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The set X is a subset of the set Y , written X ⊆ Y , if the following holds: For every x ∈ X, we have x ∈ Y . If X ⊆ Y and X = Y , then X is a proper subset of Y and we indicate this by writing X Y (some authors write X ⊂ Y ). Note that for any set Y we have both ∅ ⊆ Y (this is vacuously so), and also Y ⊆Y. 1 Example Prove that X ⊆ Y . Put X = {x ∈ Z | x ≥ 7} and Y = {y ∈ Z | y > 3}. Proof Let x ∈ X. By the definition of X, we have x ∈ Z and x ≥ 7. Therefore, x ∈ Z and x ≥ 7 > 3, implying x ∈ Y .

The first inequality of the string is due to the fact that x ≥ 4 (just subtract 1 from both sides). 30 In the next example, we make use of the fact that for a real number r the statement |r| < 3 means the same as the two statements r < 3 and r > −3. 2 Example Put X = {x ∈ R | |x − 2| < 1}. Prove that X ⊆ (0, 5). Proof Let x ∈ X. Then x ∈ R, and also |x−2| < 1, which implies x−2 < 1 and x − 2 > −1. The first inequality gives x < 3 < 5 and the second gives x > 1 > 0. This, together with the observation that x ∈ R, shows that x ∈ {y ∈ R | 0 < y < 5} = (0, 5).

Discussion: We first note that A, B ⊆ Z so the method applies with Z playing the role of the set X. As with all strings involving ( ⇐⇒ ) it should be checked that the argument makes sense in both directions. It is for this reason that it is not enough to write “for some n ∈ Z” in the first line only. If we had omitted this phrase in the second line, for instance, the letter n would have been undefined at that step working in the backwards direction (⇐). The usual method for showing that two sets A and B are equal is to show that each set is a subset of the other, that is, A ⊆ B and B ⊆ A.