The bijection could be defined from the set of numbers and it is proved.
We can define a bijection f from S to T as follows: for any integer x in S, let f(x) = x + 1.
To show that f is a bijection, we need to prove that it is both injective and surjective.
To prove that f is injective, suppose f(x) = f(y) for some x, y in S. Then x + 1 = y + 1, which means x = y. Thus, f is injective.
To prove that f is surjective, let y be an arbitrary element of T. Since y is odd, we can write y = 2n + 1 for some integer n. Now consider the element x = 3n in S. We have f(x) = x + 1 = 3n + 1 = y, so f is surjective.
Therefore, f is a bijection from S to T, so |S| = |T|.
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