Alright, 2+1/2 problems here:(adsbygoogle = window.adsbygoogle || []).push({});

First:

36. Let F be a function from the set A to the set B. Let S and T be subsets of A. Show that:

a) f(S[tex]\cup[/tex]T)=f(S)[tex]\cup[/tex]f(T)

b)f(S[tex]\cap[/tex]T)[tex]\subseteq[/tex] f(S)[tex]\cap[/tex]f(T)

Note: This must be done using a membership proof. i.e. prove that f(S[tex]\cup[/tex]T)[tex]\subseteq[/tex]f(S)[tex]\cup[/tex]f(T) AND f(S)[tex]\cup[/tex]f(T)[tex]\subseteq[/tex]f(S[tex]\cup[/tex]T)

40. Let f be a function from A to B. Let S be a subset of B. We define the inverse image of S to be the subset of A whose elements are precisely all pre images of all elements of S. We denote the inverse image of S by f[tex]^{-1}[/tex](S) so f[tex]^{-1}[/tex](S)={a[tex]\in[/tex]A| f(a)[tex]\in[/tex]S}

Prove parts a and b of 36 substituting the inverse image for f, and an equality in both parts (rather than subset)

1) (not a question from the text). This was not covered in the lecture or in the book, but how do most people go about proving that a function is surjective from the definition?

2. Relevant equations

f[tex]^{-1}[/tex](S)={a[tex]\in[/tex]A| f(a)[tex]\in[/tex]S}

f(S)={f(s)|s[tex]\in[/tex]S}

3. The attempt at a solution

Alright, I only have an idea of how to do 36, 40 I am lost. The professor said to use proof by case for 36, so here is how I thought it would go:

a) prove:f(S[tex]\cup[/tex]T)=f(S)[tex]\cup[/tex]f(T)

Proof:

1) assume x[tex]\in[/tex] f(SUT), then, by definition x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]SUT}

2) If x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]SUT}, then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S or T} by definition of union.

3) If x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}, then x[tex]\in[/tex]f(S), and since f(S)[tex]\subseteq[/tex]f(S)Uf(T), thus if x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S} then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}[tex]\cup[/tex]{f(s)|s[tex]\in[/tex]T}.

4)If x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]T}, then x[tex]\in[/tex]f(T), and since f(T)[tex]\subseteq[/tex]f(S)Uf(T), thus if x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]T} then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}[tex]\cup[/tex]{f(s)|s[tex]\in[/tex]T}.

4)Therefore if x[tex]\in[/tex]f(SUT) then x[tex]\in[/tex]f(S)Uf(T)

5 Now we must prove that f(S)[tex]\cup[/tex]f(T)[tex]\subseteq[/tex]f(S[tex]\cup[/tex]T, to do this, just invert the steps of this proof (obviously I wouldn't put this on a test but to cut length, this is reasonable)

b)f(S[tex]\cap[/tex]T)[tex]\subseteq[/tex] f(S)[tex]\cap[/tex]f(T)

proof:

1) assume x[tex]\in[/tex] f(S[tex]\cap[/tex]T), then, by definition x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S[tex]\cap[/tex]T}

2) Then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S and T}.

3) Thus s[tex]\in[/tex]S and s[tex]\in[/tex]T and so, through simplification, we can say:

4) s[tex]\in[/tex]S, if s[tex]\in[/tex]S then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}

5) We can also say s[tex]\in[/tex]T, and if s[tex]\in[/tex]T then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]T}.

6) Therefore, if s[tex]\in[/tex]S and s[tex]\in[/tex]T, which are true, we have x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S} and x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]T} (we can say through addition).

7) Note that by the definition of intersection, x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}[tex]\cap[/tex]{f(s)|s[tex]\in[/tex]T}.

8) Therefore, if x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S[tex]\cap[/tex]T} then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}[tex]\cap[/tex]{f(s)|s[tex]\in[/tex]T}.

9)if x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S[tex]\cap[/tex]T} then x[tex]\in[/tex]{f(s)|s[tex]\in[/tex]S}[tex]\cap[/tex]{f(s)|s[tex]\in[/tex]T}.

10)This means f(S[tex]\cap[/tex]T)[tex]\subseteq[/tex]f(S)[tex]\cap[/tex]f(T).

For Problem 40, I am wondering whether the proofs follow the same steps, or is there something else I have to worry about?

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# Homework Help: Function on sets and subsets

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