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\centerline{\bf Assignment 2}
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Let $N$ be an infinite countable set.
Let $K$ be the abstract simplicial complex consisting
of all finite subsets of $N$.
The collection of all {\sl finite\/} subsets of a countable set is countable
so $K$ has only countably many simplicies.
Since a finite complex is a finite union of simplicies, $K$ also
has only countably many finite subcomplexes.
On the other hand, any finite complex $L$ is simplicially isomorphic to
a finite subcomplex of $K$: just embed the vertices of $L$ in $N$
and observe that this embedding provides a simplicial isomorphism
between $L$ and its image as a subcomplex of $K$.
This proves that the collection of finite simplicial complexes is
at most countable.
The $0${--}dimensional simplicial complexes, $L_n$, where $L_n$
has $n$ vertices, is a countably infinite collection of distinct
complexes, so the collection of finite simplicial complexes is
exactly countably infinite.
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