Open Subsets of an Simplicial Complex

Question

By an simplicial complex, I mean a finite collection of simplexes in some Euclidean space satisfying the well known conditions. So I don't mean an abstract simplicial complex, which is purely combinatoric, but its geometric realization.

For a simplex $\sigma$, denote $\mathring\sigma$ to be its open simplex, that is, what is left of when one removes every proper faces of $\sigma$ from itself. If $K$ is a simplicial complex, it is well known that the collection of sets $\mathring\sigma$ (where $\sigma$ varies in $\sigma\in K$) partitions $|K|$, the underlying space of $K$. Moreover, if $\sigma\in K$ is a simplex having the largest dimension among $K$, we have that $\mathring\sigma$ is open in $|K|$. The analogous statement involving arbtrary simplex of $K$ is however false in general.

My question is : Given an nonempty open subset $U$ of $|K|$, does there exist a simplex $\sigma\in K$ such that $W=U\cap\mathring\sigma$ is again an nonempty open subset of $|K|$?

This is intuitively very clear, but I don't see how one could prove this. The textbook I'm using (Rotman, An Introduction to Algebraic Topology) writes that this follows from the fact that "the stars of vertices form an open cover of $|K|$". Now, I have succeeded in proving that last "fact", but I still don't understand why that fact implies the result. (The star of a vertex $p\in\text{Vert}(K)$ is, by definition, the union of all $\mathring\sigma$'s where $\sigma\in K$ has $p$ as one of its vertices, i.e. $p\in\text{Vert}(\sigma)$)

$\mathring\sigma$ is, in general, NOT open in $|K|$, so even though the star of p (denoted by $\text{st}(p)$) is an open subset of $|K|$, each $\mathring\sigma$ constituting $\text{st}(p)$ need not be an open subset... I don't see how one should proceed any further. Any suggestions?

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P.S. I faced this problem while going through the proof of the Invariance of Dimension for simplicial complexes (Theorem 7.1 of [Rotman]). This says : If $K$ and $L$ are simplicial complexes and if there exists a homeomorphism $f:|K|\rightarrow |L|$, then $\text{dim}(K)=\text{dim}(L)$. Any advice in proving this fact directly would also be appreciated.

Thank you.

Answer 1

Edited to conform more to the question as asked, and to clarify further after comments.

If $\Delta$ is a simplex of $K$ which is not contained in any higher-dimensional simplex, then the corresponding open simplex will be open in $|K|$.

Let $U$ be an open subset of $K$. Let $\Delta$ be a closed simplex of $K$ not contained in any higher-dimensional simplex of $K$, such that $\Delta$ intersects $U$.

Let $u$ be a point in $U\cap \Delta$. The fact that $U$ is open says that there is some radius $r$ such that all points of $\Delta$ within a distance of $r$ from $u$ are in $U$. I claim that it is obvious that this includes points in the open simplex of $\Delta$.

One way to see this is to say that if the simplex is $d$-dimensional, we can imagine it as embedded in $\mathbb R^d$, with the faces that $u$ lies on, lying along some of the co-ordinate hyperplanes.

Then, since the intersection of U with the open simplex of $\Delta$ is non-empty and open since it is the intersection of two open sets, it satisfies the question posed.

Answer 2

Eventually I figured out a way to prove this.

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Since $|K|$ is the disjoint union of $\mathring\sigma$s where $\sigma$ varies in $K$, there are (finitely many) $\sigma$s satisfying $U\cap\mathring\sigma\ne\emptyset$. Let $\sigma$ be one having maximal dimension among them, and fix a point $p\in U\cap\mathring\sigma$. We will show that there exist some open subset of $|K|$ that contains $p$ and is still a subset of $U\cap\mathring\sigma$. This will imply that $U\cap\mathring\sigma$ is an nonempty open subset of $|K|$, since $p$ was arbitrary.

Claim : $\text{dist}(p,\mathring\tau)>0$ for all $\tau\in K$ with $\tau\ne\sigma$. ($\text{dist}$ here means the distance in the ambient Euclidean space)

Proof : If $p\notin\tau$, closedness of $\tau$ (it is compact) gives $\text{dist}(p,\tau)>0$ which implies the result immediately. We will show that $p\in\tau$ is actually impossible. For contradiction, assume that $p\in\tau$. Now, $p$ is a point in $\mathring\sigma$, so $p\in\tau$ is possible only if $\sigma$ is a face of $\tau$. Since $\tau\ne\sigma$, $\sigma$ is a proper face of $\tau$, showing that $p$ also lies in a proper face of $\tau$. Since $U$ is an open subset of $|K|$ and $p\in U$, there exist some $\epsilon >0$ such that every point of $|K|$ less than $\epsilon$ away (in the Euclidean distance) from $p$ also belongs to $U$. But it is easy to see that we can choose a point $q\in\mathring\tau$ arbitrary close to $p$, and choosing such $q$ with $\text{dist}(p,q)<\epsilon$ implies that $U\cap\mathring\tau\ne\emptyset$ while the dimension of $\tau$ is greater than that of $\sigma$. This contradicts the maximality of $\sigma$ and the proof of the claim is complete.

From the claim above, we have that $\text{dist}(p,|K|-\mathring\sigma)>0$, since $|K|-\mathring\sigma$ is a disjoint union of (finitely many!) $\mathring\tau$s where $\tau\in K, \tau\ne\sigma$. Now choose $r>0$ small enough so that

• Every point of $|K|$ less than $r$ away from $p$ also belongs to $U$;
• $r<\text{dist}(p,|K|-\mathring\sigma)$.

By our construction, any point $q$ of $|K|$ less than $r$ away from $p$ must actually belong to $\mathring\sigma$. Let $V$ be the set of all such $q$s. Than $V$ is an open subset of $|K|$, contains $p$, $V\subset U$, and $V\subset\mathring\sigma$. We have found our desired open subset of $|K|$. Q.E.D.

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