Let $X$ be a topological space and $x \in X$ a point. Let $\Omega$ be the set of open sets (viꝫ. the topology) of $X$, and $\Omega\_x$ the set of *germs* around $x$ of open sets, that is, $\Omega\_x = \Omega/{\sim}$ where $U\sim V$ means there exists a neighborhood $W$ of $x$ such that $U\cap W = V\cap W$. (Note that $U$ and $V$ need not be neighborhoods of $x$.)


Clearly, $\Omega$, partially ordered by inclusion, is a distributive lattice, with join and meet operations being given by $\cup$ and $\cap$. This structure passes to the quotient in the obvious way, so $\Omega\_x$ is a distributive lattice.


But $\Omega$ is more than a distributive lattice: it is a **frame** meaning that it has arbitrary joins (given by $\bigcup$), and that finite meets distribute over arbitrary joins. (A frame also has arbitrary meets, but they are not considered part of the structure and need not be preserved by homomorphisms.) Now this structure *does not* pass to the quotient $\Omega\_x$ in the sense that the canonical surjection $\Omega \to \Omega\_x$ does not, in general, define a frame structure on its target so that it is a frame homomorphism. (Indeed, it is easy to construct families $U\_i$ and $V\_i$ of open sets such that $U\_i \sim V\_i$ for every $i$ but $\bigcup\_i U\_i$ and $\bigcup\_i V\_i$ do not have the same germ, e.g., take $U\_i = \varnothing$ in $X = \mathbb{R}$ and $V\_i$ the complement of the closed interval $[-\frac{1}{i}, \frac{1}{i}]$ for $i\geq 1$ around $x=0$. But note that in this example, the join of the classes $[U\_i] = [V\_i]$ under $\sim$ still exists in $\Omega\_x$, and is $[\varnothing]$. So this counterexample does not answer the following question:)


**Question:** Is $\Omega\_x$ itself a frame (albeit not a quotient frame of $\Omega$)? In other words:


* Does the sup of an arbitrary family of germs of open sets exist?
* If so, do finite intersections (=meets) distribute over these arbitrary joins?


(In case the answer is negative, a counterexample with $X=\mathbb{R}$ would be most appreciated.)



Do germs of open sets around a point form a frame?
I just cooked up a counterexample in my head. Joseph Van Name beat me, but still here it is - to make things completely explicit. I hope I did not make any mistakes.


I will reason with closed sets and intersections; to get the answer with open sets and unions, just take the complement of everything.


So let $x = 0$, and let $V\_i = \{\frac{1}{2^i n} \;|\; n \in \mathbb{N}^\*\} \cup \{0\}$. Clearly the intersection of all $V\_i$ is $\{0\}$.


However, it is easy to find a family of closed sets $V'\_i$ that locally coincide with $V\_i$, but whose intersection is larger. Namely set $V'\_i = \{1, \frac{1}{2}, \ldots, \frac{1}{f(i)}\} \cup V\_i$, where $f: \mathbb{N}^\* \to \mathbb{N}^\*$ is some (let's say increasing) function. Then one easily sees that


$$\bigcap\_{i \in \mathbb{N}} V'\_i = \left\{ \frac{1}{k} \;|\; k \in \mathbb{N}^\* \text{ with } f(v\_2(k)) \geq k \right\} \cup \{0\}.$$


This set can be made arbitrarily "thick" by choosing a function $f$ that grows fast enough. So if there were a largest such set, it would have to be at least all of $V\_0$. But it is clear that the intersection of sets locally equal to $V\_i$ can never be all of $V\_0$.


