Sequences and Summations Proof (Pre-imaging)

• Oct 16th 2008, 02:51 PM
Saint22
Sequences and Summations Proof (Pre-imaging)
Hey, could anyone give me some assistance on this proof?

Let f be a function from A to B. Let S and T be subsets of B. Show that

f-1(S U T) = f-1(S)U f-1(T).

Where the f to the negative one denote the pre-image or inverse image.

I think by comparing the inverse images of two subsets in a function we might be comparing if they are equivalent, but not sure.

This comes from the Basic Structures: Sets, Functions, Sequences, and Sums from Sequences and Summations chapter.

Thanks!
• Oct 16th 2008, 05:43 PM
poutsos.B
Quote:

Originally Posted by Saint22
Hey, could anyone give me some assistance on this proof?

Let f be a function from A to B. Let S and T be subsets of B. Show that

f-1(S U T) = f-1(S)U f-1(T).

Where the f to the negative one denote the pre-image or inverse image.

I think by comparing the inverse images of two subsets in a function we might be comparing if they are equivalent, but not sure.

This comes from the Basic Structures: Sets, Functions, Sequences, and Sums from Sequences and Summations chapter.

Thanks!

To prove that \$\displaystyle f^-1\$(SUT) = \$\displaystyle f^-1\$(S)U\$\displaystyle f^-1\$(T) WE must show that :

.........xε{\$\displaystyle f^-1\$(SUT)} <=====>xε{ \$\displaystyle f^-1\$(S)U\$\displaystyle f^-1\$(T)}.

So :

xε{\$\displaystyle f^-1\$(SUT)}<======> xεA & ( f(x)ε(SUT))<======>xεA& (f(x)εS v f(x)εT) <=====> (xεA & f(x)εS) v ( xεA & f(x)εT)

<=====> xε\$\displaystyle f^-1\$(S) v xε\$\displaystyle f^-1\$(T) <=====>xε{\$\displaystyle f^-1\$(S)U\$\displaystyle f^-1\$(T)}

Hence :

............................ \$\displaystyle f^-1\$(SUT) = \$\displaystyle f^-1\$(S)U\$\displaystyle f^-1\$(T) .

This a proof using the double implication procedure,where each step of the forward proof can be reversed;

Also note that by definition \$\displaystyle f^-1\$(S) = { xεA: f(x)εS } e.t.c,e.t.c

and if we put ........xεΑ=p..............f(x)εS=q............... ..f(x)εT = r,then

.........xεA& (f(x)εS v f(x)εT) becomes p&( q v r) which by the distributive property of propositional calculus is equal (p&q) v ( p & r) =

(xεA & f(x)εS) v ( xεA & f(x)εT)...e.t.c,e.t.c
• Oct 16th 2008, 08:48 PM
Saint22
Hey thanks for the response, but just wondering if "ε" is meant to be denoted as "an element of" or a "subset."

Thanks again!
• Oct 17th 2008, 05:06 AM
poutsos.B
Quote:

Originally Posted by Saint22
Hey thanks for the response, but just wondering if "ε" is meant to be denoted as "an element of" or a "subset."

Thanks again!

ε ..........is the 1st letter of the Greek word ,i think, ειναι which means belongs to i.e "is an element of"