30
$\begingroup$

One distinct difference between axioms of topology and sigma algebra is the asymmetry between union and intersection; meaning topology is closed under finite intersections sigma-algebra closed under countable union. It is very clear mathematically but is there a way to think; so that we can define a geometric difference? In other words I want to have an intuitive idea in application of this objects.

$\endgroup$
  • 1
    $\begingroup$ The best way to think about the difference is to imagine some examples. The canonical example of a topology is the collection of open sets on $\mathbb{R}$, while the canonical example of a $\sigma$-algebra is the collection of Lebesgue measurable sets on $\mathbb{R}$. $\endgroup$ – Jim Belk Jun 18 '15 at 20:08
  • 1
    $\begingroup$ Another clear difference is that the $\sigma$-algebra demands that the complement of some set of the $\sigma$-algebra is also contained. A topology does not demand that $\endgroup$ – IceFire Jul 13 '16 at 11:32
13
$\begingroup$

I would like to mention that in An Epsilon of Room, remark 1.1.3, Tao states:

The notion of a measurable space (X, S) (and of a measurable function) is superficially similar to that of a topological space (X, F) (and of a continuous function); the topology F contains ∅ and X just as the σ-algebra S does, but is now closed under arbitrary unions and finite intersections, rather than countable unions, countable intersections, and complements. The two categories are linked to each other by the Borel algebra construction.

Later, in example 1.1.5:

given any collection F of sets on X we can define the σ-algebra B [ F ] generated by F , defined to be the intersection of all the σ-algebras containing F , or equivalently the coarsest algebra for which all sets in F are measurable. (This intersection is non-vacuous, since it will always involve the discrete σ-algebra 2^X). In particular, the open sets F of a topological space ( X, F ) generate a σ-algebra, known as the Borel σ-algebra of that space.

$\endgroup$
18
$\begingroup$

Your question is a little vague, but here is something to consider: Topology is normally discussed as its own subject while $\sigma$-algebras are typically just used as a tool in measure theory. One reason why finite intersections are needed in a topology is that it preserves what we think of as "openness" in a metric space. For instance, the finite intersection of any intervals of the form $(a,b) \subseteq \mathbb{R}$ still has the property of containing a ball around each point. This property is not shared with $\sigma$-algebras. For instance we can consider the countable intersection

$$ \bigcap_{n \in \mathbb{N}} \left(a - \frac{1}{n}, b+ \frac{1}{n} \right ) \;\; =\;\; [a,b] $$

which doesn't preserve this "openness" property we would like a topology to preserve. We can see that every neighborhood around either points $a$ or $b$ contain elements outside the interval $[a,b]$.

$\endgroup$
  • $\begingroup$ Good explanation. But sigma-algebra is closed under completion and finite unions. Would $A \cap B = \bar{\bar{A} \cup \bar{B}}$? So they are still equivalent? $\endgroup$ – Albert Chen Jun 4 '17 at 19:54
  • $\begingroup$ @AlbertChen I don't really understand your question. If the bar stands for complement then that equation is true. Sigma algebras are closed under arbitrary unions, arbitrary intersections and complements, while topologies are only closed under finite intersections, arbitrary unions, and must contain the empty and total sets. $\endgroup$ – Mnifldz Jun 4 '17 at 21:09
  • $\begingroup$ The little bar over $\cup$ is the complement of all. sigma-algebra also needs to include empty set. I think the difference only lies on whether the intersection is finite or countable? $\endgroup$ – Albert Chen Jun 5 '17 at 0:20
  • 1
    $\begingroup$ I'm sure there are a number of ways to define sigma algebras that are equivalent to each other, but the finite-vs-arbitrary intersections is a key difference as well as complements. A topology is in general not closed under complements. $\endgroup$ – Mnifldz Jun 5 '17 at 0:40
6
$\begingroup$

An easy way to get a feeling for this is to consider basic examples.

For example, let $X=\{1, 2, 3\}$.

A topological space $(X, 𝓸)$ could be for constructed by choosing for example $𝓸=\{∅,\{1, 2\},\{2\},\{2,3\},X\}$.

But this is as far from a $σ$-algebra as you can get since in fact no complement of any set in $𝓸$ is in 𝓸 except for $X$ and $∅$.

Have a look at some examples of topologies, some examples of $σ$-algebras and try to compare them. Start easy (like this) and move on to some harder ones and you will develop an intuition after hand.

$\endgroup$
4
$\begingroup$

The geometry of $\sigma$-algebras is in general badly understood. Proofs involving topologies often work directly on the topology, proving sets are open directly. The proof of Arzela-Ascoli, for example, works in two topologies and proves convergence directly. A great many proofs start with pick $U$ a neighborhood of $x$. Working with $\sigma$-algebras is somewhat more complicated. Often the approach is to build a sequence of approximations to the desired $\sigma$-algebra. Even the definition of the Borel algebra generated by some sets $\mathfrak{B}$ is either very abstract or build through approximations. I.E. the intersection of all $\sigma$-algebras containing $\mathfrak{B}$, or transfinite induction on stages of approximations to the full Borel algebra.

$\endgroup$
0
$\begingroup$

I just wanted to add a remarkable comment on this as I was reading through a great book on probability theory by Achim Klenke on p.8

Differences between topologies and sigma-algebra

In contrast with σ -algebras, topologies are closed under finite intersections only i.e. why the answer of the previous question applies as a case in point. Furthermore, the topologies are also closed under arbitrary unions while σ-algebras need not be closed under arbitrary unions, only under countable unions.

I hope this clarifies the matter

$\endgroup$
  • 2
    $\begingroup$ "Like sigma algebras" is wrong; $\sigma$-algebras need not be closed under arbitrary unions, only under countable unions. $\endgroup$ – Andreas Blass Mar 13 '18 at 13:48

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.