Question

A subset $S \subseteq \mathbb{R}$ is called open if for every $x \in S$, there exists a real number $\epsilon > 0$ such that $(x-\epsilon,x \epsilon) \subseteq S$. A subset $T \subseteq \mathbb{R}$ is called closed if $\mathbb{R} \setminus T$ is open. (a) Show that an open interval is open and that a closed interval is closed. (b) Show that $\emptyset$ and $\mathbb{R}$ are the only subsets of $\mathbb{R}$ that are both open and closed. (This is very hard. At least try to convince yourself that $\emptyset$ and $\mathbb{R}$ are both open and closed. Showing there is no other set that is both open and closed is quite difficult.) (c) Show that an arbitrary union of open intervals is open. (d) Show that an arbitrary union of closed intervals need not be closed. (Hint: in light of the definition of closed, this is the same thing as showing that an arbitrary intersection of open intervals need not be open.)

Answer 1

Answer:

Step-by-step explanation:

REcall that given sets S,T if we want to prove that $S\subseteqT$, then we need to prove that  for all x that is in S, it is in T.

a) Let (a,b) be a non empty interval and $x\in (a,b)$. Then a<x <b. Let $\varepsilon = \min{\min\{b-x, x-a\}}{2}$ Consider $y \in (x-\varepsilon,x+\varepsilon)$, then

$y$ and

$y>x-\varepsilon>x-(x-a) = a$.

Then $y\in (a,b)$. Hence, (a,b) is open.

Consider the complement of [a,b] (i.e $(a,b)^c$).

Then, it is beyond the scope of this answer that

$(a,b)^c = (-\infty,a) \cup (b,\infty)$.

Suppose that $x\in (a,b)^c$ and without loss of generality, suppose that x < a (The same technique applies when x>b). Take $\varepsilon = \frac{a-x}{2}$ and consider $y \in (x-\varepsilon,x+\varepsilon)$. Then

$y$

Then y \in (-\infty,a). Applying the same argument when $x \in (b,\infty)$ we find that [a,b] is closed.

c) Let I be an arbitrary set of indexes and consider the family of open sets $\{A_i\}_{i\in I}. Let [tex]B = \bigcup_{i\in I}A_i$. Let $x \in B$. Then, by detinition there exists an index $i_0$ such that $x\in A_{i_0}$. Since $A_{i_0}$ is open, there exists a positive epsilon such that $(x-\varepsilon,x+\varepsilon)\subseteq A_{i_0} \subseteq B$. Hence, B is open.

d).  Consider the following family of open intervals $A_n = (a-\frac{1}{n},b+\frac{1}{n})$. Let $B = \bigcap_{n=1}^{\infty}A_n$. It can be easily proven that

$B =[a,b]$. Then, the intersection of open intervals doesn't need to be an open interval.

b) Note that for every $x \in \mathbb{R}$ and for every $\varepsilon>0$ we have that $(x-\varepsilon,x+\varepsilon)\subseteq \mathbb{R}$. This means that $\mathbb{R}$ is open, and by definition, $\emptyset$ is closed.

Note that the definition of an open set is the following:

if for every $x \in S$, there exists a real number $\epsilon > 0$ such that $(x-\epsilon,x \epsilon) \subseteq S$. This means that if a set is not open, there exists an element x in the set S such that for a especific value of epsilon, the subset (x-epsilon, x + epsilon) is not a proper subset of S. Suppose that S is the empty set, and suppose that S is not open. This would imply, by the definition, that there exists an element in S that contradicts the definition of an open set. But, since S is the empty set, it is a contradiction that it has an element. Hence, it must be true that S (i.e the empty set) is open. Hence $\mathbb{R}$ is also closed, by definition. If you want to prove that this are the only sets that satisfy this property, you must prove that $\mathbb{R}$ is a connected set (this is a topic in topology)