a problem in Abstract Algebra?

let A be a closed subset of R. show that A has a countable subset whose closure is A.

Comments

  • For a positive integer n, let

    Pn ={ [k/n, (k+1)/n) | integer k},

    ie. Pn is the partition of R into half open intervals of length 1/n with left end points k/n, where k varies over the integers. Fix n, and for each integer k such that the intersection of [k/n, (k+1)/n) and A is non empty, choose an element that interesection; and let An be the set of all such elements chosen. Clearly, An is countable. Then let B be the union of all An's over all positive integers n.

    Then B is countable subset of A which is dense in A; and since A is closed, the closure of B is A. QED.

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