# Math Help - Set theory proof?

1. ## Set theory proof?

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.

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.

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.

2. 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 $A$ and $B$.
The union of two sets $A$ and $B$ is the set of elements that are in $A$ or in $B$. Using Set and Logic notation, this means that
$(x \in A \cup B) \Leftrightarrow (x \in A \lor x \in B)$
Next, note that to prove two sets $A$ and $B$ are equal, we usually show that $A \subseteq B$ and $B \subseteq A$.

So here you'll need to show that
(i) $A \cup \oslash \subseteq A$
and
(ii) $A\subseteq A \cup \oslash$
Now we prove that $A \subseteq B$ by proving that $x \in A \Rightarrow x \in B$. So we start the proof of (i) with:
$x \in A \cup \oslash$
$\Rightarrow (x \in A) \lor (x \in \oslash)$, from the definition of union, above

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

So we have now proved (i) that:
$A \cup \oslash \subseteq A$
For (ii), for all sets $B$:
$x \in A \Rightarrow (x \in A) \lor (x \in B)$
$\Rightarrow x \in A \cup B$, again using the definition of union, above
$\Rightarrow A \subseteq A \cup B$
In particular, when $B = \oslash$:
$A \subseteq A \cup \oslash$
And that completes the proof.