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Klaus Janich's Topology, Page 97:

The decomposition of a simplicial complex into its open simplices is a CW-decomposition.

No further explanation of this proposition is provided after that,so I'm trying to work it out.

There's not much difficulty in proving the axioms of "characteristic maps" and "closure finiteness", yet I have no idea about dealing with the axiom of "weak topology", i.e. $A\subset X$ is closed if and only if every $A\cap \bar e$ is.($e$ is a cell of the decomposition).

Any tips please?

The definition of simplicial complex in Janich's book:

Definition(Simplicial Complex or Polyhedron).A set $K$ of simplices in $R^n$ is called a simplicial complex or a polyhedron if the following three conditions are satisfied:

  1. If $K$ contains a simplex it contains all faces of this simplex.
  2. The intersection of two simplices of $K$ is either empty or a common face.
  3. (In case $K$ is infinite), $K$ is locally finite, i.e. every point of $R^n$ has a neighborhood that intersects only finitely many simplices of $K$.
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    $\begingroup$ How is the topology on a simplicial complex defined? $\endgroup$ – Lee Mosher Aug 8 '15 at 13:31
  • $\begingroup$ Here the topology of a simplicial complex is considered as a subspace of $R^n$, nothing special... $\endgroup$ – Ire Shaw Aug 8 '15 at 13:35
  • $\begingroup$ In that case you are not using the standard definition of a simplicial complex. What definition are you using? Ordinarily, not every simplicial complex is a subspace of $\mathbb{R}^n$. In fact, not every simplicial complex is homeomorphic to a subspace of $\mathbb{R}^n$. $\endgroup$ – Lee Mosher Aug 8 '15 at 13:58
  • $\begingroup$ Oh, I'm a beginner in algebraic topology,so I'm only acquainted with the definition by Janich.I'll add his definition in the question as a complement. $\endgroup$ – Ire Shaw Aug 8 '15 at 14:09
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The key to this is the local finiteness condition.

Suppose $x_i$ is a sequence in $A$ that converges to a point $p \in X$. By local finiteness, we may choose an open neighborhood $U$ of $p$ which intersects only finitely many closed cells $\bar e_1,\ldots,\bar e_K$. Then we may choose $I$ so that if $i \ge I$ then $x_i \in U$ and therefore $x_i \in \bar e_1 \cup\cdots\cup \bar e_K$. Since there are infinitely many $i>I$ but only finitely many $k=1,\ldots,K$, there exists an infinite subsequence $i_1<i_2<i_3<\cdots$ and a $k \in \{1,\ldots,K\}$ such that $x_{i_j} \in \bar e_k$ for all $j$. Since the limit of a sequence equals the limit of any subsequence, it follows that $p \in \bar e_k$. But $x_{i_j} \in A \cap \bar e_k$ which is a closed subset of $\bar e_k$, so $p \in A \cap \bar e_k \subset A$.

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Any closed simplex in $\mathbb R^n$ is closed subset, so it will be closed subset in $X\subset \mathbb R^n$. So desired statement becomes tautological, because induced topology on $\Delta^k$ from $\mathbb R^n$ coincides with induced one from $X$.

In fact, both $CW$-complexes and simplicial complexes has factor-topology, obtained by gluing $k$-cells/simplexes to the $(k-1)$-skeleton.

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  • $\begingroup$ You seem to have answered a question different from the one asked. $\endgroup$ – Mariano Suárez-Álvarez Aug 8 '15 at 16:40

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