Question

Let FS R be a finite set. (a) Prove that F is closed (b) Prove that every point in F is an isolated point.

Answer 1

Answer:

a) Suppose that F is ordered in ascending order: $F = \{x_1,\ldots, x_n\}$. Then, the complement of F can be written as

$F^c = (-\infty,x_1)\cup (x_1,x_2)\cup (x_2,x_3)\cup \cdots \cup (x_{n-1}, x_{n})\cup (x_n,+\infty)$

which is the union of a finite number of open intervals, then $F^c$ is an open set. Thus, F is a closed subset of the real numbers.

b) Take an arbitrary element of F, let us say $x_k$. Now, choose a real number $\epsilon$ such that

$0$ there are not other element of F, because $\epsilon$ is less that the minimum distance between $x_k$ and its neighbors.

In case that $k=1$ we only consider $0$, and if $k=n$ we only consider $0$.

Then, all points of F are isolated.

Step-by-step explanation: