>>606
アンタね、その無限グラフってやね、もし連結性を仮定しなかったらサ、例えば
assume that the vertex set is fixed to be ${\Bbb Z}$ due to the fact
that you work on the category of infinite graphs. Then the set of all
possible choice of the edges with this fixed vertex set is restricted
in the following simple way as:
\[
\{ (n)\mbox{---}(n+1) : n\in {\Bbb Z} \} \simeq {\Bbb Z}
\]
is clearly a countable set. This is the same as to consider the 1-dimensional
lattice model together with the assumption that you just pay attention to the
nearest neighbors assigned by the edges of the graphs. Then {\bf the
set of all possible configurations corresponds to the power set of such},
namely the cardinality of what you are saying is nothing but:
\[
2^{\Bbb Z}
\]
so that, your statement seems to be quite obvious.
I really do not understand {\bf what is that good for}.
>>606
You should remind the fact that the graph $\Gamma$ consists of the pair
$(V,E)$ of the set of all vertex $V$ together with the subset $E\subset
V\times V$. Therefore if we allow $V$ to be the (countable) infinite set,
the choice of all possible edges $E$ is clearly the set with its cardinality
{\bf One Up}. This means, in particular, that {\bf Your claim is logically
guaranteed but nothing difficult to prove}. If you are the first grade
student of Komaba, it should be okay. However, if you want yourself to
be okay as a professional researcher to publish something, you may need
to re-enter the (under)graduate course of any theoretical subjects. You
need some basic training on logical way of using your brain. Or otherwise
you will never be recognized as a professional researcher that I'm very
much afraid.
