When every element of set A is also in set B, you can write the implication “if x is in A, then x is in B”. The smaller set gives the condition, and the larger set gives the result.
This lesson is part of proving or disproving an everyday mathematical claim. It applies writing implications and their converse to a Venn diagram.
What does a subset say as an implication?
A subset relationship is an “all” statement in disguise. “A is inside B” means “all elements of A are elements of B”, which is exactly an implication.
| Sets | Implication |
|---|---|
| A is inside B | if x is in A, then x is in B |
| A and B do not overlap | if x is in A, then x is not in B |
| A and B are equal | if x is in A then x is in B, and the reverse |
Reading the diagram, start from the inner circle. The arrow of the implication always leaves the smaller set.
Worked example: a school club
Let R be the set of students in the robotics club. Let S be the set of students in the science society. Every robotics member also belongs to the science society, but some science society members are not in robotics.
Step 1. Draw R inside S. R is the smaller set.
Step 2. Write the implication from R to S: “If a student is in the robotics club, then the student is in the science society.” This is true for the situation described.
Step 3. Write the converse: “If a student is in the science society, then the student is in the robotics club.” This is false, because a science society member outside robotics is a counterexample.
Step 4. State the contrapositive: “If a student is not in the science society, then the student is not in the robotics club.” This is true, because anyone outside S is outside R as well.
The diagram answers all three at once, so you can check each sentence against the picture.
The mistake that costs marks
The common slip is reading the implication from the bigger set to the smaller one.
| Step | Wrong | Right |
|---|---|---|
| Diagram | R inside S | R inside S |
| Sentence written | if in S, then in R | if in R, then in S |
| Test with a science-only member | member is in S but not in R, so the sentence fails | the sentence says nothing about this member |
| Verdict | false implication | true implication |
A quick test is to pick an element from the outer ring. If it breaks your sentence, the direction is wrong.
Check yourself
Let P = {10, 20, 30, 40, 50} and Q = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50}. Write the implication that P ⊂ Q gives, and the converse with a counterexample.
Answer
Every element of P is in Q, so the implication is “If x is in P, then x is in Q.”
The converse is “If x is in Q, then x is in P.” It is false, because 15 is in Q and is not in P.
The number 15 is a counterexample to the converse. It shows P is a proper subset of Q.
What to study next
Move on to explaining when a diagram suggests a result but does not establish it. Then try the integrated practice set.
The algebra step repair trainer can help with the list work in set questions. For a teacher to check your sentences as you write them, see online one-to-one Mathematics tuition.