Simplicial Complex as a topological space I am new to topological data analysis. I just do not get the idea that a simplicial complex can be considered as a topological space when in the first place, it does not satisfy the definition of a topology.
For example, if the Vertex Set V = {a,b,c,d} and we have a simplicial complex K = {{a}, {b}, {c}, {d}, {a,b},  {a,c}, {b,c}, {c,d}, {a,b,c}}. And since arbitrary union of open sets are open, {a,c} $\cup$ {c,d} = {a,c,d} which is obviously not in K,  how come K can be considered as topological space? I don’t know if my question makes sense but thank you in advance for answering.
 A: There is confusion here over what the topological space associated to the combinatorial data of a simplicial complex actually is.  Suppose our vertex set is finite, say $\{a,b,c,d\}$  Consider a vector space spanned by vectors corresponding to those vertices, say $v_a, v_b, v_c$, and $v_d$.  Associated to any subset $S$ of the vertex set, we can define the open simplex $V_S=\{\sum_{\alpha \in S} x_\alpha v_\alpha \mid 0<x_\alpha<1, \sum_{\alpha\in S} x_\alpha=1\}$.  This is just an algebraic way of expressing the (interior of the) smallest convex set that contains all the vertices.  If we had $\leq$ instead of $<$, this would give us the whole convex set instead of its interior.
Now, for every subset in your simplicial complex, you associate the geometric simplex in Euclidean space, and you take the union of all of them, together with the single points corresponding to the vertices (as the open simplices were empty). This is a subspace of Euclidean space, and inherits the subspace topology.
The condition to be a simplicial complex, namely that we are downward closed, says that if we have a simplex, then we have its face, and is essentially saying that we are the union of closed simplices.
