Answer:
a) Suppose that F is ordered in ascending order: [tex]F = \{x_1,\ldots, x_n\}[/tex]. Then, the complement of F can be written as
[tex] F^c = (-\infty,x_1)\cup (x_1,x_2)\cup (x_2,x_3)\cup \cdots \cup (x_{n-1}, x_{n})\cup (x_n,+\infty)[/tex]
which is the union of a finite number of open intervals, then [tex]F^c[/tex] is an open set. Thus, F is a closed subset of the real numbers.
b) Take an arbitrary element of F, let us say [tex]x_k[/tex]. Now, choose a real number [tex]\epsilon[/tex] such that
[tex] 0<\epsilon<\min\{|x_k-x_{k-1}|, |x_k-x_{k+1}|\}.
Notice that in the interval [tex](x_k-\epsilon, x_k+\epsilon) [/tex] there are not other element of F, because [tex]\epsilon[/tex] is less that the minimum distance between [tex]x_k[/tex] and its neighbors.
In case that [tex]k=1[/tex] we only consider [tex]0<\epsilon<|x_1-x_2|[/tex], and if [tex]k=n[/tex] we only consider [tex]0<\epsilon<|x_n-x_{n-1}|[/tex].
Then, all points of F are isolated.
Step-by-step explanation:
