# Miscellaneous pointset topology and measure theory

1. Apr 6, 2004

### marcus

In a quantum gravity discussion ("Chunkymorphism" thread) some issues of basic topology and measure theory came up. Might be fun to have a thread for such discussions.

for instance the statement was made, apparently concerning the real line (or perhaps more generally) that a countable set must consist entirely of isolated points

this is a purely topological matter, separate from measure theory (although it came up when the talk was also about some basic measure theory issues as well)

At least two of us IIRC made the point that this is mistaken, a countable subset of the reals may have no isolated points at all. Indeed the rational numbers are an example: they are countable and have no isolated points.

This brings up basic definitions in topology, like what is an isolated point, about which we could have a refresher thread if people want.

It seemed to me after I (for one) had mentioned the rationals and said you could have a countable set with no isolated points, that I was being asked "well, what does that have to do with measure theory? nothing, right?"
Right. It's purely a pointset topology thing.

However some measure theory questions came up at the same time.

Very often in measure theory you get sets of measure zero arising as exceptional sets and some statements were made or implied about sets of measure zero. It wasnt always clear what was being proposed but IIRC statements like

sets of measure zero cannot be connected
sets of measure zero can be at most countably infinite
sets of measure zero consist of isolated points

It might be fun to consider statements like these: either to find counterexamples so as to see why they arent true in general or perhaps to
figure out in which special circumstances they might be true!

there are some more or less standard measures on Rn which
people usually assume are meant, nothing said to the contrary, but
one is free to invent unusual measures---can you think of a topology on the reals and a measure such that any set of measure zero must consist solely
of isolated points?

Another basic question, a kind of beginning exercise, might be to prove that the rationals do in fact have measure zero----with an ordinary measure on the real line.

I was hoping someone else would start this thread. It is a utility for discussing basic topology and measure theory if people want. And if they dont thats fine with me!

2. Apr 6, 2004

### matt grime

A set of measure zero must be countable: no, the cantor set is your counter example there.

As for zero measure of any countable subset of R: let (x_i) be an enumeration of the countable subset. Let U_i be the open ball of radius e/(2^i) centred on x_i, then the union U_i contains all the x_i and has measure at most e, e was arbitrary.

For a measure you don't need a topology, you need a sigma algebra (which can be generated by a topology, admittedly, but does create serious differences).

3. Apr 6, 2004

### marcus

your proof of countable implies zero measure (ordinary Lebesgue on the line one may assume) is just what I had in mind except you said it more concisely

about measure zero does not imply countable, what I had in mind was
the x axis in the plane (or the unit interval in the unit square). Judging from your reply I think we see eye-to-eye

It is not at all sure that people will want to discuss basic questions like this
but if a desire emerges for some kind of tutorial thread like this
then I would nominate you to be in charge. you seem to know the stuff
and get things across concisely

I will see if I can come up with some more questions or maybe some will come in from the others

4. Apr 6, 2004

### matt grime

I thought you wanted non-trivial constructions. Any n-1 dimensional object in R^n has measure zero. That's just a simlpe volume argument, not even measure theoretic. The examples you give are also trivially connected sets too.

5. Apr 6, 2004

### NateTG

Consider that for any measure

$$m(\{\frac{1}{n}: n \in \mathbb{N}\}) = m(\bigcup_{n \in \mathbb{N}} \{\frac{1}{n}\}) = \sum_{n \in \mathbb{N}} m(\{\frac{1}{n}\})$$
So a measure on the reals can only have the property that you desire if there are some points with non-zero measure. Now, by a similar argument, you can see that the set of points with individual measure zero contains none of its own limit points. This leads to a whole slew of undesirable properties.

6. Apr 6, 2004

### marcus

I confess to extreme laziness, but your example of the Cantor set is
interesting. Not everyone may be familiar with it and I'm curious to know how you would describe it to a stranger, and say why it's uncountable and measure zero.

the recipe in halmos says take the unit interval and chop out the middle third which leaves two intervals and chop out the middle third of each of those
and so on

so I guess (2/3)n goes to zero qed
and pretty clearly uncountable too putting ternary expansion into correspondence with binary expansion of irrational numbers in unit interval

hey that was not too hard

what are some other things from basic topology and measuretheory
anything especially cool come to mind?

7. Apr 6, 2004

### matt grime

Most of the cool things might mean lots of repeated application of the axiom of choice. Pretty much everything behaves as you want and the only way to get pathological behaviour is to construct things using the axiom of choice and transfinite induction. I quite like a proof using the fact that the there is no uncountable subset of R that is well ordered by < to show that a continuous function on a compact subset of R is uniformly continuous. Then there are functions that are continuous on a countable subset of [0,1] and also discontinuous almost everywhere. And who can ignore the Banach Tarski paradox (strictly speaking that requires unmeasurable sets, but you only get those if there are measurable ones). Or two functions on [0,1] may have the same fourier series yet not be the same (they'll differ by something zero in the L^2 norm), something that even fourier didnt' realize.

8. Sep 20, 2005

### benorin

card{Lebesgue measurable subsets of R} > card(R)

card{Borel subsets of R} = card(R)

This result can be found in Section 5, Exercise (9) in Halmos's book MEASURE THEORY

these are quoted from this web page (it long, so Crtl+F 'halmos' to find it)