Every weakly compact cardinal is Mahlo This is a reference question. Does anyone know any book or paper that has the proof that every weakly compact cardinal is Mahlo, using only combinatorics?
I know the definition of weak compactness has a lot of equivalent forms. In this particular case I am referring to a proof that is based only on the combinatorial definition: $$\kappa \text{ is weakly compact iff }\kappa\to (\kappa)^2_2,$$
i.e. for every partition of the 2-element subsets of $\kappa$ into two sets, there is a homogeneous set of size $\kappa$.
 A: This may help:
Hajnal, Kanamori, and Shelah analyzed regressive partition relations for infinite cardinals in their paper
Regressive partition relations for infinite cardinals. Trans. Amer. Math. Soc. 299 (1987), no. 1, 145–154.
They characterize Mahlo cardinals (even n+1-Mahlo cardinals for each $n<\omega$) in terms of these regressive partition relations:
For example, Theorem 3.4 tells us that $\kappa$ is Mahlo if and only if
For any closed unbounded $C\subseteq\kappa$ and regressive coloring $f$ of $[C]^4$, there is a min-homogeneous set of size $\omega$.
There are some comments in the paper about how weakly compact cardinals satisfy these regressive partition relations, but it's not clear to me (yet) if there's a direct proof or if the proofs need to go through the tree property.  I feel like Asaf --- I need to look at it after I get some other work done!  
Edit:  One can prove directly that if $\kappa\rightarrow (\kappa)^5_2$ then $\kappa$ satisfies the regressive partition relation mentioned above that is equivalent to being Mahlo. 
This is still unsatisfying, but what I'm wondering is if this can be put together with the result from the paper to get something like:
If $\kappa$ is not Mahlo, then there is a 2-coloring of the 5-tuples from $\kappa$ with no homogeneous set of size $\kappa$.
Another edit based on another strategy:
If $\kappa>\omega$ is not $\omega$-Mahlo, then $\kappa\nrightarrow(\kappa)^2_2$ by virtue of results on negative square-brackets partition relations: If $\kappa$ has a non-reflecting stationary subset this follows from Todorcevic's result that $\kappa\nrightarrow[\kappa]^2_\kappa$ for such $\kappa$. Otherwise, $\kappa$ must be weakly inaccessible with a stationary subset that does not reflect in an inaccessible cardinal, and by results in Shelah's Cardinal Arithmetic, we have $\kappa\nrightarrow[\kappa]^2_\theta$ for every $\theta<\kappa$.
In either case, we certainly have $\kappa\nrightarrow[\kappa]^2_2$, which is equivalent to $\kappa\nrightarrow(\kappa)^2_2$, so $\kappa$ cannot be weakly compact.
So the explicit "bad coloring" of pairs is there to see. It's just really really complicated.
Final Edit
Suppose $\kappa$ is weakly inaccessible and not Mahlo.  Let $\langle C_\delta:\delta<\kappa\rangle$ be a $C$-sequence in the sense of Todorcevic, so $C_\delta$ is club in $\delta$ of order-type ${\rm cf}(\delta)$. Let $c(\alpha,\beta)$ be the length of the minimal walk from $\beta$ down to $\alpha$ (so $c:[\kappa]^2\rightarrow\omega$).  The fact that $\kappa$ isn't Mahlo implies that the $C$-sequence is non-trivial in the sense of Todorcevic, and his work shows that for any $H\subseteq\kappa$ of size $\kappa$ the range of $c$ restricted to the pairs from $H$ is infinite.   In particular, $\kappa\nrightarrow(\kappa)^2_{\omega}$ and so $\kappa$ is not weakly compact.
Still not as simple as we'd like, but I don't see how to do better!  This is why the tree property is a good thing...
A: Here is a way to get the tree property directly from the partition
property, and this gets you to Mahloness from the references in the links provided in the comments.
Assume that $\kappa$ is an uncountable cardinal and that the
partition property $\kappa\to(\kappa)^2_2$ holds.
First, show that $\kappa$ is regular. If not, there is a short
unbounded sequence $\kappa_0\lt\kappa_1\lt\cdots\kappa_\xi<\cdots$
for $\xi\lt\text{cof}(\kappa)$, which is unbounded in $\kappa$.
Define a coloring $f(\alpha,\beta)$ is $1$, if they are in the
same interval determined by these $\kappa_\xi$, and $0$ otherwise.
Since each interval has size less than $\kappa$, there cannot be a
size $\kappa$ homogeneous set with value $1$; and since there are
fewer than $\kappa$ many intervals, there cannot be a size
$\kappa$ homogeneous set with value $0$. So $\kappa$ must be
regular.
Next, show $\kappa$ is a strong limit and hence inaccessible. If
$2^\beta\geq\kappa$ for some $\beta\lt\kappa$, then let $\langle
a_\alpha\mid\alpha\lt\kappa\rangle$ be a $\kappa$-sequence of
distinct subsets of $\beta$. Let $F(\alpha,\beta) = 0$ if
$a_\alpha$ precedes $a_\beta$ in the lexical ordering of
$2^\beta$, and otherwise 1. It is not difficult to prove that
there is no monotone sequence of length $\beta^+$ in the lexical
order on $2^\beta$. Consequently, there can be no monochromatic
set for $F$ of size $\kappa$, and so $\kappa$ is inaccessible.
Finally, show the tree property. Suppose that $T$ is a
$\kappa$-tree. We may assume that the underlying set of $T$ is
exactly $\kappa$. For any two nodes $\alpha$ and $\beta$ in $T$,
let us say that $\beta$ is to the right of $\alpha$ in $T$, if
$\alpha\perp_T\beta$ and $\alpha'\lt\beta'$, where
$\alpha'\leq_T\alpha$ and $\beta'\leq_T\beta$ are least in $T$
such that $\alpha'\perp_T\beta'$. Define $F(\alpha,\beta)=0$ if
$\beta$ is above or to the right of $\alpha$, and otherwise $1$.
Suppose that $H\subset\kappa$ is a monochromatic set of size
$\kappa$. Suppose first that the monochromatic value is $0$. Thus,
whenever $\alpha\lt\beta$ in $H$, then $\beta$ is above or to the
right of $\alpha$ in $T$. By applying the partition property once
more, we may further assume that only one of these answers arises.
If any $\alpha\lt\beta$ in $H$ has $\alpha\lt_T\beta$, then the
elements of $H$ are linearly ordered, and $T$ has a
$\kappa$-branch, as desired. Otherwise, we assume that whenever
$\alpha\lt\beta$ in $H$, then $\beta$ is to the right of $\alpha$
in $T$. We may therefore follow the ``right-most'' path through
(the $T$-predecessors of elements of) $H$. Specifically, using the
fact that the levels of $T$ have size less than $\kappa$ and
$\kappa$ is regular, there is on each level $\xi$ a right-most
node $\zeta$ that occurs as a $T$-predecessor of all sufficiently
large elements of $H$. The set of such nodes is clearly linearly
ordered, and therefore forms a $\kappa$-branch through $T$. The
final case, when the monochromatic value of $F$ on $H$ is $1$, is
similar, except that we follow the left-most branch through $H$ to
provide a $\kappa$-branch through $T$. So we've got the tree property as desired.
(I'm sorry I don't have a direct argument from the partition property to Mahloness, but I suspect there may be one if one could find a clever coloring.)
