The Archive

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Problem

Our next problem comes from Penn State Math 403, the upper-level undergraduate real analysis course. We have selected a problem which does not require advanced theorems developed in the course. Yet, its solution is challenging, as it requires a carefully constructed bijection.

Prove that and have the same cardinality.

Solution

By definition. two sets and have the same cardinality if there exists a bijection . Thus it will suffice to find a bijection from to .

To get a handle on how to deal with the endpoints when constructing the bijection, a related problem is considered first.

Show that the sets and have the same cardinality. Define by . It is straight forward to verify that has inverse . Thus, since a function is invertbile if and only if it a bijection, is a bijection. More generally if , then defined by is a bijection Hence and have the same cardinality.

To put this to use in the current problem, consider the sets and and the corresponding sets and The sets , and form a partition of ; and the sets , and form a partition of . This allows for a bijection to be defined piecewise. Define by , by , and by . Finally, define by As noted earlier, and are invertible, while is the identity map on , so it too is invertible. Consequently is invertible, with Therefore is a bijection. Hence it has been shown that and have the same cardinality.

Although the problem has been solved at this point, the it would be nice to have an explicit form for . To this end, note that