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Is this true?

cl(A ∩ B) = cl(A) ∩ cl(B)

No — it is false.

The counterexample

$A = Q, B = R$ ∖ $Q$ in $R$

cl(A ∩ B) = cl(∅) = ∅ but cl A ∩ cl $B = R$ .

The kind of mistake this is

Hypothesis dropped

A theorem applied without checking one of its preconditions.

Drill statements like this

Others that fail the same way

From Topology › Topological spaces, bases, subspace/product/quotient

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