Prove the sentences about limit point

Question

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 $\lim_{n\rightarrow \infty}f(a_n)=c$.

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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?

Answer 1

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)) $.

Answer 2

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 $\lim\limits_{x\to x_0}f(x)= c$.