Functions with Very Low Symmetry and the Continuum Hypothesis

A function from \mathbb{R} to \mathbb{R} is called even if for all h\in\mathbb{R}, f(-h) = f(h).  We might call it even about the point x if, for all h\in\mathbb{R}, f(x-h) = f(x +h).

Conversely, we can call a function strongly non-even if for all x\in\mathbb{R}, h>0, f(x-h)\ne f(x+h).

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 \mathbb{R} to \mathbb{N}.  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 f\colon \mathbb{R}\to\mathbb{N} be non-even of order n if, for all x, |\{h>0\mid f(x-h) = f(x + h)\}|\leq n.  Thus, a strongly non-even function is non-even of order 0, and a function being non-even of order n implies that it’s non-even of order m for all m\geq n.

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 2^{\aleph_0} = \aleph_1).

Thus, if we assume that there is a non-even function of order 1, then we can conclude that 2^{\aleph_0} = \aleph_1.  Can we weaken the hypothesis and still conclude something interesting?  We can, as they also proved:

For any n, if there is a non-even function of order n, then 2^{\aleph_0}\leq \aleph_{\lceil\log_2{(n+1)}\rceil}.

They showed this by showing the following (the statement above follows directly from this, given just a bit of thought):

For any vector space V over \mathbb{Q} of cardinality \aleph_n, and any function f from V to \mathbb{N}, there is an x\in V and a set S of unordered pairs (y,z) such that y+z = 2x and f(y) =f(z), where the cardinality of S is 2^n - 1.

I’ll show here how to prove the weaker statement obtained by replacing 2^n - 1 with 2^{n-1}.

Fix V and f.  We will construct a set of basis elements b_1^0,\ldots,b_n^0,b_1^1,\ldots,b^1_1 with the property that for every set S\subseteq \{1,\ldots,n\}, f(\sum_{i\in S}b_i^0 + \sum_{i\notin S}b_i^1) = f(\sum_{i\in S}b_i^1 + \sum_{i\notin S}b_i^0).  Taking 2x to be \sum_{i\in\{0,1\}, j\in\{1,\ldots,n\}} b^i_j, this will provide the required number 2^{n-1} of unordered pairs.  (As a slight bit of notational convenience, if S is a set of basis elements, I will write f(S) for f(\sum S).)

Let \{c_\alpha\}_{\alpha<\omega_n} be a basis for V. For 1\leq i\leq n, let \{c_\alpha \mid \omega_{i-1} < \alpha < \omega_i\} be called the ith slice of basis elements.  For each i between 1 and n and j\in\{0,1\}, we will pick b_i^j to be in the ith slice of basis elements.

Given i, suppose that we have defined b^0_{i'} for i + 1 \leq i' \leq n and we will define b^0_i as follows: pick it so that it is in the ith slice of basis elements but is not any singleton set of the form: \{x \mid f(s_1\cup \{x\}\cup\{b_{i+1}^0,\ldots, b_{n}^0\}) = m_1,\ldots, f(s_t\cup \{x\}\cup\{b_{i+1}^0,\ldots,b_n^0\})=m_t\} for any finite sets s_1,\ldots,s_t\subset \omega_{i-1} and m_1,\ldots,m_t\in \mathbb{N}.  Since there are only \aleph_{i-1} such singleton sets, and \aleph_i elements of the ith slice of basis elements, we can find such a b_i^0.

Now we define the b^1_i.  Given i, assume that b^1_1,\ldots, b^1_{i-1} have already been defined and we will define b^1_i: pick it from the ith slice so that for all subsets S of \{b^0_1, b^1_1,\ldots, b^0_{i-1},b^1_{i-1}\}, f(S \cup \{b^0_i\}\cup \{b^0_{i+1},\ldots,b^0_n\}) = f(S\cup \{b^1_i\}\cup \{b^0_{i+1},\ldots,b^0_n\}) (this is possible by the defining condition of b^0_i.

It is now easy to prove the following proposition by induction on m:

For any set S\subseteq \{1,\ldots,m\} (where m\leq n), f(\{b^0_i \mid i\in S\}\cup \{b^1_i \mid i\notin S\} \cup \{m_1,\ldots, n\}) equals f(\{b^1_i\mid i \in S\}\cup \{b^1_i\mid i\notin S\}\cup \{m_1\ldots,n\}).

(The proof simply uses the defining property of the b^1_i.)  Now, taking m=n, the result follows.

About these ads

5 Comments

Filed under Uncategorized

5 responses to “Functions with Very Low Symmetry and the Continuum Hypothesis

  1. That’s a fun construction! It’s always neat to have a result where the cardinality \aleph_n actually means something about something in some way n dimensional.

    But not having looked at the Komjáth and Shelah paper, does this tell us anything interesting and new about these cardinalities? These strongly non-even functions seem sort of silly as a sort of non-symmetry, since they just are the injections, while being non-even of order n feels a bit more meaningful, but perhaps too complicated to be intuitively interesting. Is there any motivation for considering these non-symmetric functions beyond the connection to the continuum?

  2. The couple of places that I’ve seen these functions come up have been in the context of the study of symmetrically continuous functions, which is defined to be true of f\colon\mathbb{R}\rightarrow\mathbb{R} at x if \lim_{h\to 0} (f(x-h) - f(x+h)) = 0, but I’ve only seen those discussed in the context of the set theory of the real line.

    However, that probably says more about my lack of breadth of knowledge than it does about the applications of the context. I’ve seen the book “Symmetric Properties of real functions” referenced as a analytical basis for studying these types of things, and you can see a preview at:

    http://books.google.com/books?id=BMWk0X8rl_YC&lpg=PP1&ots=SF0LuDF6P8&dq=thomson%20symmetric%20properties%20of%20real%20functions&pg=PR11

    In any case, I could easily be wrong, but as far as I know, this result is only entertainment; I don’t know that it tells anything fundamentally new either about cardinalities or real functions.

    • That notion of symmetric continuity sounds fun too, though I don’t know what a non-logician would make of it. Anyway, it still sounds like fun stuff, even if mainly recreational.

  3. This is a very cool post.

    thanks

  4. But now, it is just as easy to show that there are no strongly non-even functions.

    Just because &Ropf; is too big?

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s