Infinite Set is Equivalent to Proper Subset

Theorem
Every infinite set is equivalent to one of its proper subsets.

Proof
Let $$S = \left\{{a_1, a_2, a_3, \ldots}\right\}$$ 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 $$S_1 = \left\{{a_1, a_3, a_5, \ldots}\right\}, S_2 = \left\{{a_2, a_4, a_6, \ldots}\right\}$$.

We can establish a bijection between $$S$$ and $$S_1$$, by letting $$a_n \leftrightarrow a_{2n-1}$$.

We can extend this to a bijection between $$S \cup \left({T \setminus S}\right) = T$$ and $$S_1 \cup \left({T \setminus S}\right) = T \setminus S_2$$ by assigning each element in $$T \setminus S$$ to itself.

So we have demonstrated a bijection between $$T$$ and one of its proper subsets $$T \setminus S_2$$, which shows that $$T$$ is equivalent to one of its proper subsets.

Comment
This is one of the ways that infinite sets can be defined, i.e. $$S$$ is infinite iff it is equivalent to a proper subset of itself.