A ⊆ B — Then A ∪ B = B and A ∩ B = A

You are staring at a problem that says "Given A \subseteq B, simplify A \cup B," and your gut whispers that the answer should be tidier than "a set containing elements from both." It is. There is a genuine shortcut here, and once you see why it works you will spot it in every subset-flavoured problem for the rest of your life.

The punchline: whenever A \subseteq B, both the union and the intersection simplify dramatically.

A \subseteq B \iff A \cup B = B \iff A \cap B = A

All three statements are equivalent. The subset relation is the statement that union and intersection collapse. No magic, no extra hypothesis — just a direct consequence of what "subset" means.

Why the union collapses to B

If every element of A is already inside B, then putting A and B together does not add anything new. The union A \cup B contains every element that is in A or in B. Since "x \in A" already forces "x \in B" (by the subset assumption), every element of the union is in B.

Proof. Show A \cup B = B by two inclusions.

A \cup B \subseteq B. Take x \in A \cup B. Either x \in A or x \in B. If x \in A, then x \in B because A \subseteq B. If x \in B, then x \in B. Either way, x \in B.
B \subseteq A \cup B. Take x \in B. Then x is in the union A \cup B because it is in B. \blacksquare

Both inclusions hold, so A \cup B = B.

When $A \subseteq B$, the Venn diagram shows $A$ entirely inside $B$. The union is the whole region of $B$, and the intersection is the region of $A$. There is no "crescent of $A$" sticking out — that would violate the subset assumption.

Why the intersection collapses to A

The intersection A \cap B is everything in both. If every element of A is already in B, then every element of A qualifies for the intersection — so the intersection picks up all of A. And elements of B not in A fail the "in A" half of the intersection, so they are excluded.

Proof. Show A \cap B = A by two inclusions.

A \cap B \subseteq A. Take x \in A \cap B. Then x \in A and x \in B. In particular x \in A.
A \subseteq A \cap B. Take x \in A. Since A \subseteq B, also x \in B. So x \in A and x \in B, which means x \in A \cap B. \blacksquare

Both inclusions hold, so A \cap B = A.

The three-way equivalence

Often the biggest conceptual payoff is realising that the three statements A \subseteq B, A \cup B = B, and A \cap B = A are not just consequences of each other — they are equivalent. Any one implies the other two.

You already saw A \subseteq B \Rightarrow A \cup B = B and A \subseteq B \Rightarrow A \cap B = A.
Conversely, suppose A \cup B = B. Take x \in A. Then x \in A \cup B = B, so x \in B. This shows A \subseteq B.
Similarly, A \cap B = A implies A \subseteq B: take x \in A = A \cap B, which means x \in B.

So the one subset relation carries two algebraic shadows, and either shadow restores the original relation. When a problem gives you any one of the three, you get the other two for free.

Where the shortcut actually saves work

In simplification questions. A problem might ask you to simplify (A \cap B) \cup (A' \cap B) given A \subseteq B. The normal approach uses distributivity: (A \cap B) \cup (A' \cap B) = (A \cup A') \cap B = U \cap B = B. Fine. But once you know A \cap B = A, you can also write it as A \cup (A' \cap B), which is the disjoint union of A and "the part of B outside A," and that sum is clearly B. Two routes, same answer.

In inclusion-exclusion. If A \subseteq B, then |A \cup B| = |B| and |A \cap B| = |A|, so the IE formula becomes |B| = |A| + |B| - |A|, a trivial identity. This is a sanity check: whenever your subset hypothesis holds and your cardinality formula does not collapse like this, you have made an error.

In set-identity proofs. When proving (A - B) \cup (B - A) = A \triangle B, having the side-hypothesis A \subseteq B simplifies the first term: A - B = \varnothing, and the statement collapses to B - A = A \triangle B, which is the symmetric difference in the special case. Used this way, A \subseteq B is a simplifying assumption that kills off one of the two terms.

What subset does not simplify

Not every operation collapses. For instance, if A \subseteq B, the set difference A - B becomes the empty set (because nothing in A is outside B), but B - A does not simplify to B — it is genuinely smaller, missing the part of B that equals A. And the complement A' has no nice expression in terms of B unless you know the universe. So the shortcut applies specifically to \cup and \cap, plus A - B = \varnothing.

Summary of what subset gives you:

Operation	Under A \subseteq B
A \cup B	= B
A \cap B	= A
A - B	= \varnothing
B - A	\subseteq B, size $

A concrete mini-example

A = \{2, 4, 6\} and B = \{1, 2, 3, 4, 5, 6, 7\}. Verify A \subseteq B: yes, every element of A (namely 2, 4, 6) is in B.

A \cup B = \{1, 2, 3, 4, 5, 6, 7\} = B. \checkmark

A \cap B = \{2, 4, 6\} = A. \checkmark

A - B = \varnothing, B - A = \{1, 3, 5, 7\}, |B - A| = 4 = |B| - |A| = 7 - 3. \checkmark

Every formula checks. The shortcut is not a suggestion — it is an identity.

How to recognise the trigger in an exam

When a problem states "A \subseteq B" or says "A is contained in B" or "every element of A is in B," three pieces of algebra immediately become available:

A \cup B simplifies to B.
A \cap B simplifies to A.
A - B simplifies to \varnothing.

Write these three substitutions on your rough sheet before tackling the algebra, and any expression involving A, B, and their operations collapses. Many JEE-style "simplify the set expression" questions are engineered around exactly this trigger, and spotting it saves a full minute of grinding.

The student's instinct that "there's a shortcut here, right?" is correct. The shortcut is that the subset relation is secretly the two statements A \cup B = B and A \cap B = A rolled into one — and in any problem where you are given one of the three, you can swap to whichever form makes the next step shorter.