1

In metric space, we have the notion of distance, which enables us to have a general idea on the "size" of an open set, say, by measuring its diameter. If we now have 2 open sets $A,B$ in a topological space, we tends have the impression that $A$ is "smaller than" $B$ if $A\subset B$. But what if we are given 2 disjoint open set? Is there a general way for us to compare "size" of open set in a topological space? (Say, $A$ is smaller than $B$, or $A$ and $B$ have the similar size) How can we get the impression of the size of an open set in a topological space? Thanks.

P.s. Cardinality does not count as a solution because it does not rely on the topology anyway.

JSCB
  • 13,456
  • 15
  • 59
  • 123
  • 1
    Not in general. – Brian M. Scott May 03 '16 at 14:47
  • 2
    You have a poset, but nothing more. – Michael Burr May 03 '16 at 14:49
  • What about Borel measures? – Augustin May 03 '16 at 14:51
  • @Augustin : What makes you think every topological space admits a (nontrivial) measure on its Borel algebra? Random example: Say a space has no notrivial compact sets, then (strong) inner regularity would imply all sets have measure zero, so the only Radon measure on such a space is the trivial one, which would contradict strict positivity (measure of a nonempty open set is positive). There are a number of properties one would like a measure to have, but the topology of the underlying space can make such properties fail. See http://math.stackexchange.com/questions/43601 for more. – Eric Towers May 03 '16 at 15:07

2 Answers2

7

The comments and answer here seem to be a little depressing - topology can still measure interesting properties of open sets! Michael Burr is probably right in that you can't hope for anything more than a poset, but let's at least construct an interesting one!

Let's try the following: $$\text{Define }\;A \le B\text{ if there is an embedding }\;A \stackrel{f}\hookrightarrow B.$$

We can check that this gives us a partial ordering on topological spaces, hence also topological subspaces. In addition, if $A \subset B$ are subspaces of $X$ then $A \le B$. (Note that if $A \le B$ it may not be the case that $A \subset B$.)

This is a purely topological characterization, and it turns out to be interesting, because we can use topological tools to study it!

One immediate question that you might have is, "Can you produce two spaces $X, Y$ such that $X \lneq Y$?" It turns out this is pretty easy: take $X = \mathbb R^2$, and take $Y = \mathbb{RP}^2$. Or if we want spaces that are incomparable elements consider $\mathbb{RP}^2$ and $S^2$. All these results rely on interesting algebraic topology.

So it seems this is an interesting way to characterize "size" in terms of topological complexity.

A few fun exercises for interested readers, in roughly increasing order of difficulty:

  1. Given a space $X$, what subspace(s) is/are minimal in this order? What set is a maximal element?
  2. Construct a space $X$ and subsets $A_i \subset X$ such that $$A_1 \lneq A_2 \lneq \cdots \lneq X.$$
  3. Give a version of the previous construction where $X$ and the $A_i$ are all connected.
  4. Prove that $\mathbb{RP}^2$ and $S^2$ are incomparable elements.
  5. Prove that $\mathbb{R}^2 \lneq \mathbb{RP}^2$.
  6. Construct a topological space $X$ and subspaces $\{A_r\}_{r \in \mathbb{R}_{\ge 0}}$ with $|X| = \mathfrak{c}$ (the cardinality of the continuum) and $A_r \lneq A_{r'}$ for all $r < r'$. (I'm not 100% sure my answer here is correct, or that this is possible.)
Thurmond
  • 1,103
  • I enjoyed your answer. – Michael Burr May 03 '16 at 15:53
  • This has the negative property that any two open intervals in $\Bbb{R}$, $A, B$ satisfy both $A \leq B$ and $B \leq A$, so in the simplest setting these do not capture "size". The same thing happens for analytic disks in $\Bbb{C}$ (i.e., any two non-empty simply connected open subsets are the same "size").

    Note also that the question asks for two open set in the same ambient space (which you have weakly ignored)...

     – Eric Towers May 04 '16 at 16:40
  • Of course it has this drawback. But if you have disjoint open sets (as in the OP's question) then the only topological tools you have to study them come from maps between them, and that's what my answer addresses. I can only try to provide a positive answer to the question that's been asked :). – Thurmond May 04 '16 at 17:04
1

You cannot. One of the consequences of "rubber sheet geometry" is that distance, hence area, is a fluid thing.

M.C. Escher represented hyperbolic geometry many times in his Circle Limit works, for instance Circle Limit IV. The referenced image is a representation of an infinite metric space. Each demon is congruent (same area) and each angel is congruent in the original space. However, topology allows us to (continuously) magnify and demagnify various parts of a space with impunity. Consequently, Escher's work is topologically equivalent to the original tiling of a hyperbolic space. If we were to imagine translating the space (or what is equivalent, translating the point at the center of our magnifying window into the space), the largest figures would shrink as they moved to the side and smaller figures would grow as they moved to the center of the view.

To borrow words of Yoda, in topology "size matters not."

Eric Towers
  • 67,037