Can all open sets in R^n be expressed as countable union of open cubes?

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In summary, the problem requires knowing this fact: for notational purposes, we can express an open set as the union of open intervals. However, if the number of points in the set is uncountably infinite, then the union is overcountably many.
  • #1
CantorSet
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Hi everyone,

I came across a problem that requires knowing this fact.

But can any open set in [tex] R^n [/tex] be expressed as the countable union of "cubes". That is subsets of the form [tex] (a_1,b_1) \times ... \times (a_n, b_n) [/tex].
 
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  • #2
Hi CantorSet! :smile:

The answer is yes! For notational purposes, I'll do this is one dimension, but multiple dimensions is quite analogous.

So, take G open, then around every [itex]x\in G[/itex], we can find an open interval such that

[tex]x\in ]a,b[\subseteq G[/tex]

By shrinking the interval, we can assume the endpoints to be rational. Thus

[tex]G=\bigcup\{]a,b[~\vert~a,b\in \mathbb{Q},~]a,b[\subseteq G\}[/tex]

We can even take the intervals/cubes to be disjoint, but that's somewhat more difficult...
 
  • #3
Thanks for the response, micromass!

I see that we can express any open set as the union of open intervals. But can we express it as a countable union of open intervals. What if the number of points in G is uncountably infinite?

You mention that we can even use disjoint intervals. Would this guarantee the union is over countably many?
 
  • #4
CantorSet said:
What if the number of points in G is uncountably infinite?

Doesn't the same proof apply regardless on the number of elements in G?
 
  • #5
CantorSet, what micromass didn't mention is that when you write an open set as a union of open cubes with rational endpoints, then the result is a union of countably many sets. This is because there are only countably many cubes with rational endpoints (something you can prove for yourself probably). In his procedure, you have your open set [itex]G[/itex], a point [itex]x \in G[/itex], and then you choose an open interval, [itex]x \in ]a_x, b_x[ \in G[/itex]. Then you shrink [itex]]a_x, b_x[[/itex] to an interval with rational endpoints, say [itex]]p_x, q_x[[/itex]. If [itex]G[/itex] is uncountable, and you do this for every [itex]x \in G[/itex], then since there are only countably many intervals with rational endpoints, it means that for many different [itex]x \neq y \in G[/itex], you'll have ended up choosing the same interval: [itex]p_x = p_y,\ q_x = q_y[/itex].
 
  • #6
Got it. Thanks for the help, guys.
 

1. What is an open set?

An open set is a set of points in a topological space that does not contain its boundary. In other words, for any point in an open set, there exists a small enough neighborhood around that point that is completely contained within the set. This is in contrast to a closed set, which contains its boundary points.

2. What is R^n?

R^n refers to n-dimensional Euclidean space, which is a mathematical space where each point is represented by a set of n real numbers. In other words, R^n is the set of all possible n-tuples of real numbers.

3. What does it mean for an open set to be expressed as a countable union of open cubes?

This means that the open set can be broken down into a countable number of smaller open sets, each of which is in the shape of a cube. The union of all of these smaller open cubes forms the original open set. This is a useful concept in mathematics for simplifying geometric and topological problems.

4. Why is it important to determine if all open sets in R^n can be expressed as a countable union of open cubes?

This question is important because it relates to the concept of measurability in mathematics. If all open sets in R^n can be expressed as a countable union of open cubes, then this means that they have a measure, or size, that can be calculated. This has implications in areas such as analysis and probability theory.

5. Has this question been answered definitively?

No, this question is still an open problem in mathematics and has not been definitively answered. It is known that all open sets in R^1 (one-dimensional Euclidean space) can be expressed as a countable union of open intervals, but it is still unknown if this is true for higher dimensions. Mathematicians continue to work on this problem and have made progress in certain cases, but a general solution has not yet been found.

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