Prove that $x$ is an accumulation point of $A$ if and only if there exists $a_n$ in $A \setminus\{x\}$ such that its limit is $x$. Let $( E , \| . \|)$ be a vector space and $A \subset E$. Prove that $x$ is an accumulation point of $A$ if and only if there exists a sequence $a_n$ in $A\setminus \{x\}$ such that $\lim_{n\to\infty}a_n = x$.
I am also stuck in this: Let $x_n$ be a sequence in a vector space $( E, \| . \| )$ and $l$ in $E$, then we have $\lim_{n\to\infty}x_n = l$ is equivalent to $\lim_{n\to\infty} \|x_n-l\|=0$.
 A: Suppose $x$ is an accumulation point of $A$. Then we can define $A_n=\{y\in A \mid \|x-y\|<\tfrac{1}{n}\land y\neq x\}$ for each $n\in\Bbb N$, knowing each $A_n$ is not empty because $B_\epsilon(x)\cap(A\setminus\{x\})\neq\emptyset$ for every $\epsilon$ (taking $\epsilon=\tfrac{1}{n}$). Then we can construct the sequence $(a_n)_{n\in\Bbb N}$ choosing $a_n$ as one element from $A_n$ for each $n$ (we need the Axiom of Choice here). This sequence clearly is in $A\setminus\{x\}$ and converges to $x$ by construction.
Now suppose we have some sequence $(a_n)$ in $A\setminus\{x\}$ such that it converges to $x$. Take $\epsilon>0$. Then there has to be some $n_0\in\Bbb N$ such that $\|x-a_{n_0}\|<\epsilon$. Since every $a_n$ is in $A\setminus\{x\}$,  $a_{n_0}$ has to be distinct from $x$, so taking $y=a_{n_0}$ we have $y\in B_\epsilon(x)\cap(A\setminus\{x\})$. Therefore $x$ is an accumulation point of $A$.
The second result is a natural consequence of the definition of limit: $l$ is the limit of a sequence $(x_n)$ if given any $\epsilon>0$ there exists some $n_0\in\Bbb N$ such that $x_n\in B_\epsilon(x)$ for every $n\ge n_0$. On the other hand, $\|l-x_n\|\to0$ as $n\to\infty$ means the following: given $\epsilon>0$ there exists some $n_0\in\Bbb N$ such that $\|l-x_n\|<\epsilon$ for every $n\ge n_0$. It's easy to see the equivalence from these definitions.
