Setting this simple example aside, a complex need not fit into real space. Imagine three triangles sharing a common side. In the plane, two of the three triangles would overlap, and that is not an accurate representation. (The points in each triangle, outside the common edge, are distinct.) You can get around this problem by embedding the complex in 3 space. Now the three triangles share a side, like the pages of a book.
To start, a complex consists of a list of all its simplexes, using arbitrary symbols to label the vertices. One might use points in real space, but every n-simplex is homeomorphic to every other n-simplex, so the precise shape in real space doesn't actually change the topology. We may as well label the vertices of a triangle a b and c.
Labels may repeat, though not within the same simplex. In our example of a complex K, with many triangles sharing a side, each triangle has two vertices labeled a and b. The third vertex is given its own label, distinct from the others.
Technically, the complex includes the faces of each simplex, recursively, all the way down to vertices. So the triangle abx implies the sides ab ax and bx, and the corners a b and x. As we shall see, this really doesn't change the resulting topological space. It is a convenience, for the following definition.
Let R be a simplex defined by a collection of labeled vertices. For instance, R might be ab, that is, the segment joining a and b. Identify every instance of R in S. These points are clumped together in equivalence classes. There will only be one point labeled a, and one point labeled b, and one line segment joining a and b. A similar clumping occurs for the triangle abx, wherever it occurs. Do this for all shared simplexes in parallel, and take the quotient space. The complex K is the quotient space of S.
Let's look at an open set O in K. Assume it includes part of the interval from a to b. Pull back to S, and the preimage has to be open in every instance of ab. Of course the same portion of the segment is replicated again and again. If it is open in one, it is open in all. Therefore the portion of O that intersects ab has to be open in ab.
Apply the above reasoning to a triangle abx. The intersection of O with this triangle is open. Using the subspace topology, this implies O intersect ab is open. In other words, it is enough to say O intersects each maximal simplex in an open set.
Taking complements, a set is closed iff each piece of that set is closed in its simplex.
Note that K need not be connected. Two separate triangles is a perfectly good complex.
A base open set in Ej is a unit ball about a point. Intersect this with a simplex and find an open set in that simplex. In other words, the map from the disjoint union onto K, in Ej, is continuous. The converse can be a problem, even when all simplexes are 1 dimensional. Let infinitely many intervals share a common endpoint x. Think of x as 0 in the interval [0,1]. Let the nth interval have the open set [0,1/n). The union becomes open in the quotient space, and should be open in K. However, there is no open ball about x whose radius is smaller than every 1/n simultaneously. The image is not open in the topology of Ej. I will solve this problem by assuming each simplex is contained in finitely many simplexes. The simplicial complex K, or the abstract complex K, is locally finite. I will make this assumption whenever K is embedded in real space, or Ej.
A point p belongs to a minimal simplex, which is part of finitely many simmplexes, hence every point p belongs to finitely many simplexes in K.
We have to be a bit careful when embedding K into real space in some way that is not homeomorphic to the standard embedding. Let the integers and reciprocal integers, plus and minus, be vertices, and let edges connect adjacent vertices. This is K embedded in the real line, and every point belongs to finitely many simplexes. In fact p belongs to at most 3. Pull the origin back to every simplex, and get nothing, or the origin itself. Each preimage is open, hence the origin should be open, yet it is closed. The embedding is not a valid representation of K. However, if we make the origin its own separate component, open and closed, the topology is correct again.
Note that L could be disconnected, even if K is connected. In fact, L could be the vertices of K, and nothing more.
Focus on a particular simplex V in L. The subcomplex K could contain V, or it could contain one or two faces of V, or it could contain a few edges, or a couple of vertices, or nothing at all. In each case, V∩L is closed. Therefore L is closed in K.