# Show symmetric difference is subset

#### J_Rod

Joined Nov 4, 2014
109
Good day,
I have a problem to show that the symmetric difference of sets A and C is a subset of the union of the symmetric differences of sets A and B and sets B and C.

$$A \Delta C \subseteq (A \Delta B) \cup (B \Delta C)$$

From the definition of the symmetric difference, really rewritten

$$(A \cap C^c) \cup (A^c \cap C) \subseteq (A \cap B^c) \cup (A^c \cap B) \cup (B \cap C^c) \cup (B^c \cap C)$$

Using distributive laws

$$(A \cap C^c) \cup (A^c \cap C) \subseteq (B^c \cap (A \cup C)) \cup (B \cap (A^c \cup C^c))$$

At this point I am unsure how to proceed. My proof strategy is attempting to show for some element in the subset necessarily implies that is an element of the superset, by rewriting the superset in a form which will make this apparent.

#### WBahn

Joined Mar 31, 2012
26,055
MOD NOTE: Moved from Math to Homework Help.

#### WBahn

Joined Mar 31, 2012
26,055
You might try looking at an example in which you have three sets, A, B, and C, each of which has four elements: one element that is only in that set, two elements that are also in one of the other sets, and one element that is in all three sets. That requires six distinct elements, so define your universe to have a seventh element that is not in any of the sets.

Now work through your reasoning with this group of sets and see how things evolve.

#### J_Rod

Joined Nov 4, 2014
109
$$\Omega = \{1, 2, 3, 4, 5, 6, 7\}$$
$$A = \{1, 2, 3, 4\}$$
$$B = \{5, 2, 3, 4\}$$
$$C = \{6, 2, 3, 4\}$$

$$A \Delta C = \{1, 6\}$$
$$A \Delta B = \{1, 5\}$$
$$B \Delta C = \{5, 6\}$$
$$(A \Delta B) \cup (B \Delta C) = \{1, 5, 6 \}$$

(For example above, $$x = 1$$)
Suppose $$x \in A$$ and $$x \not\in C, x \not\in B$$.
Then $$x \in A \Delta C$$ and $$x \in A \Delta B$$, since $$A \Delta C = \{x \in \Omega \,| \, x \in A, x \not\in C$$ OR $$x \not\in A, x \in C \}$$.
Since $$x \not\in B$$ and $$x \not\in C$$, $$x \not\in B \Delta C$$, hence $$x \in (A \Delta B) \cup (B \Delta C)$$.

(If, considering the above, $$x = 6$$).
Let $$x \in C$$ and $$x \not\in A, x \not\in B$$. By definition, $$x \in A \Delta C$$ and $$x \in B \Delta C.$$.
Because $$x \not\in A , x \not\in B$$, $$x \not\in A \Delta B$$. So $$x \in (A \Delta B) \cup (B \Delta C)$$.

Do you think this will be sufficient to show? Also, in the example given by Mod. Bahn, why is it necessary to make the sets $$A, \, B, \, C$$ not form a partition of the universe $$\Omega$$ ?

#### WBahn

Joined Mar 31, 2012
26,055
Also, in the example given by Mod. Bahn, why is it necessary to make the sets $$A, \, B, \, C$$ not form a partition of the universe $$\Omega$$ ?
The idea is to try to cover all the possible bases. If you want to then consider special cases, such as the sets partitioning the universe, then you simply only look at those elements that constitute the special case.