Question

Let c be a number and let S be a set. We will say that c is S-magic if and only if there is a sequence of terms in S, none of which is c, that converges to c. Let L be the set of S-magic numbers and let x be L-magic. Prove that x is in L.

Answer 1

If x is L-magic, then there is a sequence in L converging to x. Since the terms in L are S-magic and have sequences in S converging to them, x can be approached by terms from S. Hence, x is S-magic and therefore in L.

Step-by-step explanation:

If x is L-magic, then by definition, there exists a sequence of terms in L (none of which is x), that converges to x. Since L is the set of S-magic numbers, each term in the sequence that converges to x is S-magic. This means that for each term l in the sequence that converges to x, there is a sequence of terms in S (none of which is l), that converges to l. Since the sequence in L converges to x, and each term in L corresponds to a converging sequence in S, we can say that x can be approached arbitrarily closely by terms from S (but not equal to x). Therefore, x fulfills the criterion of being S-magic, and hence x is in L.