A function from to is called even if for all , . We might call it even about the point if, for all , .
Conversely, we can call a function strongly non-even if for all , , .
Finding strongly non-even functions is easy, as any injective function provides a trivial example. We can make things harder for ourselves by considering only functions from to . But now, it is just as easy to show that there are no strongly non-even functions.
Therefore, let’s make the following definition: Let a function be non-even of order if, for all , . Thus, a strongly non-even function is non-even of order , and a function being non-even of order implies that it’s non-even of order for all .
In this paper, the set theorists Peter Komjáth and Saharon Shelah proved:
The existence of a non-even function of order 1 is equivalent to the Continuum Hypothesis (i.e., the statement that ).
Thus, if we assume that there is a non-even function of order 1, then we can conclude that . Can we weaken the hypothesis and still conclude something interesting? We can, as they also proved:
For any , if there is a non-even function of order , then .
They showed this by showing the following (the statement above follows directly from this, given just a bit of thought):
For any vector space over of cardinality , and any function from to , there is an and a set of unordered pairs such that and , where the cardinality of is .
I’ll show here how to prove the weaker statement obtained by replacing with .
Fix and . We will construct a set of basis elements with the property that for every set , . Taking to be , this will provide the required number of unordered pairs. (As a slight bit of notational convenience, if is a set of basis elements, I will write for .)
Let be a basis for . For , let be called the th slice of basis elements. For each between and and , we will pick to be in the th slice of basis elements.
Given , suppose that we have defined for and we will define as follows: pick it so that it is in the th slice of basis elements but is not any singleton set of the form: for any finite sets and . Since there are only such singleton sets, and elements of the th slice of basis elements, we can find such a .
Now we define the . Given , assume that have already been defined and we will define : pick it from the th slice so that for all subsets of , (this is possible by the defining condition of .
It is now easy to prove the following proposition by induction on :
For any set (where ), equals .
(The proof simply uses the defining property of the .) Now, taking , the result follows.