# If $A$ is a finite set in $(\mathbb R, \mathfrak T_U)$ then $A' = \emptyset$.

If $A$ is a finite set in $(\mathbb R, \mathfrak T_U)$ then $A' = \emptyset$.

My knowledge: $\mathfrak T_U$ is the usual topology

$A'$ is the set of all limit points and my definition for this is: Let $(X, \mathfrak T)$ be a topological space with $A \subseteq X$. A point $x$ in $X$ is said to be a limit point of $A$ provided that every open set containing $x$ contains a point $A$ different from $x$.

Does the set $A$ have to be closed in order for the set of limit points to be empty? That is my thought right now? So I am thinking I need to be looking for counterexample for this false conjecture?

• Be careful with your definition. "....contains a point different from $x$" should be "contains a point * of $A$ * different from $x$. – WillO May 13 '15 at 13:51
• @Will0 thanks. I edited it. – user219081 May 13 '15 at 13:54

The claim is true. If $A$ is finite then we can list its elements as $A = \{a_1,a_2,\ldots,a_n\}$. WLOG let's assume $a_1<a_2<\ldots<a_n$. Now consider the open set $U = (a_1-1,a_n+1)$. It should be clear that $A\subset U$. Next use the fact that singletons are closed, hence a finite union of singletons is closed, like $\{a_1\}\cup \{a_2\}\ldots \cup \{a_n\}=A$. If $A$ is closed then there can be no limit points outside of $A$. However, notice that $\{a_2\}\cup \{a_3\}\ldots \cup \{a_n\}$ is closed which means $V = U\setminus (\{a_2\}\cup\{a_3\}\cup\ldots \cup \{a_n\})$ is open, $a_1 \in V$ and $$\left[V\setminus \{a_1\}\right]\cap A = \emptyset$$ so $a_1 \in A$ is not a limit point of $A$. We can repeat this process for each $a_i$ to show no point inside of $A$ is a limit point of $A$. Since there are no limit points inside of $A$ or outside of $A$, we conclude that $A' = \emptyset$.
If there is someone in $A'$, say $a$, prove that $A$ must contain infinitely mane elements by taking the balls:
$$D_\frac{1}{n}(a)$$
for $n\in \Bbb N$ (intuitively, since you always can find someone in $A$ that is as close as you want to $a$, there must be an infinite number of such elements in $A$).