Filling in the blanks proof, having some issues Set Theory Unions/Subsets

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SUMMARY

The discussion focuses on a proof in set theory demonstrating that for any sets A and B, if A is a subset of B, then the union of A and B (A U B) is also a subset of B. The proof requires filling in specific blanks related to the definitions of union and subset. Key definitions provided include the formal definition of union, which states that an element x is in the union of sets X and Y if and only if x is in X or x is in Y. The participants clarify the logical steps needed to complete the proof accurately.

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  • Familiarity with mathematical proofs and logical reasoning.
  • Knowledge of the subset symbol and its notation.
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mr_coffee
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Hello everyone. Our book has a problem where we are to fill out the missing spots and its quite confusing, I'm not sure if i got this right or not.
Any help would be great!

Here is the question/directions:
The following is a proof that for all sets A and B, if A is a subset of B,
then A U B subset B. Fill in the blanks.

note: i didn't nkow how to write the subset symbol, so if you look at (a) u will see what is to be proved

http://suprfile.com/src/1/3onjybq/lastscan.jpg
If you can't read it, here is what the book has:

Proof: Suppose A and B are any sets and A is a subset of B. We must show that (a). Let x be in (b). We must show that (c). By Definition of union, x in (d) (e) x in (f). In case x in (g), then since A is a subset of B, x in (h). In case x in B, then clearly x in B. So in either case, x in (i) as was to be shown.The definition of union out of this book is the following, which didn't match up to will with th efill in the blanks (d), (e), and (f).

The union of sets x and Y, X U Y, is deifned as X U Y = {x | x in X or X in Y}

This means that any time you know an element x is in X U Y, you can concludde that x must be in X or x must be in Y. Conversely, any time you know that a particular x is in some set X or is in some set Y, you can conlucde that x is in X U Y. Thus, for any sets X and Y and any element x,

x in X U Y if, and only if, x in X or X in Y.Thanks.
 
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the proof looks good
 

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