Question #4551

Prove that every infite set is equivalent to one of its proper subsets....

Expert's answer

Let S={a1,a2,a3,…} be a countably infinite subset of an infinite set T. Such a subset can always be constructed by Infinite Set has Countable Subset.

Partition S into S1={a1,a3,a5,…},S2={a2,a4,a6,…}.

We can establish a bijection between S and S1, by letting a(n) and a(2n−1).

We can extend this to a bijection between SU(T\S)=T and S1U(T\S)=T\S2 by assigning each element in T\S to itself.

So we have demonstrated a bijection between T and one of its proper subsets T\S2, which shows that if T is infinite, it is equivalent to one of its proper subsets.

Now, let T0 that strictly belongs to T be a proper subset of T, and f:T→T0 be a bijection. It follows from Subset of Finite Set No Bijection that T must be infinite.

Partition S into S1={a1,a3,a5,…},S2={a2,a4,a6,…}.

We can establish a bijection between S and S1, by letting a(n) and a(2n−1).

We can extend this to a bijection between SU(T\S)=T and S1U(T\S)=T\S2 by assigning each element in T\S to itself.

So we have demonstrated a bijection between T and one of its proper subsets T\S2, which shows that if T is infinite, it is equivalent to one of its proper subsets.

Now, let T0 that strictly belongs to T be a proper subset of T, and f:T→T0 be a bijection. It follows from Subset of Finite Set No Bijection that T must be infinite.

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