Closed set representation as union of closed intervals

[dimitri151]

There the well known theorem that every open set (I'm talking about R here with standard topology) is the union of disjoint open intervals. Now, looking at the geometry, it seems that between any two adjacent open intervals which are in the union constituting our open set there is a closed interval. Therefore the union of all these disjoint closed sets that are between the open intervals constitute a closed set. Since every closed set is the complement of an open set, and every open set is the countable union of disjoint open intervals, every closed set is the countable union of disjoint closed intervals.

Any errors?

 

[micromass]

So, how would that work for the Cantor set?

 

[disregardthat]

every closed set is the countable union of disjoint closed intervals.

Any errors?

Yes, this is in fact not true. It can without much difficulty be proven that R (the real numbers in its standard typology), by definition being a closed set in its topology, cannot be written as a countable union of disjoint closed (finite) intervals.

A quick sum-up of the proof: assume on the contrary that there is such a countable union. Give the set of closed intervals in this union the order where one interval A is larger than another B, if any element in A is larger than any element in B. The topology induced is easily proved to be hausdorff, and to satisfy the least upper bound property. Hence any interval in this topology is compact (see Munkres Thm 27.1). Furthermore, it is easily proved that no singleton set is an open set in this topology. Hence for any interval in this topology, it will be compact, hausdorff and without singleton open sets, and is hence uncountable (see Munkres Thm 27.7). But this is of course our desired contradiction.

Without trying to derail this thread too much, I find it an interesting question as to whether R can be covered with a countable union of (at least two) non-empty disjoint closed sets. Any finite such union is of course not possible, since such a union would give us a separation of R. But for a countably infinite union I am not too sure.

 

[dimitri151]

I know it's not true. But where is the gap in the reasoning?
There's no way to do it with the cantor set since it's nowhere dense, and you couldn't do it with singletons since it's uncountable.

 

[disregardthat]

For starters, there are no adjacent (meaning that no open intervals come between I suppose) open intervals in a covering of R of disjoint open intervals. If you do not mean this, you must define what you mean by two disjoint open intervals in such a union being adjacent.

EDIT: Sorry, I did not realize that you did not require the intervals to be finite.

 

[jbunniii]

I see at least two problems:

1. A union of infinitely many closed sets is not necessarily closed. Consider the subset of R consisting of the points {1/n} for all positive integers n. This set is the countable union of singletons (each of which is closed), but the union is not closed because it does not contain the limit point 0.

2. You presume that it's possible, given one of the open intervals in the disjoint union, to identify its neighbors ("adjacent intervals"). This is not necessarily possible. Consider the disjoint union of intervals of the form

I_n = (1/(n+1), 1/n)

for all positive integers n, along with the interval I_0 = (-infinity, 0).

Then which of the I_n is adjacent to I_0? Answer: none of them. So your algorithm for forming the union of closed sets is ill-defined.

 

[dimitri151]

Thanks for the replies. I figured it out today while at work. There's generaly no adjacent interval. Cantor set is best example that comes to mind.

Then if an open set has to be a finite union of open intervals for the closed set that is its complement to be a finite union of closed intervals.