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

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<\epsilon<\min\x_k-x_(k+1).</p><p>Notice that in the interval [tex](x_k-\epsilon, x_k+\epsilon)` 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<\epsilon<|x_1-x_2|`, and if
`k=n` we only consider
`0<\epsilon<|x_n-x_(n-1)|`.

Then, all points of F are isolated.

Explanation: