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To show closure of A is the smallest closed set containing A

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What is your definition of Cl(A)? Not really sure what yours is from the proof. Anyways, One way is simply to define it as the smallest closed subspace containing A. Another way is to define it as A union A’ where A’ is the set of limit points of A. From this definition, we have:

Let F be a closed set containing A. Then F contains cl(A). The reason is immediate. If F contains A and is closed, then it must contain every limit point of A, hence it contains A’.

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