# Every finite subset of the real numbers is closed

Prove: Every finite subset of $\mathbb{R}$ is closed

definition of closed: A set $A$ is closed if it contains all it accumulation or limit points.

definition of accumulation point: Let $A$ be a subset of $\mathbb{R}$. A point $p\in \mathbb{R}$ is an accumulation or limit point if and only if every open set $G$ containing $p$ contains a point of $A$ different from $p$.

proof: Let $A$ be a finite subset of $\mathbb{R}$ with elements $a_{1},a_{2},\ldots,a_{n}$ where each $a_{i}\in \mathbb{R}$, $i = 1,2,\ldots,n$.

I am lost where to go from here, I could say that there is an accumulation point in $A$, but I am confused what I do next.

• Can you prove it when $|A| = 1$? Commented Mar 11, 2015 at 13:21
• No, I dont think I could lol Commented Mar 11, 2015 at 13:23
• Just to make sure, what definition of accumulation or limit point are you using? Commented Mar 11, 2015 at 13:25

Let $$A=\{x_1,\ldots, x_n\}$$. We can express $$A$$ as $$\bigcup_{i=1}^n\{x_i\}$$. Each singleton is closed so the union of finitely many closed sets is still closed.

(We know that singletons are closed since their complement is open. For $$c\in\mathbb{R}\setminus\{x\}$$, simply choose $$\epsilon=|x-c|$$ and note that $$x\notin B_{\epsilon}(c)$$. So $$B_{\epsilon}(c)$$ is not a subset of {x}. This means that $$B_{\epsilon}(c)\subset\mathbb{R}\setminus\{x\}$$. So $$\mathbb{R}\setminus\{x\}$$ is open and so its complement $$\{x\}$$ must be closed.)

• @MorganWeiss: I just added it in. Commented Mar 11, 2015 at 14:04

We will show $\mathbb{R} \setminus \{x_1,x_2,\dotsc,x_N\}$ is open.

Let $y \in \mathbb{R} \setminus \{x_1,x_2,\dotsc,x_N\}$.

Let $r(i) = d(y,x_i)$

Let $r'= \min \{ r(i) \}$

Let $r = r'/4$.

$B(y,r)$ is a subset of $\mathbb{R} \setminus \{x_1,x_2,\dotsc,x_N\}$

Hence $y$ is an interior point of $\mathbb{R} \setminus \{x_1,x_2,\dotsc,x_N\}$

Hence $\mathbb{R} \setminus \{x_1,x_2,\dotsc,x_N\}$ is open.

Hence $\{x_1,x_2,\dotsc,x_N\}$ is closed.

Hint: If $a_1,\dotsc,a_n$ are distinct points of $A$, then choose $\varepsilon=\frac{1}{3}\min\{|a_i-a_j|:1\le i,j\le n,i\neq j\}$. Show that $B(a_i,\varepsilon)\cap A=\{a_i\}$, for each $1\le i\le n$. Hence all points of $A$ are isolated.

If you take any sequence $(s_i)$ in A that tends to a limit it suffices to show that the limit is also in A.

If a sequence does tend to a limit that means that eventually it gets close (and stays close) to that limit and close here is defined as close as you like. If we suppose that a given sequence does have a limit and that limit is called $l$ then formally:

for any $\epsilon > 0$ there must be a natural number $N$ so that for every $n \in \mathbb{N}$ with $n > N$

$$|s_n - l| < \epsilon$$

Suppose that $l$ is not a member of the set $A$ then each element of $A$ is some positive distance from $l$. Let $d$ be a quarter the smallest of these distances.

However, if we try and get to with $d$ of the limit we find that we are always too far away whichever of the $a_i$ we choose.

It must be that any sequence tending to a limit tends to one of the elements $a_i$ and it can only do this because eventually every element of the sequence is $a_i$

If you can use continuous functions, here is an argument:

• If $f:\mathbb R \to \mathbb R$ is continuous then its zero set is closed. The zero set of $f$ is $\{ x \in \mathbb R : f(x)=0 \}$.

• Every finite set $\{ a_{1},a_{2},\ldots,a_{n} \}$ is the zero set of the continuous function $f(x)=(x-a_{1})(x-a_{2})\cdots(x-a_{n})$ and so is closed.