Q.Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B. (Hints A = A ∩ ( A ∪ X ) , B = B ∩ ( B ∪ X ) and use Distributive law )

Concept: Set equality via distributive laws and the given union/intersection conditions.

We're given $A \cap X = B \cap X = \phi$ and $A \cup X = B \cup X$. The hint tells us to express $A$ and $B$ using their unions with $X$.

Start with $A$. By the absorption identity, $A = A \cap (A \cup X)$. Since $A \cup X = B \cup X$ by hypothesis, we have:

Apply the distributive law $A \cap (B \cup X) = (A \cap B) \cup (A \cap X)$. But $A \cap X = \phi$, so:

$A = (A \cap B) \cup \phi = A \cap B$ …

The heart of this problem lies in recognizing that if $A$ and $B$ behave identically with respect to $X$ — neither shares elements with $X$, yet both produce the same union — then $A$ and $B$ cannot differ from each other. The hint guides us to rewrite each set using the identity property and distributivity.

Why this approach works

Any set $S$ can be written as $S = S \cap U$ when $U$ is a universal set containing $S$. Here, since $A \subseteq A \cup X$, we have $A = A \cap (A \cup X)$. The same holds for $B$. But we're given that $A \cup X = B \cup X$, so both sets intersect with the same union. The disjointness condition $A \cap X = \phi$ then allows the distributive law to separate out the unwanted $X$ term, leaving only $A$ (or $B$).

Proof

1. Express $A$ using the identity property. Since $A \subseteq A \cup X$, we can write:

2. Substitute the given equality $A \cup X = B \cup X$. Replace $A \cup X$ with $B \cup X$:

3. Apply the distributive law for intersection over union. The distributive property states $(P \cap (Q \cup R)) = (P \cap Q) \cup (P \cap R)$. Applying this:

4. Use the disjointness condition $A \cap X = \phi$. Since $A$ and $X$ share no elements:

Therefore, $A \subseteq B$.

5. Repeat the argument for $B$. By symmetry, start with $B = B \cap (B \cup X)$, substitute $B \cup X = A \cup X$: …

Method: Set Equality via Distributive and Identity Laws

We prove $A = B$ by showing each is a subset of the other, using the given conditions and set algebra.

Steps

Step 1: Express $A$ using the hint

Given: $A = A \cap (A \cup X)$

But we know $A \cup X = B \cup X$ (given). So:

Step 2: Apply Distributive Law

Step 3: Use the given condition $A \cap X = \phi$

Step 4: Identity Law

This means $A \subseteq B$.

Step 5: Repeat symmetrically for $B$

Start with $B = B \cap (B \cup X) = B \cap (A \cup X)$ (since $A \cup X = B \cup X$).

Given $B \cap X = \phi$:

Thus $B \subseteq A$. …

Here are the common mistakes students make when proving A = B using set difference and the given conditions, along with how to avoid each.

Mistake 1: Misapplying the Distributive Law

The Error:

Students often write something like:

but then incorrectly simplify $A \cap A$ to $A$ (which is correct) but forget to handle the $A \cap X$ term using the given condition.

Why it happens:

They rush to expand without first checking which terms vanish due to the given conditions ($A \cap X = \phi$).

How to avoid:

• Always list the given conditions first: $A \cap X = \phi$, $B \cap X = \phi$, and $A \cup X = B \cup X$.
• Before expanding, note which intersections are empty. For example, $A \cap X = \phi$ means that term will disappear.

Correct approach:

This is trivial — the real work comes when you use $A \cup X = B \cup X$.

Mistake 2: Forgetting to Use the Equality $A \cup X = B \cup X$

The Error:

Students prove $A = A \cap (A \cup X)$ and then stop, thinking they’ve shown $A = B$.

Why it happens:

They don’t see that the hint is a bridge — you must replace $A \cup X$ with $B \cup X$ (since they are equal) to connect $A$ and $B$.

How to avoid:

• Write the chain of equalities explicitly:

Now apply distributive law:

• Similarly, show $B = A \cap B$, hence $A = B$.

Key insight: The condition $A \cup X = B \cup X$ lets you swap the union inside the intersection.

Mistake 3: Assuming $A \cap X = \phi$ Means $A$ and $X$ Are Disjoint in All Contexts

The Error:

Students think $A \cap X = \phi$ implies $A$ and $X$ have no relation, so they ignore the possibility that $A$ might still be a subset of $B \cup X$.

Why it happens:

They treat “disjoint” as “completely unrelated,” but set operations still apply.

How to avoid:

• Remember: $A \cap X = \phi$ only tells you that no element is in both $A$ and $X$.
• It does not prevent $A$ from being a subset of $B \cup X$ — in fact, that’s exactly what you use when you write $A \cap (B \cup X)$.

Example to internalize:

Let $A = \{1,2\}$, $X = \{3,4\}$, $B = \{1,2,5\}$.

Here $A \cap X = \phi$, and $A \cup X = \{1,2,3,4\}$, $B \cup X = \{1,2,3,4,5\}$ — they are not equal, so the proof fails. This shows why the equality $A \cup X = B \cup X$ is crucial.

Mistake 4: Not Using the Symmetric Argument for $B$

The Error:

Students prove $A = A \cap B$ and then conclude $A = B$ without showing $B = A \cap B$.

Why it happens:

They assume $A = A \cap B$ automatically implies $B = A \cap B$, which is false (e.g., $A = \{1\}$, $B = \{1,2\}$ gives $A \cap B = A$ but $B \neq A$).

How to avoid:

• Always repeat the same steps for $B$:

• Now you have both $A = A \cap B$ and $B = A \cap B$, so $A = B$.

Pro tip: Write the proof in two symmetric halves — it’s cleaner and prevents missing the second half.

--- …