1. ## Accumulation point

Prove that a real number s is an accumulation point of a set S if and only if there exists some sequence {a} in S such that a is not equal to s for every natural number n where the limit of a as n approaches infinity is equal to s.

I'm not really sure how to start. I know since it's an if and only if statement we have to prove it both ways.

2. $s\in S$ is a accumulation point if for every $\epsilon > 0$ there exists a $x_0\in S$ with $a-\epsilon < x_0 < a+\epsilon$ and $x_0\neq a$

Choose a $\epsilon > 0$.

For every $n\geq 1$ you can choose a $x_n\in S$ with $x_n\neq a$ such that: $a-\frac{\epsilon}{n} < x_n < a+\frac{\epsilon}{n}$

Now $x_n\to a$ as $n\to \infty$

I'll leave the other implication to you.

3. Originally Posted by bethh
Prove that a real number s is an accumulation point of a set S if and only if there exists some sequence {a} in S such that a is not equal to s for every natural number n where the limit of a as n approaches infinity is equal to s.
You did not tell us what sort of top-space we are considering.
I will assume that you are working in a metric space (so the statement is true).
If $r>0$ then $\mathcal{B}(a;r)=\{x:d(a,x) is a ball.
Using the definition of accumulation point we get:
$\left( {\exists x_1 } \right)\left[ {s \ne x_1 \,\& \,d\left( {x_1 ,s} \right) < 1} \right]$.
$\left( {\exists x_2 } \right)\left[ {s \ne x_2 \,\& \,d\left( {x_2 ,s} \right) < \min(d\left( {x_1 ,s} \right),\frac{1}{2}} )\right]$.

Now you have a way to generate your sequence by induction.

4. Originally Posted by bethh
Prove that a real number s is an accumulation point of a set S if and only if there exists some sequence {a} in S such that a is not equal to s for every natural number n where the limit of a as n approaches infinity is equal to s.

I'm not really sure how to start. I know since it's an if and only if statement we have to prove it both ways.
Let $\left\{x_n\right\}_{n\in\mathbb{N}}$ be a sequence in $S$ such that $x_n\ne x,\text{ }\forall n$ and $x_n\to x$. Now, let $B_{\varepsilon}(x)$ be arbitrary. Since $x_n\to x$ there exists an $N\in\mathbb{N}$ such that $N\leqslant n\implies x_n\in B_{\delta}(x)$ and since $x_N\ne x$ and $x_N\in B_{\delta}(x)$ we see there exists a point of $S$ distinct from $S$ in this ball. The conclusion follows from the arbitrariness of the open ball.

Conversely, choosing any arbitrary point in $B_{\frac{1}{n}}(x)-\{x\}$ finishes the job. There is no need for them to be all distinct.