Set of 0-cells of a simplicial complex and discreteness When given a simplicial complex, does its set of $0$-cells have to be a discrete set? In particular, can the set $\{0, 1/n\}_{n=1,...}$ be a simplicial complex (consisting of $0$-cells)? From the definition of a simplicial complex, it's not obvious to me set of 0-cells has to be a discrete set, but I heard "in a simplicial complex, the set of $0$-cells form a discrete subset" is a key fact to show "a real tree $T$ is simplicial if and only if the set of singular points of $T$ is discrete in $T$". Can someone explain a little about it?
Note here $\{0, 1/n\}_{n=1,...}$, I mean $\{1/n\}_{n=1,...}\cup \{0\}$.
 A: What you heard is correct: for any simplicial complex $X$ endowed with the simplicial topology, if we let $X^{(0)}$ denote its set of vertices endowed with the subspace topology, then $X^{(0)}$ is indeed a discrete topological space. This is an almost immediate consequence of the definition of the simplicial topology.
Popular accounts of simplicial complexes, such as the one in that wolfram link you gave in a comment, might not contain any formal definition of the simplicial topology. You can find a description of the simplicial topology at this wikipedia link, although perhaps a good textbook description would be better.
Proving that $X^{(0)}$ is discrete should be a straightforward exercise once one understand that definition. Here's a brief outline. In the simplicial topology on $X$, a subset of $X$ is open if and only it its intersection with each individual simplex is an open subset of that simplex. Using this, it's not hard to directly construct, for each $x \in X^{(0)}$, an open subset $U_x \subset X$ such that $X^{(0)} \cap U_x = \{x\}$. It follows that $\{x\}$ is an open subset of the subspace topology on $X^{(0)}$.
A: The definition you found in Wolfram MathWorld describes simplicial complexes as a collection $K$ of simplices in some $\mathbb R^n$ such that

*

*Every face of a simplex of $K$ is in $K$.

*The intersection of any two simplices of $K$ is a face of each of them.

This is a quotation from §2 Simplicial Complexes and Simplicial Maps, p.7, of

Munkres, James R. Elements of algebraic topology. CRC press, 2018.

Wolfram does not say anything about topology; and in fact, a simplicial complex is not a topological space. Yes, it is tempting to identify $K$ with the subset $\lvert K \rvert = \bigcup_{\sigma \in K} \sigma \subset \mathbb R^n$, but the set $\lvert K \rvert$ does not allow to reconstruct the collection $K$. In particular the set $K^0$ of all $0$-simplices of $K$ does not have a topology, thus it does not make sense to say it is a discrete set. $K^0$ is a set of single point subsets of $\mathbb R^n$, and we must not confuse it with the set $\lvert K^0 \rvert = \bigcup_{\sigma \in K^0} \sigma \subset \mathbb R^n$. As an example consider any subset $S \subset \mathbb R^n$ and define a simplicial complex
$$K_S = \{ \{x\} \mid x \in S\}$$
which has only $0$-simplices. We have $K_S^0 = K_S$ and $\lvert K_S^0 \rvert = S$. This applies in particular to your set $\{0, 1/n\}_{n=1,...}$.
Concerning topology let us read what Munkres writes:


Using the above topology of $\lvert K \rvert$, it is easy to see that the subspace $\lvert K^0 \rvert \subset \lvert K \rvert$ is always discrete. Our above example nicely illustrates that the "simplicial topology" on $\lvert K \rvert$ is in general different from the subspace topology induced from $\mathbb R^n$ on the subset $\lvert K \rvert \subset \mathbb R^n$.
