Let $X$ be a set, $X = \bigcup_{i\in I} X_i$, where $X_i$ is equipped with the topology $\mathcal{T}_i$. The coherent topology (also called weak topology) is the topology on $X$ defined as $$\mathcal{T} = \{U\subset X: U\cap X_i\in \mathcal{T}_i, \forall i\in I\},$$ which is the finest topology on $X$ such that $\mathcal{T}|_{X_i}\subset \mathcal{T}_i$ for all $i\in I$, where $\mathcal{T}|_{X_i}$ is the subspace topology of $\mathcal{T}$ on $X_i$.

Now, I'm interested in the case where $\mathcal{T}|_{X_i} = \mathcal{T}_i$ for all $i\in I$. It is easy to see that a necessary condition for that is $\mathcal{T}_i|_{X_i\cap X_j} = \mathcal{T}_j|_{X_i\cap X_j}$ for all $i,j\in I$, which means that $\{\mathcal{T}_i\}$ agree on the common part of $\{X_i\}$.

But is this also sufficient? If $\mathcal{T}_i|_{X_i\cap X_j} = \mathcal{T}_j|_{X_i\cap X_j}$ for all $i,j\in I$, can we say $\mathcal{T}|_{X_i} = \mathcal{T}_i$ for all $i\in I$, that is: for every $i_0\in I$, $U_{i_0}\in \mathcal{T}_{i_0}$, there exists $U\subset X$ such that:

$\bullet$ for every $i\in I$, $U\cap X_i\in \mathcal{T}_i$;

$\bullet$ $U\cap X_{i_0} = U_{i_0}$?

I cannot prove it or find a counterexample. Thank you in advance for any help.

NOTE: We don't have the condition $X_i\cap X_j$ is closed in $X_i$,$X_j$ for all $i,j\in I$ like in this question. This seems to restrictive and is not a necessary condition.

EDIT: An easy example is given by Eric Wofsey below, thanks a lot! But feel free to post anything if anyone have some ideas to characterize the necessary and sufficient conditions!


1 Answer 1


Here is a simple counterexample. Let $X=\{0,1,2\}$ and $X_i=X\setminus\{i\}$ for $i=0,1,2$. Give $X_i$ the topology where $\{i-1\}$ is open and $\{i+1\}$ is not (taken mod 3). These trivially satisfy your pairwise compatibility condition since the intersections are singletons, but the coherent topology is the indiscrete topology.

(If you want an example with Hausdorff spaces, you can easily modify the idea of this one. For instance, you can replace each of $0,1,$ and $2$ with an infinite sets $A_0,A_1,A_2$. Then give $X_i=X\setminus A_i$ a topology where $A_{i-1}$ and $A_{i+1}$ are each discrete, $A_{i+1}$ is closed, and $A_{i-1}$ is dense. Then in the coherent topology on $X$, each $A_i$ will be dense, and so not closed in the subspace topology on $X_{i-1}$.)

  • $\begingroup$ I like this example, thanks a lot! ($\{i-i\}$ should be $\{i-1\}$, right?) $\endgroup$ Commented Mar 21, 2022 at 6:37
  • 1
    $\begingroup$ Just out of curiosity: in the second example, is the coherent topology on $X$ itself Hausdorff? I found that topology very hard to describe, apart from the fact that every nonempty open set of $X$ must intersect all of $A_0$, $A_1$, and $A_2$. $\endgroup$ Commented Mar 21, 2022 at 10:29
  • $\begingroup$ That is an interesting question. It's easy to see the topology on $X$ is $T_1$, but I don't know about Hausdorff. I wouldn't be surprised if whether it is Hausdorff depends on the exact topologies chosen on each $X_i$. $\endgroup$ Commented Mar 21, 2022 at 14:11
  • $\begingroup$ Of course. Thanks! $\endgroup$ Commented Mar 22, 2022 at 10:12

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