can not answer to them.
So, if it is possible , please help me.
Thank you

Question 1) Proof that def1 and def2 are equal. ( using convex fuzzy sets.) (Question 1 is Fuzzy sets Question)
def1= Fuzzy sets is convex iff For all of x, 0<x<=1 , x-cut is convex.
def2= x1 <= For all of x <= x2 , Membership function(x) >= minimum( Membership function(x1) , Membership function(x2) )
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Question 2) A is a subset of X and B is a subset of Y : (Question 2 is Function Question)
1)proof that A is a subset of f-inverse(f(A))
2)proof that B is a superset of f(f-inverse(B))
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Question 3) A=[1,4] , B=[7,10]
proof that A+B={for C member of R|C=a+b , For all a member of A , For all b member of B}=[8,14]
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Question 4)(function Question) If f: x --> y , g: y --> z and f,g are Bijection(Bijection means One-to-One and Onto at a same time)
proof that gof is Bijection.
note that gof means g(f(x)). (Question 4 is function Question)
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I have problem with these proof.
Please help me.
Thank you(Clapping)

 

[Barioth]

I don't have a lot of time, but I'll try to answer Q4.

let h=goy :x-->z

say $$h(a)=h(b) ==> goy(a)=goy(b)$$ Since y is a bijection we have that y(a)=y(b) if and only if a=b. Say a is not equal to b...

then
$$y(a)=c$$ and $$ y(b) = d$$ then we have $$g(d) = g(c)$$ Now since d is not equal to d and since g is a bijection we have that g(d) is not equal to g(c). Witch end in a contradiction

 

[zack95]

Thank you for answering Q4.

Any help for Questions 1,2,3

 

[MarkFL]

...Any help for Questions 1,2,3

You will more likely get help if you post what you have done so far so our helpers know where you are stuck and how best to try to get you unstuck. :D

 

[Deveno]

OK, zack, let's look at question 2. Fair warning, I'm not going to do the heavy lifting, YOU are.

We'll start with the basics: we have a function:

$f:X \to Y$

and we have two sets, $A$, which is a subset of $X$; and $B$, which is a subset of $Y$.

Given these two sets, we want to look at how $f$ interacts with them.

The first thing we will look at is: what happens to $A$ when $f$ acts upon it?

The set in question is usually called the image of $A$ under $f$:

$f(A) = \{y \in Y: y = f(a)$ for some $a \in A\}$, which is where all the stuff in $A$ winds up after it gets hit by $f$.

The other set we are interested in is called the pre-image of $B$:

$f^{-1}(B) = \{x \in X: f(x) \in B\}$ which are all the things in $X$ that $f$ takes to $B$.

So the first part of your question is to show that:

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

Well, how do we do that?

We need to show that if we start with any old $a \in A$, that it is in the set $f^{-1}(f(A))$.

Now, it's your turn...what do you make of this?

 

[zack95]

I proof Question number 3. Please somebody proof Question number 4
Question 3) A=[1,4] , B=[7,10]
proof that A+B={for C member of R|C=a+b , For all a member of A , For all b member of B}=[8,14]

To prove two sets are equal, prove each is a subset of the other. That is, here, first prove A+ B is a subset of [8, 14] the prove that [8, 14] is a subset of A+ B. To prove set X is a subset of set Y, start "if x in X" then use the definitions of X and Y to conclude "x in Y".

Here, suppose x in {A+ B}. Then x=a+ b for some a in A, some b in B. The smallest a can be is 1: 1< a. The smallest b can be is 7: 7< b. Therefore 1+ 7< a+ b= x so x is larger than 8. The largest a can be is 4: a< 4. The largest b can be is 10: b< 10. Therefore x= a+ b< 4+ 10= 14. Since x must be between 8 and 14, x is in [8, 14]. That is, A+ B is a subset of [8, 14]

Now, suppose x in [8, 14]. Let a be any number in [1, 4]. Then 8\le x\le 14 and 1\le a\le 4. Then -4\le -a\le -1 so that 8- 1= 7\le x- a\le 14- 4= 10. That is, if b= x- a, we have b in [7, 10] and x= a+ b so that x is in A+ B. That is [8, 14] is a subset of A+ B.

Putting those together, A+ B= [8, 14].

Am I correct?

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Q4)Suppose that g is a function from A to B and f is a func*tion from B to C. a) Show that i f both f and g are one-to-one functions, then f o g is also one-to-one. b) Show that if both f and g are onto functions, then f o g is also onto.

any answer ?

 

[Evgeny.Makarov]

Here, suppose x in {A+ B}. Then x=a+ b for some a in A, some b in B. The smallest a can be is 1: 1< a.

The last claim is not correct: it is not true that 1 < 1. A similar remark can be made about the rest of this paragraph.

Now, suppose x in [8, 14]. Let a be any number in [1, 4]. Then 8\le x\le 14 and 1\le a\le 4. Then -4\le -a\le -1 so that 8- 1= 7\le x- a\le 14- 4= 10.

You have
\begin{align*}
8&\le x\le 14\\
-4&\le -a\le -1
\end{align*}
Yet by adding them you get $8-1\le x-a\le 14-4$. This is incorrect.

There are several ways to recover $1\le a\le 4$ and $7\le b\le 10$ from $8\le x\le 14$ in such a way that $a+b=x$. One is to consider $b=\max(7,x-4)$ and $a=x-b$. Prove that $a$ and $b$ thus defined satisfy the required inequalities. Consider two cases: $8\le x\le 11$ and $11<x\le 14$.

Q4)Suppose that g is a function from A to B and f is a func*tion from B to C. a) Show that i f both f and g are one-to-one functions, then f o g is also one-to-one.

This is literally shown in two easy steps applying the definition of a one-to-one function. Recall that some function $h$ is one-to-one (injective) if
\[
h(x)=h(y)\implies x=y.
\]