Hello matthayzon89 Originally Posted by

**matthayzon89** IS my proof correct?

Prove: For all sets A, A U (empty set)= A

This statement is true.

__Proof:__

Suppose that A is a set where x is an element of A U (empty set).

So, x is an element of A OR x is an element of (empty set).

If x is an element of A then A=A because every set is a subset of itself. Grandad says: this doesn't really make sense.

If x is an element of (empty set) then x is STILL an element of A b/c (empty set) is an element of all sets, namely, set A. Grandad says: this doesn't either. And note that the empty set is a subset of all sets, not an element.

There fore if x is an element of A or x is an element of empty set it is always an ELEMENT of A. Therefore the original statement is TRUE and for all sets A U (empty set)= A.

First, let's define first what we mean by the union of two sets $\displaystyle A$ and $\displaystyle B$.The union of two sets $\displaystyle A$ and $\displaystyle B$ is the set of elements that are in $\displaystyle A$ *or* in $\displaystyle B$. Using Set and Logic notation, this means that$\displaystyle (x \in A \cup B) \Leftrightarrow (x \in A \lor x \in B)$

Next, note that to prove two sets $\displaystyle A$ and $\displaystyle B$ are equal, we usually show that $\displaystyle A \subseteq B$ and $\displaystyle B \subseteq A$.

So here you'll need to show that(i) $\displaystyle A \cup \oslash \subseteq A$

and(ii) $\displaystyle A\subseteq A \cup \oslash $

Now we prove that $\displaystyle A \subseteq B$ by proving that $\displaystyle x \in A \Rightarrow x \in B$. So we start the proof of (i) with:$\displaystyle x \in A \cup \oslash$$\displaystyle \Rightarrow (x \in A) \lor (x \in \oslash)$, from the definition of union, above

$\displaystyle \Rightarrow x \in A$, since $\displaystyle x \in \oslash$ is False for all $\displaystyle x$

So we have now proved (i) that:$\displaystyle A \cup \oslash \subseteq A$

For (ii), for all sets $\displaystyle B$:$\displaystyle x \in A \Rightarrow (x \in A) \lor (x \in B)$$\displaystyle \Rightarrow x \in A \cup B$, again using the definition of union, above

$\displaystyle \Rightarrow A \subseteq A \cup B$ In particular, when $\displaystyle B = \oslash$:$\displaystyle A \subseteq A \cup \oslash$

And that completes the proof.

Grandad