# Prove the sentences about limit point

1. Let $$x_0\in D$$ be a limit point of $$D$$. Show that there is a sequence $$(a_n)$$, $$n\in \mathbb{N}$$, in $$D$$ that converges to $$x_0$$.

2. Let $$f:D\rightarrow \mathbb{R}$$ be a function and $$x_0$$ a limit point of $$D$$. Show that the below sentences are equivalent.

i) The function has a limit $$c\in \mathbb{R}$$ for $$x\rightarrow x_0$$.

ii) For each sequence $$(a_n), n\in \mathbb{N}$$ in $$D$$ that converges to $$x_0$$, it holds that $$\lim_{n\rightarrow \infty}f(a_n)=c$$.



I have done the following:

1. Per definition we have that a number $$a$$ is a limit point of a sequence $${\displaystyle (a_{n})_{n\in \mathbb {N} }}$$, if there is a subsequence $${\displaystyle \left(a_{n_{k}}\right)_{k\in \mathbb {N} }}$$ of the sequence $${\displaystyle (a_{n})_{n\in \mathbb {N} }}$$, that converges to $$a$$.

So doesn't the part that we have to show trivially follow from the definition?

2. We suppose that the function has a limit $$c\in \mathbb{R}$$ for $$x\rightarrow x_0$$. So $$\lim_{x\rightarrow x_0}f(x)=c$$.

Let $$\epsilon >0$$. By the definition of the limit $$\exists\delta>0$$ : $$|f(x)-c|<\epsilon$$ for $$0<|x-x_0|<\delta$$.

Let $$(a_n)$$ be a sequence that converges to $$x_0$$. From the defintion we have that $$\exists N : \ \forall n>N : \ 0<|a_n-x_0|<\delta$$.

This mean then that $$|f(a_n)-c|<\epsilon$$ for all $$n>N$$. This means that $$\lim_{n\rightarrow \infty}f(a_n)=c$$.

Is the proof of i) $$\rightarrow$$ ii) correct and complete? Could we improve something?

For the other direction, we suppose that for each sequence $$(a_n), n\in \mathbb{N}$$ in $$D$$ that converges to $$x_0$$, it holds that $$\lim_{n\rightarrow \infty}f(a_n)=c$$.

We assume that the function hasnot the limit $$c$$ for $$x\rightarrow x_0$$.

So $$\exists \epsilon >0$$ : $$\forall\delta>0$$ : $$|f(x)-c|\geq \epsilon$$ for $$0<|x-x_0|<\delta$$.

Is this correct so far? How can we continue?

For the first one, I thought that $$x_0$$ is a limit point of $$(a_n)$$ if, for any $$c > 0$$ there is an $$n(c)$$ such that $$|a_n-x_0| < c$$ for all $$n > n(c)$$.

From this definition, you have to construct a sequence $$(a_{n_k})|_{k=1}^{\infty}$$ that converges to $$x_0$$.

Hint: choose $$n_1 = n(1), n_{k+1} =\max(n_{k}+1, n(1/2^k))$$.

• We know that $x_0$ is a limit point that means that there is a subsequence of the sequence $a_n$ that converges to $x_0$, right? Why do we have to construct such a subsequence? And could you explain to me further the hint? I haven't really understood why we take these values. Jan 22, 2020 at 23:55
• Please quote the definition of limit point that you have been given. I do not think the one you have is correct. For example, if $a_{2n} \to x_0$ and $x_{2n+1} \to x_0+1$, then your definition implies that $x_0$ is a limit point when it clearly is not. Jan 23, 2020 at 0:07

Let $$B(x,r)=\{y\in D: |y-x|\lt r\}$$ for given $$x\in D$$ and $$r\gt0.$$

For the first part, I rather use this definition.

$$x_0$$ is a limit point of $$D$$ iff for all $$r>0$$, $$(B(x_0,r)-\{x_0\})\cap D \neq \emptyset$$.

Now suppose $$x_0$$ is a limit point of $$D$$. then by definition for all $$n \in \mathbb Z^+$$, $$(B(x_0,\frac{1}{n})-\{x_0\})\cap D \neq \emptyset$$.

So for all $$n \in \mathbb Z^+$$, you can pic an $$x_n \in D$$ s.t $$x_n\in (B(x_0,\frac{1}{n})$$. Now you can easily prove that $$\{x_n\}$$ converges to $$x_0$$.

For the reverse implication of second part,

Suppose if $$\{a_n\}$$ is a sequence in $$D$$ which converges to $$x_0$$ then $$\lim\limits_{n\to \infty}f(a_n)=c$$.

Now to get a contradiction assume $$\lim\limits_{x\to x_0}f(x)\neq c$$

Then, for all $$n\in \mathbb Z^+$$ there is a $$x_n \in D$$ s.t $$|x_n-x_0|\lt \frac{1}{n}$$ and $$|f(x_n)-c|\gt1.$$

Now again you can see $$x_n \to x_0$$ as $$n\to \infty$$, but $$\lim\limits_{n\to \infty}f(x_n)\neq c$$ which contadict the supposition. thus $$\lim\limits_{x\to x_0}f(x)= c$$.

• Don't know if OP knows what $B(x, r)$ is. Jan 23, 2020 at 0:09
• @martycohen edited. Never occurred to me. Thanks. Jan 23, 2020 at 0:15
• I like to also have an informal statement like "all points within $r$ of $x$." I think it makes it easier to understand. Jan 23, 2020 at 0:22
• Ok! So is the first implication of the second part correct? Jan 23, 2020 at 6:08
• @MaryStar yes it is Jan 23, 2020 at 8:12