0

Let $X$ be a topological space and denote $O(X)$ the set of open sets of $X$. Then I read that "$O(X)$ is a complete lattice since the union of any family of open sets is again open". I don't understand the justification why do we care only about union and why we don't care about the intersection of any family of open sets not being necessarely open.

palio
  • 11,466
  • $O(X)$ is in general a frame, hence a suplattice, and this implies that it has all meets as well as all joins. Infinite meets of opens are obtained as the interior of their intersection. – Naïm Camille Favier May 16 '24 at 17:44
  • See here: https://math.stackexchange.com/questions/3156278/does-the-frame-of-open-sets-in-a-topological-space-or-locale-really-have-all-mee – Naïm Camille Favier May 16 '24 at 17:45
  • 2
    If a partial order has all suprema, then it has all infima. The argument is that if $\mathcal{U}$ is a family of open sets then $\bigcup \mathcal{U} = \sup \mathcal{U}$. Hence it has all suprema, and so it has all infima as well (note you don't need to check that $O(X)$ is a lattice beforehand, like the answer below is trying to suggest). – Jakobian May 16 '24 at 18:21
  • Note that $\inf \mathcal{U} = \sup {x\in L : \forall_{y\in \mathcal{U}} x\leq y}$ verifies the definition of the infimum of $\mathcal{U}$ if your lattice $L$ has all the suprema. – Jakobian May 16 '24 at 18:27

1 Answers1

2

It is clear that $O(X)$ is a lattice. For the fact that it is complete lattice, take any $\mathcal U \subset O(X)$. Then show then $\bigcup \mathcal U$ is the least upper bound of the set $\mathcal U$ and $\bigcup \mathcal U'$ is greatest lower bound of the set $\mathcal U$ where $\mathcal U'$ is the subset of $O(X)$ whose members are contained in every member of $\mathcal U$.

This only uses the fact the arbitrary union of the open sets is open which your reference explicits mentions.

ash
  • 4,113
  • what do you mean by $\bigcup \mathcal U$ and least upper bound of what? remember that a complete lattice is a lattice $L$ in which each subset of $L$ has a least upper bound and a greatest lower bound. – palio May 16 '24 at 17:59
  • @palio "$\bigcup\mathcal{K}$" is standard notation for "the union of the elements of the set-of-sets $\mathcal{K}$." – Noah Schweber May 16 '24 at 18:04