Editing Freiling's axiom of symmetry (section)

==Relation to the (Generalised) Continuum Hypothesis==
Fix <math>\kappa\,</math> an infinite cardinal (''e.g.'' <math>\aleph_{0}\,</math>). Let <math>\texttt{AX}_{\kappa}</math> be the statement: ''there is no map <math>f:\mathcal{P}(\kappa)\to[\mathcal{P}(\kappa)]^{\leq\kappa}\,</math> from sets to <math>\leq\kappa</math>-sized collections of sets such that for any <math>{x,y\in\mathcal{P}(\kappa)}\,</math> either <math>x\in f(y)\,</math> or <math>y\in f(x)\,</math>.''

'''Claim:''' <math>\texttt{ZFC}\vdash 2^{\kappa}=\kappa^{+}\leftrightarrow\neg\texttt{AX}_{\kappa}\,</math>.

'''Proof:'''
''Part I'' (<math>\Rightarrow\,</math>):

Suppose <math>2^{\kappa}=\kappa^{+}\,</math>. Then there exists a bijection <math>\sigma:\kappa^{+}\to\mathcal{P}(\kappa)\,</math>. We will exploit the well-ordering of <math>\kappa^+</math> to build an ascending chain in <math>[\mathcal P(\kappa)]^{\leq\kappa}</math>.

Define a function <math>f:\mathcal{P}(\kappa)\to[\mathcal{P}(\kappa)]^{\leq\kappa}\,</math> by <math>\sigma(\alpha) \mapsto \{\sigma(\beta) : \beta \leq \alpha\}\,</math>. Given this function, it is straightforward to see that this demonstrates the failure of Freiling's axiom. The map <math>\sigma</math> induces a well-ordering <math>\preceq</math> on <math>\mathcal P(\kappa)</math> (sometimes called the ''pushforward'' of the standard ordering on <math>\kappa</math>). Picking any two <math>x,y \in \mathcal P(\kappa)</math>, we can state without loss of generality that <math>x \preceq y</math>. But then, noting the definition of <math>f</math>, we see that this implies <math>x \in f(y) = \{t \in \mathcal P(\kappa) : t \preceq y \}</math>. Thus we have found a function <math>f</math> witnessing <math>\neg\texttt{AX}_\kappa</math>.

''Part II'' (<math>\Leftarrow\,</math>):

Suppose that Freiling's axiom fails. Then fix some <math>f\,</math> to verify this fact. Define an order relation on <math>\mathcal{P}(\kappa)\,</math> by <math>A\leq_{f} B</math> iff <math>A\in f(B)</math>. This relation is total and every point has <math>\leq\kappa</math> many predecessors. Define now a strictly increasing chain <math>(A_{\alpha}\in\mathcal{P}(\kappa))_{\alpha<\kappa^{+}}</math> as follows: at each stage choose <math>A_{\alpha}\in\mathcal{P}(\kappa)\setminus\bigcup_{\xi<\alpha}f(A_{\xi})</math>. This process can be carried out since for every ordinal <math>\alpha<\kappa^{+}\,</math>, <math>\bigcup_{\xi<\alpha}f(A_{\xi})\,</math> is a union of <math>\leq\kappa\,</math> many sets of size <math>\leq\kappa\,</math>; thus is of size <math>\leq\kappa<2^{\kappa}\,</math> and so is a strict subset of <math>\mathcal{P}(\kappa)\,</math>. We also have that this sequence is ''cofinal'' in the order defined, ''i.e.'' every member of <math>\mathcal{P}(\kappa)\,</math> is <math>\leq_{f}\,</math> some <math>A_{\alpha}\,</math>. (For otherwise if <math>B\in\mathcal{P}(\kappa)\,</math> is not <math>\leq_{f}\,</math> some <math>A_{\alpha}</math>, then since the order is total <math>(\forall{\alpha<\kappa^{+}})A_{\alpha}\leq_{f} B\,</math>; implying <math>B\,</math> has <math>\geq\kappa^{+}>\kappa\,</math> many predecessors; a contradiction.) Thus we may well-define a map <math>g:\mathcal{P}(\kappa)\to\kappa^{+}\,</math> by <math>B\mapsto\operatorname{min}\{\alpha<\kappa^{+}:B\in f(A_{\alpha})\}</math>.
So <math>\mathcal{P}(\kappa)=\bigcup_{\alpha<\kappa^{+}}g^{-1}\{\alpha\}=\bigcup_{\alpha<\kappa^{+}}f(A_{\alpha})\,</math> which is union of <math>\kappa^{+}\,</math> many sets each of size <math>\leq\kappa\,</math>. Hence <math>2^{\kappa}\leq\kappa^{+}\cdot\kappa=\kappa^{+}\,</math>. {{NumBlk|1=|2=|3=<math>\blacksquare</math> (Claim)|RawN=.}}

Note that <math>|[0,1]|=|\mathcal{P}(\aleph_{0})|\,</math> so we can easily rearrange things to obtain that <math>\neg\texttt{CH}\Leftrightarrow\,</math> the above-mentioned form of Freiling's axiom.

The above can be made more precise: <math>\texttt{ZF}\vdash(\texttt{AC}_{\mathcal{P}(\kappa)}+\neg\texttt{AX}_{\kappa})\leftrightarrow \texttt{CH}_{\kappa}\,</math>. This shows (together with the fact that the continuum hypothesis is independent of choice) a precise way in which the (generalised) continuum hypothesis is an extension of the axiom of choice.