Will give brainliest for the first and the correct

Ответ:

SkyDoki

21.11.2022

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

$y<x+\varepsilon < x+b-x =b$ and

$yx-\varepsilonx-(x-a) = a$ .

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

$y<x+\varepsilon < x+(a-x) = a$

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 $\varepsilon0$ 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)

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