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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.)

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    $\begingroup$ Perhaps I am being dense, but isn't $\sim$ trivial? Given $x \in U$ and $x \in V$, take $W = U \cap V$, which is a neighborhood of $x$, to conclude that $U \cap W = V \cap W$ and hence $U \sim V$? $\endgroup$ Commented Mar 11, 2023 at 11:51
  • $\begingroup$ @AndrejBauer I'm not assuming $x$ belongs to the open sets ($U,V$) in question. Indeed, all those containing $x$ are mapped to the top element of $\Omega_x$, but there are many different classes not containing $x$. I will try to rephrase to clarify. $\endgroup$
    – Gro-Tsen
    Commented Mar 11, 2023 at 13:32
  • $\begingroup$ @AndrejBauer I changed “germs of open sets around $x$” to “germs around $x$ of open sets”: was that the cause of the confusion? $\endgroup$
    – Gro-Tsen
    Commented Mar 11, 2023 at 13:58
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    $\begingroup$ I added a remark which would have made it clear for me. $\endgroup$ Commented Mar 11, 2023 at 14:20
  • $\begingroup$ One could also say "germs of open sets near $x$"... $\endgroup$ Commented Apr 17, 2023 at 20:02

2 Answers 2

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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 $V_0$.

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  • $\begingroup$ Thanks! I like your answer better not just because it is more explicit and self-contained, but also because it satisfies the requirement at the end of my question of providing a counterexample with $X=\mathbb{R}$ (although in reality Joseph Van Name's answer can probably also be made such). $\endgroup$
    – Gro-Tsen
    Commented Mar 11, 2023 at 22:13
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This is generally not a frame nor is it a complete lattice. For example, consider $\omega+1$ which is the one-point compactification of the countable discrete space $\omega$. Then the set of all germs in $\Omega_\omega$ which do not contain $\omega$ is just the Boolean algebra $P(\omega)/\text{Fin}$ which is not complete (here $\text{Fin}$ is the ideal of finite subsets of $\omega$); the completion of $P(\omega)/\text{Fin}$ has size $2^{2^{\aleph_0}}$ since $\omega$ has an almost disjoint family $p$ of size continuum, and the mapping $\iota:P(p)\rightarrow\overline{P(\omega)/\text{Fin}}$ defined by $\iota(R)=\bigvee R$ is an injective Boolean algebra homomorphism.

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