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 $x_0$.
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:
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?
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?