Proof that every open set in R^n is the union of an at most countable collection of...?

mm1239

New member
I can't figure this 2 part question out- I think the first part involves using open balls but I'm not sure. It's straightforward to prove this for R^1 with disjoint segments but I'm a little lost as to what to do when it comes to R^n.

(1) Prove that every open set in R^n is the union of an at most countable collection of disjoint segments
Hint: You need to replace disjoint segments with the appropriate objects (which I'm thinking are open balls)

(2) Generalize the statement to a separable topological space
Hint: you may need some extra assumption

Any help would be much appreciated! The more detail, the better- thanks!

caffeinemachine

Well-known member
MHB Math Scholar
I can't figure this 2 part question out- I think the first part involves using open balls but I'm not sure. It's straightforward to prove this for R^1 with disjoint segments but I'm a little lost as to what to do when it comes to R^n.

(1) Prove that every open set in R^n is the union of an at most countable collection of disjoint segments
Hint: You need to replace disjoint segments with the appropriate objects (which I'm thinking are open balls)

(2) Generalize the statement to a separable topological space
Hint: you may need some extra assumption

Any help would be much appreciated! The more detail, the better- thanks!
I can help you on part (1) but I have no idea what a separable topological space is.

Lets prove the result for $\mathbb{R}^2$. It can be easily generalized. Let $x_1<x_2$ and $y_1<y_2$ be four real numbers.

Consider the region $D(x_1,x_2;y_1,y_2)=\{(x,y) \in \mathbb{R}^2:x_1<x<x_2,y_1<y<y_2\}$. This $D(x_1,x_2;y_1,y_2)$ will be called open box with end points $x_1,x_2$ and $y_1,y_2$.
Now let $U$ be any open set in $\mathbb{R}^2$ .

Let $x \in U$. Its easy to see that there exists an open box containing $x$ with all four of its end points rational which is contained in $U$. In other words there exist rational numbers $x_1<x_2,y_1<y_2$ such that $x \in D(x_1,x_2;y_1,y_2) \subseteq U$.

We choose one such open box for each $x$ in $U$ and denote it by $D(x)$.

Note that $\displaystyle{\bigcup_{x \in U}{} D(x)}=U$.

Also there are only countably many such open boxes (this easily follows from the countability of rationals).

So we have countably many open boxes whose union is $U$.

But we had to show for open balls and not open boxes. This doesn't seem to be difficult. We could have chosen our open boxes $D(x)$ such that for each $x \in U$ there exist an open ball $B(x)$ such that $x \in D(x) \subset B(x) \subseteq U$.

The desired result follows.

MHB Math Scholar

Deveno

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MHB Math Scholar
(2) is easy once you have (1):

if X is a separable space, there is a sequence of elements {xi} such that every non-empty open set U of X contains at least one element of the sequence.

(such a sequence is called a countable dense subset).

now if X is a metric space, we can derive that:

X has a countable base.

for each point xi in {xi}, we can define the family of open balls {B(xi,ε)} where ε is rational. this is a countable family, since Q is countable. the set of all such families (for all i in N) is a countable union of countably many sets, so is itself countable.

now suppose U is an open set in X containing x. if x = xi for some i in our sequence, we have B(xi,ε) ⊆ U for some sufficiently small ε, which we may take to be rational (for if ε is irrational there is some rational ε' between 0 and ε, since the rationals are dense in the reals, and we may use this rational ε' instead).

on the other hand if x is a point of U NOT in our sequence, then for any ball B(x,ε), we have some xi in the open set B(x,ε)∩U, since the xi are dense in X. in particular, if we choose rational ε so small that B(x,ε) ⊆ U, then we have some xi in B(x,ε/2).

so consider B(xi,ε/2). this is a member of our countable family that contains x, and is contained in U. thus every point x of U is contained in some element of:

B = {U{B(xi,ε)}: i in N, ε in Q}. which means that any open set U is a union of elements of B, that is: B is a countable base for X.

(i DO hope i've done this right: it's been quite a while).

mm1239

New member
Thanks a lot!

I'm still not sure how to do this for R^n though- can you help? Following from your proof for R^2, I did this for R^1 using open intervals and that's as far as I got. Would it help at all to define some interval with endpoints negative and positive infinity?

I can help you on part (1) but I have no idea what a separable topological space is.

Lets prove the result for $\mathbb{R}^2$. It can be easily generalized. Let $x_1<x_2$ and $y_1<y_2$ be four real numbers.

Consider the region $D(x_1,x_2;y_1,y_2)=\{(x,y) \in \mathbb{R}^2:x_1<x<x_2,y_1<y<y_2\}$. This $D(x_1,x_2;y_1,y_2)$ will be called open box with end points $x_1,x_2$ and $y_1,y_2$.
Now let $U$ be any open set in $\mathbb{R}^2$ .

Let $x \in U$. Its easy to see that there exists an open box containing $x$ with all four of its end points rational which is contained in $U$. In other words there exist rational numbers $x_1<x_2,y_1<y_2$ such that $x \in D(x_1,x_2;y_1,y_2) \subseteq U$.

We choose one such open box for each $x$ in $U$ and denote it by $D(x)$.

Note that $\displaystyle{\bigcup_{x \in U}{} D(x)}=U$.

Also there are only countably many such open boxes (this easily follows from the countability of rationals).

So we have countably many open boxes whose union is $U$.

But we had to show for open balls and not open boxes. This doesn't seem to be difficult. We could have chosen our open boxes $D(x)$ such that for each $x \in U$ there exist an open ball $B(x)$ such that $x \in D(x) \subset B(x) \subseteq U$.

The desired result follows.

Deveno

Well-known member
MHB Math Scholar
the basic idea here is we want to "cover" an uncountable thing (the real numbers) with countably many things (open sets).

now the total number of possible open sets in the real numbers is....large. is even larger than the total number of real numbers, because sets can be quite arbitrary, and just limiting them to OPEN sets (while it does cut down on the kinds of sets we get a little) doesn't really help much.

so, in this state of affairs, it seems unlikely we would ever be able to "get a handle" on how to even describe open sets at all, except in very vague terms.

but the real numbers have some special properties that come to our rescue.

1) the rational numbers are DENSE in the reals.

what does this mean, "spatially"? it means no matter WHERE on the real line you are, there's always a rational number, nearby, waiting to take your call. how near? within any epsilon you choose! since there's an infinite number of INTEGERS, and they just keep getting bigger, eventually there's a integer bigger than 1/ε (which is finite), and for this integer (which we always call N, for some reason), 1/N < 1/ε.....and...it's rational. and if you're waiting at "real number x" there's some rational (actually a WHOLE LOTTA rationals) between you and x-1/N, or x+1/N.

so if we can describe something in terms of rationals, it's often "good enough to use" instead of describing that thing in terms of (arbitrary) real things.

in the case at hand, it means we can describe open sets in terms of unions of RATIONAL endpoint intervals (a,b). (this is good, it means we can NAME a and b. most real numbers are impossible to be named....they haven't even been computed yet!).

so we don't need EVERY open interval to describe the open sets, we just need ones with rational endpoints (and we can specifiy these by specifying the end points). that is:

(a,b) = {x in R: a < x < b}

which brings me to special property number 2 of the real numbers:

2) they have an ORDER. this order allows us to make the following definition:

|x| = x, if x > 0
|x| = x, if x = 0
|x| = -x, if x < 0

this defines a NORM on the real numbers, which then allows us to define a concept of "distance":

d(x,y) = |y-x|

************

now in Rn, we still have a dense subset (Qn), but we don't have the order anymore:

is (3,4) > (4,3)? uh.......

but it turns out all is not lost, we STILL have a norm (the norm induced by the dot product...yay pythagoras!), and so we still have a distance function:

d(x,y) = ||y-x|| = √(y-x).(y-x)

this norm gives us "open balls"...which is ONE way to generalize open intervals.

but we also have that R = RxRx...xR (n times). so we can try to define open sets "one coordinate at a time". this gives us "open boxes" (and it turns out there is a norm called the "box norm", that measures "how big our boxes are", by keeping the largest dimension length (in this norm, in R2, an "open ball" is a square with sides of length 2ε)).

both sets (bases) give us the same collection of open sets when we take all the unions of them. because it turns out that in every open box, there is an open ball that fits inside it, and in every open ball, there is an open box that fits inside IT. and all we really need (since we are going to take infinitely many unions anyways, perhaps), is a little bit of "blobby space" surrounding a point x, it doesn't really matter what the "shape of the blob" is (triangles would work in the plane, and tetrahedrons would work in 3-space. there are people called simplicial homological algebraists who appreciate this).

as long as we can describe our blob by specifying a countable number of "somethings" from a set of "somethings" that are themselves countable, we get a countable base. the technical term for such a space (set) is: "second-countable".

this is a way to work with "uncountably infinite things" in such a way that it's "manageable". and the point of this whole exercise is to convince you that:

Rn is second countable.

**********

the topology on Rn that caffeinemachine is using is called "the product topology" (well, actually the "box topology", but for a finite number of factors, these are the same, and Rn has n factors, so finite). this is a BASIC construction, and is well worth getting intimately acquainted with. the topology you are thinking of (the open balls) is called the "metric topology" and is also a useful thing, but not as widely applicable (not every space has a metric, nor is even metrizable (that is, we could PUT a metric on it if it's not there already)). as i pointed out before, it turns out that these generate the same collection of open sets in Rn, often just called simply "the usual topology".

this is entirely analogous with the fact the the same vector space might have two different basis sets. taking unions in topology is what we do because we don't have addition. in metric spaces, "changing epsilon" is what we do because we don't have scalar multiplication. mathematicians are pretty-much "one trick ponies", we have a few good ideas, and then we make analogies until they break.

Plato

Well-known member
MHB Math Helper
(1) Prove that every open set in R^n is the union of an at most countable collection of disjoint segments
Hint: You need to replace disjoint segments with the appropriate objects (which I'm thinking are open balls)

(2) Generalize the statement to a separable topological space
Hint: you may need some extra assumption

I must have miss something in all these replies. I am sure I have.
Which of the replies deal with the word disjoint?

caffeinemachine

Well-known member
MHB Math Scholar
I must have miss something in all these replies. I am sure I have.
Which of the replies deal with the word disjoint?
Well in the "Hint" it says you may need to replace 'disjoint segments' with 'open balls'. Anyways disjoint segments don't make sense when $\mathbb{R}^n$ is talked about.

Plato

Well-known member
MHB Math Helper
Anyways disjoint segments don't make sense when $\mathbb{R}^n$ is talked about.
Of course it does not make sense when $\mathbb{R}^n$ is talked about.

But would not a true generalization be "Any open in $$\mathcal{R}^n$$ is the union of countably many disjoint open sets"?

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caffeinemachine

Well-known member
MHB Math Scholar
Of course it does not make sense when $\mathbb{R}^n$ is talked about.

But would not a true generalization be "Any open in $$\mathcal{R}^n$$ is the union of countably many disjoint open sets"?
Well that doesn't seem to be true to me. Take $X=\mathbb{R}$. Now $X$ can't be expressed at a union of disjoint open segments except in the trivial way, that is when you take $\mathbb{R}$ itself as the (only) open segment whose union is $X$.
This is because $\mathbb{R}$ is connected.

Plato

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MHB Math Helper
Well that doesn't seem to be true to me. Take $X=\mathbb{R}$. Now $X$ can't be expressed at a union of disjoint open segments except in the trivial way, that is when you take $\mathbb{R}$ itself as the (only) open segment whose union is $X$.
This is because $\mathbb{R}$ is connected.
Well the trivial way works. So it is true.

caffeinemachine

Well-known member
MHB Math Scholar
Well the trivial way works. So it is true.
That worked because I took the open set as $\mathbb{R}$ itself. Consider another example. Take $X=\mathbb{R}^2$ and the open set $U$ as any open box say $U=\{(x,y):-1<x<1,-1<y<1\}$. $U$ can't be expressed a a (countable) union of disjoint open balls.