# The derived set $A'$ is closed by sequences

I am trying to prove that $$A'$$ the set of limit points is closed using different methods. For this I am using sequences and I wrote two answers that I don't know if it is right.

Let $$(x_n) \subset A'$$ such that $$x_n \to x$$. Need to show $$x \in A'.$$

Since $$x_n \to x$$, $$\forall \epsilon >0$$, $$\forall n > N \implies d(x_n,x) <\epsilon.$$ That is the ball $$B_\epsilon(x)$$ contains infinitely many points of $$(x_n)$$ (for $$\forall n > N)$$), that is $$(x_n)_{n = N}^{\infty} \subset B_\epsilon(x)$$. So $$B_\epsilon (x) \cap A \neq \emptyset$$.

Then the set $$B_\epsilon (x) - \{x\} \cap A$$ is also nonempty, a set containing infinitely many points with one point removed. So $$x$$ is a limit point of $$A$$.

Same set up as (1), but this time I think we can use some facts about $$x_n$$ being a limit point. Meaning

$$B_{\epsilon_n}(x_n)-\{x_n\} \cap A \neq \emptyset$$

My idea is to take $$\epsilon_n = 1/n$$ and argue that $$\cap B_{1/n}(x_n) -\{x_n\} \cap A$$ somehow gets to the answer (so somehow argue $$\cap B_{1/n}(x_n) -\{x_n\} = B_\epsilon (x) - \{x\}$$ I am not too sure if this will go anywhere.

• In Answer 1 you are assuming that $x_n$'s are distinct. $\{x_n: n>N\}$ can be a finite set. – Kavi Rama Murthy Aug 18 '19 at 5:48
• What is your definition of closed? Because there’s a fairly easy answer if a closed set is defined to be the complement of an open set. – Robert Shore Aug 18 '19 at 5:48
• I know we can just go and show complement is open, but I want to prove using other definitions. – Hawk Aug 18 '19 at 5:50

... That is the ball $$B_\epsilon(x)$$ contains infinitely many points of $$(x_n)$$ (for $$\forall n > N)$$), that is $$(x_n)_{n = N}^{\infty} \subset B_\epsilon(x)$$. So $$B_\epsilon (x) \cap A \neq \emptyset$$.

Your stated conclusion does not immediately follow because your sequence elements $$(x_n)_{n = 1}^{\infty}$$ are from $$A'$$ and not (necessarily) from $$A$$. So you can only conclude $$B_\epsilon (x) \cap A' \neq \emptyset$$ so far and not necessarily $$B_\epsilon (x) \cap A \neq \emptyset$$. To show the latter, you have to work harder using triangle inequality arguments.

In fact if you look closer at your proof for Answer 1, you never explicitly used the hypothesis that each sequence element $$x_n \in A'$$ is itself a limit point of $$A$$. You only used $$x_n \to x$$ so far. You did notice this unused idea in Answer 2 but this idea is not optional; it is needed to prove your result that $$x \in A'$$ in Answer 1.

Here is how:

If any of the $$x_n = x$$, then we are done because then $$x = x_n \in A'$$ directly. So assume $$x_n \neq x$$ for all natural $$n$$.

Now fix $$\epsilon > 0$$. As $$x_n \to x$$, there is a natural $$N$$ s.t. $$d(x, x_n) < \frac{\epsilon}{2}$$ for all naturals $$n \geq N$$. But note that each such $$x_n \in A'$$ is a limit point of $$A$$. And (this is a crucial simplifying trick) observe that $$U = B\big(x_n; \frac{\epsilon}{2}\big) - \{x\}$$ is an open set containing $$x_n$$ (because we are assuming $$x_n \neq x$$). So by the definition of $$x_n$$ being a limit point, you have some element $$a_n \in \big(U - \{x_n\}\big) \cap A = \Big(B\big(x_n; \frac{\epsilon}{2}\big) - \{x, x_n\}\Big) \cap A$$ Hence $$d(x_n, a_n) < \frac{\epsilon}{2}$$. And therefore by the triangle inequality $$d(x, a_n) \leq d(x, x_n) + d(x_n, a_n) = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ i.e. $$a_n \in B(x_n; \epsilon)$$. Also since $$a_n \neq x$$ by construction (because remember $$a_n \in U$$ which did not contain $$x$$) we have that $$a_n \in \big(B(x_n; \epsilon) - \{x\}\big) \cap A$$. As this demonstration holds for arbitrary $$\epsilon > 0$$, we have shown that $$x$$ is indeed a limit point of $$A$$ i.e. $$x \in A'$$ as desired.

• The trick definitely wasn't simple to observe thanks. – Hawk Aug 20 '19 at 2:36

For (2) you could say: Since $$x_n \to x$$ theres a $$k$$ such that $$|x_k-x|=\delta<\epsilon/2$$. If $$\delta=0$$ we have $$x_k=x$$ and therefore x is a limit point.

Lets consider the case with $$\delta>0$$. Since $$x_k$$ is a limit point theres a sequence in A converging to $$x_k$$. Let this sequence be $$(m_n)_{n\in \mathbb{N}}$$.

Then theres a $$N$$ such that $$n\geq N\Rightarrow |m_n-x_k|<\min({\epsilon/2,\delta})$$. This makes sure $$m_n\neq x$$.

Therefore $$\forall n\geq N, |x-m_n|\leq |x-x_k|+|x_k-m_n|<\epsilon/2+\epsilon/2=\epsilon.$$ Therefore for each $$\epsilon>0$$ there are infinitely many points $$(m_n)_{n=N}^{\infty}$$ such that $$|m_n-x|<\epsilon.$$ Therefore $$x$$ is a limit point.

I want to highlight $$m_n$$ does not converge to $$x$$ but if we were to take $$(\epsilon_k)=1/k$$ and let $$m_{N}^{(\epsilon_k)}$$ be the $$m_N$$ for given $$\epsilon$$. Then $$(m_{N}^{(\epsilon_k)})_{k\in \mathbb{N}}$$ would converge to $$x$$, hence it's a limit point. (since $$m_{N}^{(\epsilon_k)}\neq x$$)

• Subsequence of $A$? – Hawk Aug 18 '19 at 6:30
• i thought A is the sequence who's limit points are A' – A. P Aug 18 '19 at 6:32
• $A$ is any set. – Hawk Aug 18 '19 at 6:35
• oh ok sry then i misread the question. The proof should still work with minor adaptions. 0XLR answer actually uses the same idea but is more thorough and actually addresses your question, so look at that one :). – A. P Aug 18 '19 at 6:40
• i fixed my answer – A. P Aug 18 '19 at 6:52