Prove: If $A$ is a finite subset of real numbers, then the derived set $A'$ of $A$ is empty
Definition: Let $A$ be a subset of $\mathbb{R}$. A point $p\in \mathbb{R}$ is an accumulation point 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 $a_{i}\in \mathbb{R}$, $i = 1,2,\ldots,n$. Since $A\subset \mathbb{R}$, there is an accumulation point $p\in \mathbb{R}$ of $A$. Since, $p$ is an accumulation point of $A$, then there exists an open set $G$ containing $p$ that contains a point $q$ of $A$ different from $p$, i.e. $p\neq q$. Therefore, $G$ open, $p\in G \Rightarrow A\cap(G \setminus {p}) = \emptyset$, i.e. $A' = \emptyset$.
I am not sure if I am right, any suggestions would be greatly appreciated.
Best Answer
I didn't actually understand your proof. You want to prove that such set $A'$ is empty but in the proof you assume it is not empty... Are you trying to use contrapositive or some kind of absurd argument?
Anyway, the proof is rather simple:
Let $p\in \mathbb{R}$. We'll show that $p\notin A'$, studying two separate cases:
If $p\notin A$, take $r=\min\{|p-a_1|,\dots,|p-a_n|\}$. Then $B(p,r)=\{x \in \mathbb{R}; |p-x|<r\}$ is an open set containing $p$ and do not contains any point of $A$. This means that the definition of accumulation point fails por $p$. Therefore, $p\notin A'$
On the other hand, suppose $p\in A$. Then $p=a_i$ for some $i$. We just need a little adjustment of the argument oused before. Now take $r=\min\{|p-a_1|,\dots,|p-a_{i-1}|,|p-a_{i+1}|,\dots,|p-a_n|\}$. Then $B(p,r)$ is an open set containing $p$. Actually, $B(p,r)\cap A=\{p\}$ which means that this open set contains no other point of $A$. Again, the definition of accumulation point fails for $p$ and $p\notin A'$.
Therefore, no point of $\mathbb{R}$ lies in $A'$, which means that $A'=\emptyset$