2
$\begingroup$

As I learned in the beginning of Topology, if I have two topologies $\tau_1, \tau_2$ on a space $X$ such that $$ \tau_1 \subseteq \tau_2 $$ then $\tau_2$ is called finer then $\tau_1$ and $\tau_1$ is called coarser then $\tau_2$. Roughly speaking "finer" means has more elements, and "coarser" means has less elements (which makes intuitively sense cause the notion of "closeness of points" axiomatizied by a topology is finer).

Now I read about open covers and refinements. The definitions are here. Now a cover is finer if it actually contains less open sets, and coarser if it contains more, am I right? So here the meaning of finer is totatlly different? (maybe intuitively finer means for a point it could be more accurately said in which open set it falls, and if there are more the more "fuzzy" the "location" of the point is). Am I right, and maybe is there some connections between these two notions of finer/coarser?

$\endgroup$
  • $\begingroup$ The refinement property for covers is not that different necessarily. The covering sets themselves are "finer" (that is smaller) than the original sets, but there may be more (because several smaller sets are needed to replace a bigger set) or less (because a single small set may replace many big sets). However, the topology induced on the covered set by the finer cover (as subbasis) is a finer topology than that by the coarser cover. $\endgroup$ – Hagen von Eitzen Jul 10 '13 at 14:41
1
$\begingroup$

It is not true that a finer cover necessarily contains more open sets than a coarser one. Consider $\Bbb Z$ with the discrete topology: $\mathscr{U}=\{\Bbb Z\}$ is an open cover of $\Bbb Z$ with one member, $\mathscr{V}=\big\{\{n\}:n\in\Bbb Z\big\}$ is an open cover of $\Bbb Z$ with (countably) infinitely many members, yet $\mathscr{V}$ is a refinement of $\mathscr{U}$ and therefore finer than $\mathscr{U}$. In the other direction, $\mathscr{W}=\wp(\Bbb Z)\setminus\{\varnothing\}$ is an uncountable open cover of $\Bbb Z$, and $\mathscr{U}$ and $\mathscr{V}$ are both refinements of $\mathscr{W}$, so a finer cover can have fewer members. This notion of finer and coarser really has nothing at all to do with the cardinalities of the covers.

In both contexts the intuition behind the terms finer and coarser is that a finer topology or cover chops the space up more finely. If $\tau_1\subseteq\tau_2$, where $\tau_1$ and $\tau_2$ are topologies on some set $X$, then $\tau_2$ chops up $X$ into all of the (open) pieces that $\tau_1$ does and possibly more besides. In this case that actually does imply that $|\tau_2|\ge|\tau_1|$, but that cardinality inequality is a byproduct, not the basis for the terminology. Similarly, if $\mathscr{V}$ is a refinement of $\mathscr{U}$, then any member of $\mathscr{V}$ containing some $x\in X$ is at least as small as some member of $\mathscr{U}$: if $x\in V\in\mathscr{V}$, then there is a $U\in\mathscr{U}$ such that $x\in V\subseteq U$.

There are differences, of course: a finer topology always has all of the open sets of the coarser topology, and possibly more besides, while a refinement need not contain any member of the original cover. But in both cases the ‘finer’ object does in some meaningful sense divide up the underlying set in a more fine-grained fashion.

$\endgroup$
0
$\begingroup$

citing from wikipedia given two covers $\cal V$ and $\cal U$, $\cal V$ is finer than $\cal U$ if for every open set $U\in \cal U$ contains some other open set $V\in \cal V$ thus for example take covers of $\Bbb R$: $\cal U = \{\Bbb R\}$ and $\cal V = \cal T_\Bbb R$ (the topology of $\Bbb R$ is a cover of $\Bbb R$) then $\cal V $ is finer then $\cal U$ and contains more open sets then $\cal U$.

On the other extreme consider the covers $\cal V =\{ (-2,2), (\Bbb R\setminus [-1,1])\}$ and $\cal U = \{(-j,j)\}_{j=1}^{\infty}$ again we have $\cal V$ is finer than $\cal U$ because $(-2,2)\in \cal V$ and is a subset of every $U\in\cal U$

So the definitions can agree, the moral is that what matters when dealing with covers is the size rather then the number.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.