The Archive
To see problems and solutions in the fall series, which runs from October 13 through December 15 visit The Fall 2025 Featured Problem Series
Problem
Two weeks ago our problem came from Penn State Math 403, the upper undergraduate real analysis course. This week we return to that course for our problem. In fact, the problem for this week is a variation of the problem from two weeks ago. And, like the problem from two weeks ago, the solution of this week’s problem does not require any advanced theorems developed in Math 403. A student who has completed Calculus I will have most of the tools required to solve the problem this week. However, putting those tools to work will require a level of mathematical sophistication that cannot be expected of a Calculus I student.
Recall that in the Week 4 problem, a bijection between and was exhibited which proved that these sets have the same cardinality. The function was defined piecewise on the sets and , and took the explicit form
The motivation for this construction can be found in the solution of the Week 4 problem, which is in the Problem Archive linked to at the bottom of this page.
Use either the or sequential definition of continuity to show that is not continuous at and .
Prove that a continuous bijection from to does not exist. Hint: Use contradiction and the Intermediate Value Theorem (IVT).
Solution
First, part (a) is examined. Both of the suggested methods will be used. To begin, the sequential definition of continuity will be used to prove that , as defined in , is discontinuous at , and then the definition of continuity will be used to prove that it is discontinuous at .
Recall the sequential definition of continuity for real-valued function of a real variable. A function is continuous at a point in its domain if and only if for every sequence that takes values in the domain of and satisfies , .
To prove discontinuity at it suffices to exhibit a sequence that takes values in the domain of and satisfies , for which .
Returning to the question of the continuity of at , for , set and observe that that Consequently the sequence is in the domain of . Further, On the other hand, for the definition of gives . Therefore Thus is discontinuous at .
The definition of continuity for real-valued function of a real variable, which will be used to prove the discontinuity of at , is as follows. A function is continuous at a point in its domain, if for all there exists a such that for all in the domain of satisfying To prove discontinuity at it will suffice to exhibit an such for all there exists an in the domain of satisfying for which
An exploratory analysis is undertaken in order to prove the discontinuity of at . The first issue is a suitable choice for such that for all there exists an in the domain of satisfying At this point, explicit forms for and are substituted into the inequality on the right in . From equation , one has . However, the explicit form of depends on whether , or . From the inequality on the left in , the selected set must have as a limit point. As can be seen from , that every sequence in converge to , thus is not a limit point of this set, which eliminates from consideration. Next, note that any sequence in converges to zero. Thus zero is a limit point of this set, and it cannot be eliminated from consideration. Finally, for any , since contains only a countable number of points. Thus zero is also a limit point of . Therefore one may select in either or . Because the function takes a simple form on , that is the choice made here. For , gives , and the right hand inequality in becomes This is equivalent to Observe that , thus . On the other hand, if the inequality on the right in gives a meaningful upper bound on . One is free to chose any which satisfies . The choice will be used because it results in a sstraightforward calculation, but in this regard it is not unique.
It follows that for any satisfying , that Therefore is discontinuous at , as claimed.
Finally, part (b) is examined, that is it is proven that a continuous bijection from to does not exist. The proof goes by contradiction. The first step is prove that if is a continuous surjection , then it is not injective. By the definition of an injection, it will suffice to show that for some with , . This will be achieved with the help of the Intermediate Value Theorem (IVT):
Two case are considered. The first case is . In this case, obviously is not injective.
The second case is . There are two subcases to this case. The first is that and the other is . The proof of the second subcase, which is nearly identical to the proof of the first, is left as an exercise for the reader.
Suppose that . Since , one has Take satisfying . By assumption, is a surjection, thus there exist such that The above inequalities give . Hence, since is a well-defined function, . Consequently . Consider the interval . The continuity of on this interval follows from its continuity on the containing interval, thus the IVT holds on the closed interval . Hence by inequality and the IVT there exists a such that . However, , since . Therefore is not injective.
As an aside, one could have begun by selecting an satisfying , and constructed a proof along similar lines. The reader is encourage to do this.
If a continuous bijection existed it would be a continuous surjection, which has been shown implies that is not an injection; which in turn implies that is not a bijection. This is a contradiction. Thus a continuous bijection from to does not exist.