# Math Help - Open Sets on the Real Line

1. Originally Posted by Plato
There are only three basic forms of connected open subsets of real numbers: $(-\infty,a),~(a,b),~\&~(a,\infty)$.
The theorem means: Any open set of real numbers can be written as a countable union of pairwise disjoint basic open connected sets.

The set $[1,\infty )\backslash \mathbb{Z}^ + = \bigcup\limits_{n = 1}^\infty {\left( {n,n + 1} \right)}$ therefore $[1,\infty )\backslash \mathbb{Z}^ +$ is an open set.

The set $(2,3)$ is a basic open set by definition.
If you delete the integers from [1,infinity) you are left with the sets (0,1), (1,2), (2,3) ..... and you are left with the trivial observation that the union of a collection of disjoint open sets is the union of the same collection of disjoint open sets.

Personally, I vote for Kolmogorov. I think the whole point is that an open set can be expressed as the union of disjoint sets not necessarily open (on the real line). Even Taylor in his theorem does not refer specifically to E1, E2... as open sets even though conditions (i)-(iii) do. That's what makes his Theorem so maddening.

2. Originally Posted by Hartlw
Personally, I vote for Kolmogorov. I think the whole point is that an open set can be expressed as the union of disjoint sets (on the real line).
I simply gave an example. I have used that book as a textbook also.

3. Originally Posted by Plato
I simply gave an example. I have used that book as a textbook also.

"I think the whole point is that an open set can be expressed as the union of disjoint sets not necessarily open (on the real line). Even Taylor in his theorem does not refer specifically to E1, E2... as open sets even though conditions (i)-(iii) do. That's what makes his Theorem so maddening."

OK, to be even more specific, I think Taylor is wrong and youi don't. That's probably as good a place to end it as any.

4. Originally Posted by Hartlw
"THE STRUCTURE OF OPEN SETS IN R
Theorem 2-3 IV. If E is an open set in R such that $E\neq 0$, and $E\neq R$, E can be expressed as a countable (possibly finite) union of disjoint sets E1, E2,..., where each En is a non-empty open set of one of the following types:
(i) a finite open interval {x : a<x<b},
(ii) a left-semiinfinite open interval {x : x<b}, where $b\in R$,
(iii) a right-semiinfinite open interval {x : a<x}, where $a\in R$."
Originally Posted by Hartlw
"I think the whole point is that an open set can be expressed as the union of disjoint sets not necessarily open (on the real line). Even Taylor in his theorem does not refer specifically to E1, E2... as open sets even though conditions (i)-(iii) do.
OK, to be even more specific, I think Taylor is wrong and youi don't. That's probably as good a place to end it as any.
How can you say Taylor does not say the $E_n$ are open? You posted it yourself that he does.

5. Originally Posted by Plato
How can you say Taylor does not say the $E_n$ are open? You posted it yourself that he does.
You are right, I missed that. Technically, I copied it from Taylors book. OK, I don't understand Taylor's version.

For reference I include Kolmogorov's version again which makes sense to me (note reference below takes you directly to the theorem, you don't have to hunt for it):

6. Shilov: "Every open set on the real line is a finite or countable union of nonintersecting open intervals."

Rudin: "Every open set in R1 is the union of an at most countable collection of disjoint segments."

Its still a tie.

By the way, you can get almost any used book cheap at abe.com (hope its OK to mention this. If not, say so and I will edit it out).

EDIT: Just as a reminder: (0,1) and [1,2) are disjoint (non-intersecting).

7. Shilov (clearly) and Taylor (obtuseley) state:
Every open set S on the real line is a countable union of nonintersecting open intervals.
Example:
S= $(1,3)\cup{(3,8)}\cup(9,12)}$
period.

Kolmogorov and Rudin state:
Every open set S on the real line is (can be expressed as) a countable union of nonintersecting intervals.
Example:
S= $(1,3)\cup{(3,5)}\cup{[5,8)}\cup(9,12)}$
or
S= $(1,2)\cup{[2,3)}\cup{(3,8)}\cup{(9,11]}\cup{(11,12)}$
or.......

The examples make everything clear. Shilov's proof became crystal clear once I understood what the theorem was saying. Taylor and Klomogorov's proofs were'nt clear, and Rudin's was an excercise.

What confused me all along was that I thiought Taylor was saying, for example,
If S=(1,3) it can be expressed as $(1,2)\cup{(2,3)}$

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