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I been reading back and forth about lattices and topologies all day. And one thing I can't seem to get a good idea about is how a base would look in such a complete lattice. And what would be the main characteristics of this collection of sets? Would they be "connected" (not as in topological connected) or could the base be spread out on separate parts of the lattice?

Every point of the set have to belong to a least one subset and and if two sets shared a point then there would be another element in the lattice belonging to the base. This tells me that a base would like the lowest part of the lattice above the empty set shaped like a V not containing the center.

I done my best to be as precise as possible even and feel free to ask for better formulations. Note! This is not the lattice of topologies on $X$ but the complete lattice of open sets of a given topology

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Let $\mathscr{T}$ be a topology, and let $\mathscr{B}$ be a base for $\mathscr{T}$. This simply means that for each $U\in\mathscr{T}$ there is a $\mathscr{B}_U\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{B}_U$, which doesn’t tell us much about the ‘shape’ of $\mathscr{B}$. In particular, if $\mathscr{B}\subseteq\mathscr{B}\,'\subseteq\mathscr{T}$, then $\mathscr{B}\,'$ is also a base for $\mathscr{T}$.

One thing that can be said about a base $\mathscr{B}$ for $\mathscr{T}$ is that if $U$ is minimal in $\mathscr{T}\setminus\{\varnothing\}$ (i.e., if $U$ covers $\varnothing$ in the lattice sense), then $U\in\mathscr{B}$. For example, if $\langle X,\mathscr{T}\rangle$ is discrete, $\mathscr{B}\subseteq\mathscr{T}$ is a base for $\mathscr{T}$ iff $\mathscr{B}\supseteq\big\{\{x\}:x\in X\big\}$. It’s also clear that $\mathscr{B}$ must be dense in $\mathscr{T}\setminus\{\varnothing\}$ in the sense that for each $U\in\mathscr{T}$ there is a $B\in\mathscr{B}$ such that $B\subseteq U$, but of course this is not sufficient. (For example, the set of open intervals disjoint from $\Bbb Z$ is a dense subset of the Euclidean topology $\mathscr{E}$ on $\Bbb R$, but it’s obviously not a base for that topology.)

In general it’s very hard to see how a base sits in the lattice of open sets even when the base itself has a simple description; the base for of open intervals for the topology $\mathscr{E}$ is a good example of this. Still, in some very simple spaces it’s not hard. For $n\in\Bbb N$ let $I_n=\{k\in\Bbb N:k<n\}$, and let $F_n=\Bbb N\setminus I_n$. Let $\mathscr{T}_I=\{I_n:n\in\Bbb N\}\cup\{\Bbb N\}$ and $\mathscr{T}_F=\{F_n:n\in\Bbb N\}\cup\{\varnothing\}$; $\mathscr{T}_I$ and $\mathscr{T}_F$ are $T_0$ topologies on $\Bbb N$. As lattices they are very simple: both are linearly ordered by $\subseteq$. It’s not hard to see that each has a minimum base: $\mathscr{T}_I\setminus\{\varnothing,\Bbb N\}$ for $\mathscr{T}_I$ and $\mathscr{T}_F\setminus\{\varnothing\}$ for $\mathscr{T}_F$.

There is at least one interesting class of spaces that is defined by having a base that inherits a nice structure from the lattice of open sets: a space $\langle X,\mathscr{T}\rangle$ is non-Archimedean if it has a base $\mathscr{B}$ such that $\langle\mathscr{B},\supseteq\rangle$ is a tree. All ultrametrizable spaces are non-Archimedean in this sense. Apart from discrete spaces, the Cantor set is the most familiar example of an ultrametrizable space: it has a base $\mathscr{B}$ such that $\langle\mathscr{B},\supseteq\rangle$ is the complete binary tree of height $\omega$.

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The smallest open set is $\varnothing$. For each integer $k$ consider the collection of open sets $\left(k - \frac{1}{n}, k + \frac{1}{n}\right)$ where $n > 1$. For each $k$ the greatest lower bound for the family is $\varnothing$. Given open intervals in two such collections, say $\left(k_{1} -\frac{1}{n_{1}}, k_{1} + \frac{1}{n_{1}}\right)$ and $\left(k_{2} -\frac{1}{n_{2}}, k_{2} + \frac{1}{n_{2}}\right)$ they will have empty intersection as well. The structure above $\varnothing$ is more complicated than is suggested by the letter "V".

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  • $\begingroup$ Alright, I had a feeling that it wasnt that easy. I had a hard time imagine an example like this. I guess this structure is to complicated to say anything about it which hold in general $\endgroup$ – user123124 Jul 29 '13 at 18:34

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