Geometric realization of an abstract simplical complex

Let K=(V,S)K=(V,S) be an abstract simplical complex which is localy finite – each vertex belongs to finitely many simplices δ∈S\delta \in S. We define its geometric realization
|K|⊂l2(V)|K| \subset l^{2}(V) where l2(V)=H l^{2}(V) = H is a hilbert space with basisis (ev)(e_v) for each vertex (so H is a set of functions from f:V→Rf: V \rightarrow R such that (∫f2)12(\int_{}f^2)^{\frac{1}{2}} (Lebesque integral with counting measure) is finite .
Now to each simplex δ∈S;δ={v0,…,vn}\delta \in S ;\delta = \left\{ v_0,…,v_n\right\} we assign its geometric realization – a subset of H compromising all convex combinations of v0,…,vnv_0,…,v_n and we define |K||K| to be a sum of all geometric realizations of its simplices. Now my question is :
|K||K| is supposed to be close in H, right?
So let’s take abstract simplical complex K, where V={v1,..,vn,…}V= \left\{ v_1,..,v_n,… \right\} and maximal simplices are {v1};{v2,v3};{v3,v4,v5};… \left\{ v_1\right\}; \left\{ v_2,v_3\right\}; \left\{ v_3,v_4,v_5\right\};…
So geometric realization of KK would be just separable sum of spaces homeomorphic to standard singular complexes of all dimensions.
So we can always find point (for each natural n ) x∈|K|⊂Hx \in |K| \subset H such that all his coordinates are zero except for nn coordinates which are all equal to 1n\frac{1}{n}, then |x−0||x – 0 | = 1n12\frac{1}{n^{\frac{1}{2}}}
and thus 00 belongs to the closure of |K| |K| where is the flaw of my reasoning?

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Just of curiosity and ignorance: what is the context of this and what is the definition of this geometric realization? The usual one is (in a sense, by definition) just the adjoint functor to the functor that assigns to a topological space the corresponding singular simplicial set.
– Alejo
Oct 20 at 21:05

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