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Miscellaneous Exercise · Q8

Q.Show that A ∩ B = A ∩ C need not imply B = C.

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Set intersection is a "lossy" operation — it only tells you about the elements that are also in A. Two sets can have identical intersections with A while differing completely outside A. The simplest counterexample: take A={1}A = \{1\}, B={1,2}B = \{1, 2\}, C={1,3}C = \{1, 3\}; then A∩B=A∩C={1}A \cap B = A \cap C = \{1\}, but B≠CB \neq C.

The question asks us to show that A∩B=A∩CA \cap B = A \cap C does not force B=CB = C. This is a classic lesson about the limits of set intersection.

Think about what A∩BA \cap B captures: it only "sees" the part of BB that lies inside AA. Anything in BB that is outside AA is invisible to the intersection. So if two sets BB and CC differ only in elements that are not in AA, their intersections with AA will be identical. The intersection is blind to differences outside AA.

To prove that a statement is not always true, we only need one counterexample — a single concrete case where the premise holds but the conclusion fails.

  1. Choose a small universal set to keep things clear. Let's work with U={1,2,3,4}U = \{1, 2, 3, 4\}.

  2. Pick a set AA that will be our "window". Let A={1,2}A = \{1, 2\}.

  3. Construct BB and CC so that they share the same elements inside AA but differ outside AA.

    Let B={1,2,3}B = \{1, 2, 3\} and C={1,2,4}C = \{1, 2, 4\}.

  4. Compute the intersections:

    A∩B={1,2}∩{1,2,3}={1,2}A \cap B = \{1, 2\} \cap \{1, 2, 3\} = \{1, 2\}

    A∩C={1,2}∩{1,2,4}={1,2}A \cap C = \{1, 2\} \cap \{1, 2, 4\} = \{1, 2\}

    So A∩B=A∩CA \cap B = A \cap C holds.

  5. Check whether B=CB = C:

    B={1,2,3}B = \{1, 2, 3\} and C={1,2,4}C = \{1, 2, 4\}. …

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