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Norma-Jean [14]
3 years ago
7

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

Mathematics
1 answer:
Lorico [155]3 years ago
8 0

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:

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