Suppose $X$ is a simplicial space of dimension $M$ (i.e. all simplices above dimension $M$ are degenerate). The claim is:


$|X|$ is compact. iff $X\_n$ is compact for each $n$.


Suppose each $X\_n$ is compact. Then $|X|$ is by definition a quotient of a compact space (you don't have to include the simplices above dimension $M$ in the realization). I wonder, whether the converse is true.


Here is one motivating example. Equip the unit interval with the structure of a simplicial space in the following way: 


Let $X\_0$ be the Cantor-set and let the nondegenerate simplices in $X\_1$ are just all the intervals, that get removed in the construction of the cantor set.
One can regard $X\_1$ as a subspace of $[0;1]$ using the map (of sets) $X\_1\rightarrow [0;1]$ , that sends every point in the Cantor set to the corresponding point in the unit interval and that sends each of the small intervals to its barycenter. Equip $X\_1$ with the subspace topology using this map.


The geometric realization of this space is the unit interval, which is compact and $X\_0,X\_1$ are also compact ($X\_1$ is a closed subset of the unit interval).


This question arose in the context of this [question](https://mathoverflow.net/questions/19232/is-the-realization-of-a-proper-map-of-simplicial-spaces-proper). I realized, that I don't have a good criterium to say, when a subspace of the geometric realization of a simplicial space is compact. 



When is the realization of a simplicial space compact ?
Ok I checked the idea mentioned in the comments above.


As $X\_n$ is a closed subspace of $X\_M$ (use any degeneracy; it has a left inverse; composing both the other way round yields a projection and projections have closed images in the Hausdorff setting), it is enough to show, that $X\_M$ is compact.


So consider the "generalized midpoint map"


$i: X\_M\rightarrow \coprod\_{n\le M} X\_n\times \Delta^n\rightarrow |X|\qquad x\mapsto [x,c]$, where $c\in\Delta^M$ denotes the barycenter.


For example, if $M=2$ and $x$ is a nondegenerate one simplex in $X$, there are two degeneracies $s\_0,s\_1:X\_1\rightarrow X\_2$. Then one gets in |X|:


$i(s\_0(x))=[s\_0(x),\frac{1}{3},\frac{1}{3},\frac{1}{3}]=[x,\frac{2}{3},\frac{1}{3}]$


$i(s\_1(x))=[s\_1(x),\frac{1}{3},\frac{1}{3},\frac{1}{3}]=[x,\frac{1}{3},\frac{2}{3}]$


So in the realisation one really picks several midpoints of a simplex, that doesn't have maximal dimension.


But the map $i$ is still injective. Any simplex $x\in X\_M$ might be written in a unique way as $x=s(y)$, where $y\in X\_n$ is nondegenerate and $s$ is a degeneracy map (it corresponds to a surjective map $[M]\rightarrow [n]$ in $\Delta^{Op}$, $\Delta^{Op}$ can be identified with the category whose objects are the sets $\{0,\ldots,n\}$ and whose morphisms are nondecreasing maps).


Then $i(x)=[y,s(c)]=[y,\frac{s^{-1}(1)}{3},\ldots,\frac{s^{-1}(1)}{3}]$. Now the right side has normal form (There is a normal form for points in $|X|$. This can be found in every book about simplicial sets and it works the same way for simplicial spaces). Given any other $x'=s'(y')$ with $i(x)=i(x')$, we get


$[y,\frac{|s^{-1}(1)|}{M+1},\ldots,|\frac{s^{-1}(1)|}{M+1}]=[y',\frac{|s'^{-1}(1)|}{M+1},\ldots,\frac{|s'^{-1}(n)|}{M+1}]$. As this is in normal form, we get $y=y'$ and $|s'^{-1}(i)|=|s^{-1}(i)|$. But such nondecreasing maps are characterized by the size of the preimages, so $s=s'$ and hence $x=x'$. So the map $i$ is injective.


We still have to show, that $i$ is a homeo onto its image and that the image of $i$ is closed. Then $X\_M$ is a closed subspace of a compact space and hence compact. Both will follow from the next claim:


For any closed subset $A\subset X\_M$ the saturation $A'$ of $A\times \{c\}$ is still a closed subset of $\coprod\_{n\le M} X\_n\times \Delta^n$. This is an even worse calculation than the last one. One has to consider for each degeneracy map $s:X\_N\rightarrow X\_M$ the preimage $s^{-1}(A)\times \{s(c)\}$ and the set $B\_s$ of all points that can be simplified to one of those points. Then one can show, that $B\_s$ is closed and $A'=\bigcup\_s B\_s$. As there are only finitely many degeneracy maps (below dim $M$) this is a finite union and hence closed.


