I have a question about constructing a topology on a space $X$ starting from topologies defined on a family of subspaces $(X_i)_{i\in I}$ of $X$.

Assume that $X$ is a set and $(X_i)_{i\in I}$ is a collection of subsets of $X$. Assume also that for every $i\in I$, there is a topology $\mathcal{T}_i$ defined on $X_i$. We say that a topology $\mathcal{T}$ on $X$ is consistent with the family $(X_i,\mathcal{T}_i)_{i\in I}$ if the induced topology by $\mathcal{T}$ on $X_i$ is exactly $\mathcal{T}_i$ for every $i\in I$. Consistent topologies may or may not exist. And if they exist, there might be a unique consistent topology, or there might be more than one. It is easy to check the following facts:

  • If consistent topologies exist, then the finest consistent topology is given by $$\mathcal{T}=\{U\subset X:\; U\cap X_i\in\mathcal{T}_i,\;\forall i\in I\}.$$
  • A necessary condition for the existence of consistent topologies is the pairwise compatibility of the topologies $(\mathcal{T}_i)_{i\in I}$, i.e., it must be the case that for every $i,j\in I$, the topology induced by $\mathcal{T}_i$ on $X_i\cap X_j$ must be the same as the topology induced by $\mathcal{T}_j$ on $X_i\cap X_j$.

The above condition becomes sufficient if the spaces $(X_i)_{i\in I}$ are an increasing sequence, i.e., if $I=\mathbb{N}$ and $X_n\subset X_{n+1}$ for every $n\in\mathbb{N}$. Is the above necessary condition also sufficient in general for arbitrary collection of subspaces? if not, is there a known necessary and sufficient condition for the existence of consistent topologies?

Is there a necessary and sufficient condition for the uniqueness of the consistent topology (in case such one exists)?

Has this problem been studied? If yes, can someone refer me to a good reference?

  • $\begingroup$ Well-asked question! It might be possible to construct three subspaces+topologies, that satisfy the pairwise consistency you describe, but nevertheless are not (three-way) consistent. If it is possible, it's probably possible to do so on a very small space, like fewer than 10 points. $\endgroup$ – Greg Martin Apr 11 '17 at 18:08
  • $\begingroup$ have you looked at en.wikipedia.org/wiki/Direct_limit and en.wikipedia.org/wiki/Coherent_topology ? $\endgroup$ – Henno Brandsma Apr 12 '17 at 6:35
  • $\begingroup$ So the sequence condition can be generalised to being directed, e.g. which holds for most common cases. It would be nice if the directedness could be shown to be necessary. $\endgroup$ – Henno Brandsma Apr 12 '17 at 6:36

Not a complete answer, but some steps along the way and too long for a comment:

Consider the collection of sets that you have already identified as the finest induced topology: $$\mathcal T=\{U\subset X:\; U\cap X_i\in\mathcal T_i,\;\forall i\in I\}$$

It is fairly direct to show that regardless of the how the subspaces $X_i$ are chosen, $\mathcal T$ is a topology on $X$. If the $X_i$ cover $X$ and are severely in conflict with each other, then it may turn out that $\mathcal T$ is the trivial topology, but it is still a topology. And as you've noted, if there is any consistent topology, then this topology will be consistent.

So the question boils down to: when is this particular topology consistent? Is the condition $\mathcal T_i \mid_{X_i\cap X_j} = \mathcal T_j \mid_{X_i\cap X_j}$ for all $i,j$ sufficient?


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