0$ (if its negative a
similar argument works). Note that for any $t\in T$ either
\begin{itemize}
\item $t\concat 0\in T$ or
\item there exists $w>0$ such that
$t\concat w\in T$ and $t\concat -w\in T$.
\end{itemize}
In other words if a node
doesn't split it continues with 0 and if it does split, then
one way is positive and other negative. Using this
it is easy to construct a sequence
$$s=s_0\su s_1\su\cdots\su s_m\su\cdots$$
with $s_m\in T_{n+m}$ such that for $m>0$
$$s_m(n+m-1)\cdot x(n+m-1)\geq 0.$$
It follows that
if $b=\cup\{s_m:m\in\om\}$ and $(x,b)$ converges, then
$$(x,b)\geq (x\res n,s)>0,$$
and so, in any case, $x$ is not orthogonal to $b$.
\qed
\begin{figure}
\unitlength=.9mm
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\end{picture}
\caption{Kunen's perfect tree \label{kunentree}}
\end{figure}
\pfof{thm2}
For $x\in \rr^\om$ define the support of $x$:
$$\supp(x)=\{n\in\om: x(n)\not=0\}.$$
Let us say that $x$ and $y$ are strongly orthogonal iff
$\supp(x)$ and $\supp(y)$ are almost disjoint
(i.e., intersection is finite) and $(x,y)=0$.
Let $S$ be all the elements of $\rr^\om$ with infinite
support. Let $S^*$ be all elements $x$ of $S$ such that
if $n\in\supp(x)$ then $|x(n)|>1$.
Let $\po$ be any countable partially ordered set.
\begin{lemma}
Suppose $\{x_n:n\in\om\}\subseteq S^*$ are
pairwise strongly orthogonal. Let
$p\in\po$ and $\tau$ be a $\po$-name such that
$$p\forces \tau\in S \mbox{ and }\forall n \;(x_n,\tau)=0.$$
Then there exists $q\leq p$ and $y\in S^*$ such that
$$q\forces (y,\tau)\not= 0$$
and $y$ is strongly orthogonal to $x_n$ for
every $n\in\om$.
\end{lemma}
\pf
Case 1. $p\forces$``$\supp(\tau)\setminus\cup_{n