Exercises about open and closed sets and intervals.

Question

1) Prove the union of a family of open intervals which intersection is not empty is an open interval.

2) Prove every open set of real numbers can be expressed (in just one way) as a numerable union of open intervals disjoint for pairs.

3) If an open interval is the union of two disjoint open sets then one of them is empty and the other is the whole interval.

4) Let A, B be subsets of real numbers then:

1. $\mathrm{int}(A \cap B) = \mathrm{int} (A) \cap \mathrm{int}( B)$
2. $\mathrm{int} (A) \cup \mathrm{int}(B) \subset \mathrm{int}(A \cup B) $

5) Let $A$ be a subset of $X$. And let $\partial A$ be the boundary of $A$. Prove:

$$X = \mathrm{int}(A) \cup \mathrm{int}(X \setminus A) \cup \partial A$$

So, for the first i tried a direct proof, i took two diferent points(a and b) of the union and then try to proof all the points between them are in the union too. We know a and b are in one of the open intervals, let A and B be those intervals, so a belongs A and b belongs B. Suppose a < b . Let c be any point such that a < c < b. I must prove c belongs to the union. Then i think maybe i could use the hypotesis about the no empty intersection of the family but i dont know how. I am stucked.

For the second the hint is use the first exercise.

For thethird i tried contradiction so i supossed no one is empty and no one is the whole open interval. Since the union of the two open sets is the whole open interval then neither of the two open sets can be bigger than the interval, so the two open sets are two open disjoint no empty sets smaller than the interval. If we call them A and B we can suppsed that every element of A is smaller tha every element of B so the infimum(we can call it R) of B must be the supremum of A ad then must be bigger than any element of A. So R dont belongs to A and dont belongs to B. But belongs to the interval and this is the contradction we are lookng for. But i dont know if this is right and how to write it formally. For 4 and 5 i am clueless.

Answer 1

$X = int(A) \cup int(X \setminus A) \cup int(\partial A)$ is false.
Within the reals, let A = (-$\infty$,0).
$int(A) \cup int(X \setminus A) \cup int(\partial A)$
= (-$\infty$,0) $\cup$ int([0,$\infty$)) $\cup$ int($\partial$ {0}) = (-$\infty$,0) $\cup$ (0,$\infty$) $\cup$ empty set
which is not the set of real numbers.

Do you see what the correct statement should be?

Answer 2

For 1), in general the union of open sets is open (this is a basic axiom of topology) and the union of connected sets with a point in common is connected (see here Union of connected subsets is connected if intersection is nonempty for proof). So the union of open intervals with a point in common must be an open interval.

For 2), let $U\subset\mathbb R$ be open and assume it is bounded (to avoid having to use intervals of the form $(-\infty, b)$ and $(a,\infty)$. For each $x\in U$ let $(x_a,x_b)$ be the union of all open intervals in $U$ containing $x$. These intervals are disjoint as if for example $x,y\in U$ with $y_a<x_a<y_b<x_b$ then we would simply have the larger open interval $(y_a,x_b)$. Each point in $U$ is contained within some interval, and moreover, each interval contains at least one rational number, and each rational number is contained within exactly one interval. It follows that the union of these intervals is a countable union which is equal to $U$.

3) follows from an open interval being a connected set.

For 4), let $x\in \mathrm{int}(A\cap B)$, then there exists a neighborhood $U$ of $x$ contained in $A\cap B$. But $U$ is also a neighborhood of $x$ contained in $A$ so that $x\in \mathrm{int}(A)$, and similarly $x\in\mathrm{int}(B)$. Now let $y\in \mathrm{int}(A)\cap\mathrm{int}(B)$. Then there exist neighborhoods $V$ and $W$ of $y$ contained in $A$ and $B$, respectively. Hence $V\cap W$ is a neighborhood of $y$ contained in $A\cap B$, so that $y\in\mathrm{int}(A\cap B)$. Part 2 uses very similar reasoning.

For 5), I will leave some hints. Show that:

• $\mathrm{int}A\cap\partial A=\varnothing$
• $\mathrm{int}(X\setminus A)\cap\partial A=\varnothing$
• $\mathrm{int}A\cap\partial \mathrm{int}(X\setminus A)=\varnothing$
• and finally that $\mathrm{int}A\cup\mathrm{int}(X\setminus A)\cup\partial A$.