Existence of a funcion $T: [2^{\kappa}]^2 \rightarrow \kappa$, containing all partial functions $f$ with $|$dom$(f)| = \kappa$

Question

Let $\kappa$ be a cardinal. I want to prove that there is a function $T: [2^{\kappa}]^2 \rightarrow \kappa$ with the following property ($[2^{\kappa}]^2$ being the subsets of $2^{\kappa}$ that have 2 elements).
If $f: D \rightarrow \kappa$ is a function where $D \in [2^{\kappa}]^{\kappa}$, then there exists an arbitrarily large $\xi \in 2^{\kappa}$ such that $T(\{\alpha,\xi\}) = f(\alpha), \, \forall \alpha \in D$.
It should follow from the fact that cf$(2^{\kappa}) > \kappa$, so $(2^{\kappa})^{\kappa} = 2^{\kappa}$.
Thanks.

Answer 1

Let me restate your question: let $$P = \bigcup\{{^D}\kappa : D\in [2^\kappa]^\kappa\}.$$ You tried to construct $T:[2^\kappa]^2\to\kappa$ such that the set $$\{\xi < 2^\kappa \mid T(\{\alpha,\xi\} = f(\alpha)\}$$ has cardinality $\kappa$.

The first observation is $|P|=2^\kappa$; You can check that ${^D}(2^\kappa)$ has cardinality $2^\kappa$ for each $D\in[2^\kappa]^\kappa$. There are $(2^\kappa)^\kappa = 2^\kappa$ subsets of $2^\kappa$ of cardinality $\kappa$, so the result follows.

Construction of $T$ is now easy: let $\{f_\alpha : \kappa < \alpha < 2^\kappa\}$ be any enumeration of $P$. Take any $\tau : 2^\kappa\to 2^\kappa$ such that for each $\alpha<2^\kappa$ there are $2^\kappa$ ordinals $\xi$ such that $\tau(\xi)=\alpha$. (You can construct such a function by considering a bijection between $2^\kappa$ and $2^\kappa\times 2^\kappa$.

Now define $T(\{\alpha,\beta\})$ as follows: $$T(\alpha,\beta) = \begin{cases} f_{\tau(\beta)}(\alpha) &\text{if } \alpha <\sup (\operatorname{dom} f_{\tau(\beta)}) < \beta \text{ and } f_{\tau(\beta)}(\alpha) \text{ is defined,}\\ 0 & \text{otherwise} \end{cases}.$$ Then $T$ satisfies your conditions.