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 $K$.
  • 1
    $\begingroup$ 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

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 $p \in A \cap \bar e_k \subset A$.


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.