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Posted July 12th, 2009 by alpha
in

Using the properties of set , prove that
$ (A-B)UA=A $

‹ Doubts regarding symbols used in Sets Hi I'm new here and just wanted to say thanks ›

Comments

answer

On July 12th, 2009 Structure says:

To have

$$X=Y$$

you have to prove

$$ X\subset Y$$

and

$$Y\subset X$$

so

$$A\setminus B\subset A$$
$$A\subset A$$

you have

$$(A\setminus B)\cup A\subset A$$
$$A\subset (A\setminus B)\cup A$$

so

$$(A\setminus B)\cup A=A$$

My solution - take a look

On July 12th, 2009 alpha says:

$ B^c $ is the complement of B.
$ (A intersectionB^c )UA $
Using the law of distribution of sets
=$ (AUA) $ intersection$ (B^c U A) $
=$ A intersection\phi $ [$ (B^c U A)=\phi $]
=A [$ A intersection\phi=A $ ]

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

Is my method wrong?

On July 12th, 2009 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On July 12th, 2009 Structure says:

You are right, your solution is wrong.
Why

$$ A\cup B^c=\Phi ?$$
$$A\cap\Phi=A ?$$

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