2

So, as of recently, I've become interested in topology and, in trying to find a reasonable generalization of a topology, have stumbled across lattices. I've found that the join and meet operations of subsets $A \subset X$ can be neatly expressed as the supremum and infimum operations of elements of the lattice $V=\mathcal{P}(X)$, respectively.

Applying this to a topology $\mathcal{T}$on a space $X$, we can see that meet and join are well-defined on $\mathcal{T}$, and hence $\mathcal{T}$ itself becomes a lattice.

So far so understandable, but I got terribly confused when I read about complete lattices. Let $(V, \wedge, \vee)$ be a lattice. $V$ is said to be complete if and only if $\forall U \subset V \, \exists \, sup(U), \, inf(U) \in V$. In particular, it is enough to require the existence of one of these two, because $sup(U) = inf \lbrace x \in V: (\forall y \in U: x \leq y)\rbrace$.

My thought was this: Since topologies generalize so nicely to lattices, and every set of open subsets of $X$ has a supremum by the definition of topology, what should stop me from defining the infimum for any (not just a finite) set of open subsets of $X$ using the method above? $\bigcap_{j \in J} W_j = \bigcup\lbrace U \subset X \text{ open}, \forall j \in J: U \subset V_j \rbrace$

The worst that can happen is that there is no element in that set, hence the union is empty. But even in that case, by the definition of topology, the empty set is open in $X$, which makes me wonder why infinite intersections of open subsets are generally not defined to be open. Are they just not relevant? In "most of the cases", the intersection is going to be empty anyway. Or have I maybe misunderstood something about the definition of lattices? I'd be grateful if somebody took the time to explain this to me.

fklein
  • 135

2 Answers2

5

which makes me wonder why infinite intersections of open subsets are generally not defined to be open

It seems that your misunderstanding lies here: the infima of infinite collections of open subsets, in the lattice of open subsets of a topological space, are typically not the intersection: they are the interior of the intersection. For example, taking the Euclidean topology on $\mathbb R$, the intersection of the collection $\{(-\tfrac1n-1,1+ \tfrac1n)\}_{n \in \mathbb N_+}$ is $[-1, 1]$, but its infimum in the lattice of open subsets is $(-1, 1)$.

Generally, the relation between the finite meet ($\land$) and join ($\lor$) and their infinite counterparts is often not obvious.

It's very important that not all intersections of collections of open sets are also open, since this would lead to trivial structures on the majority of interesting topological spaces. (Exercise: every $T_1$ space for which this is true has the discrete topology.)

Mees de Vries
  • 26,947
  • So you are saying that the lattice of open subsets is complete if it is endowed with join as a supremum and the interior of the intersection as the infimum, whereas with the plain intersection as the infimum it would be not complete? What exactly breaks if I define the infimum to be the mere intersection? I'm sorry if this is too vague, I can't put it more precisely at the moment, as I seem to be having a mental blockade. – fklein May 03 '18 at 13:13
  • 2
    You can't define the general infimum as the intersection, because the infinite intersection is typically not open, and thus that operation lands you outside the lattice we are talking about (which is the lattice of opens). You could redefine the notion of topology to demand that all intersections of open subsets are open, but then you're just no longer talking about topology and you get much less interesting objects (in fact, you've more or less defined equivalence relations in a roundabout way). – Mees de Vries May 03 '18 at 13:17
2

Well, yes, the open sets form a complete lattice, with meet operation as you suggested, or equivalenty, $$\bigwedge_{i\in I} A_i=\mathrm{int}(\bigcap_{i\in I} A_i)\,. $$ This is just the intersection operation whenever $I$ is finite. But if $I$ is infinite, this might result in a smaller set than ordinary intersection.

In our prototype of Euclidean space $\Bbb R^n$, the intersection of concentric open balls with arbitrarily small radius will be just a one point set, and we don't want to regard that 'open'.

Berci
  • 90,745