tree property and arrow notation How can I see that for an uncountable cardinal $\kappa$ these 2 conditions are equivalent :

*

*$\kappa$ has the tree property

*$\kappa \to (\kappa)^2$
 A: Assuming that in (1), we are also assuming $κ$ is inaccessible:
$(2) ⇒ (1)$:
$κ$ is strongly inaccessible (try to show this), hence regular.
Let $T=(κ, <')$ be a tree with level of size $κ$. For $λ∈κ$ with level $≥α$, let $π_α(λ)$ be the unique $δ∈κ$ with the level $α$ and $δ≤'λ$.
Let $\prec$ be order on $κ$ defined as: $α\prec β⇔α<'β$ or $α,β$ are incomparable, and $π_γ(α)<π_γ(β)$ for the least $γ$ that $π_γ(α)≠π_γ(β)$.
This is a linear ordering.
Let $f:[κ]^2\to2$, $f(\{α,β\})=1⇔$ $∈$ and $\prec$ agree on $α,β$ (that is: $α∈β$ and $α\prec\beta$ or $α∋β$ and $α\succ\beta$)
Let $A$ be homogeneous subset of size $κ$.
By assumption, $T$ does not have levels of size $κ$, so for each $α∈κ$, there exists $\xi_α$ such that $ξ_α∈β$ implies $β$ has a level $≥α$ (by the regularity of $κ$).
Specifically, it is true for $ξ_α∈β∈A$.
If $f''[A]^2=\{1\}$, Given an $α$, the sequence $(π_α(β)\mid ξ_α∈β∈A)$ is not $\prec$-decreasing ($ξ_α∈β∈γ$ and $β,γ∈A$ implies $∈$ and $\prec$ agree with each other, so $β\prec\gamma$. If $β,γ$ are on the same branch, then $π_α(β)=π_α(γ)$, otherwise, $π_α(β)<π_α(γ)$).
Similarly, if $f''[A]^2=\{0\}$ it is not $\prec$-increasing (same argument but reverse the order).
At some point, the sequence must be constant, because $\{π_α(β)\mid ξ_α∈β∈A\}$ is a subset of the level $α$ of cardinality $<κ$. Let $b_α$ be the value this sequence eventually stabilize into.
$\{b_α\mid\alpha\in\kappa\}$ is (subset of a) branch. Indeed, let $b_α,b_β$ be 2 elements and $x$ big enough, we have $π_α(x)=b_α$ and $π_β(x)=b_β$, so $b_α,b_β<'x$ which implies $b_α<'b_β$ (or the other way around). Noting that $\{b_α\mid\alpha\in\kappa\}$ is of cardinality $κ$ finish the proof.

For $(1)\implies (2)$, let $κ$ be inaccessible with the tree property. Let $f:[κ]^2\to 2$ any function.
Let $T=(κ,<_f)$ a tree with the following properties:

*

*$α∈β⇒ α<_fβ$

*$α<_f\beta<_f\gamma\implies f(\{α,β\})=f(\{α,γ\})$

*The level $α$ has cardinality $≤2^{\max(\aleph_0,|α|)}$
(Try to show such tree exists). This tree has cofinal branch (because $κ$ is inaccessible + property (3)).
For (infinite) $λ<κ$, let $ν_λ$ be the element of the level $λ^+$ from the cofinal branch.
Now, let $g_{λ}:\{x\mid x<_f ν_{λ}\}→2$ defined as: $g_λ(x)$ equal to the value of $f(\{x,y\})$ for some $y>_f x$, this is well defined by property (2) of the tree.
$\{x\mid x<_f ν_{λ}\}$ has homogeneous subset for $g_{λ}$ of size $λ^+$, $B_λ$. ($\{x\mid x<_f ν_{λ}\}$ has regular cardinality, so either the preimage of $1$ or of $0$ are homogeneous of size $λ^+$).
$B_λ$ is also homogeneous for $f$, indeed for $α,β,γ,δ∈B_λ$, let $α<_fβ,γ<_fδ$, then $f(\{α,β\})=g_λ(α)=g_λ(γ)=f(\{γ,δ\})$.
Finally, we notice that if $λ∈μ∈κ$, we have that if $f''[B_λ]^2=f''[B_μ]^2$, $B_λ∪B_μ$ is also homogeneous (and similarly for arbitrary union).
Because either $B^0=\{B_λ \mid λ≥\aleph_0\text{ and }f''[B_λ]^2=\{0\}\}$ or $B^1=\{B_λ \mid λ≥\aleph_0\text{ and }f''[B_λ]^2=\{1\}\}$ are of size $κ$, either $\bigcup B^0$ or $\bigcup B^1$ are homogeneous subset of $κ$ for $f$ of size $κ$.
