Simple Limit Points and Sequences proof I'm just starting to learn some real analysis and I was wondering if somebody could help verify and critique my proof. If somebody could also provide a different proof that would also be nice. Thank you.
1) Show that if M is the open interval (a,b), and p is in M, then p is a limit point of M.
Let  $a<p<b$ and $\varepsilon > 0$ and $(p-\varepsilon, \ p+\varepsilon)$  be any open interval where $p\in(p-\varepsilon, \ p+\varepsilon)$. If c is a point in M such that $c\not=p$ and $p-\varepsilon<c<\ p+\varepsilon$. Then $(p-\varepsilon, \ p+\varepsilon)$ contains p and c and since $c\not=p$ this implies that p is a limit point.
 A: You need to actually show that for every $\epsilon > 0$ such a $c$ exists. You could take $c = \frac{p + p - \epsilon}{2}$. Then $p-\epsilon<c<p$ so $c\neq p$ and $c\in (p-\epsilon,p+\epsilon)$.
We need to be a bit careful about the values of $\epsilon$ we work with though. If the $\epsilon$ we're given is too large, then it's possible for $c =  \frac{p + p - \epsilon}{2} \leq a$, which would mean $c\not\in (a,b)$.
But since we have that $p\in (p-\delta,p+\delta)\subseteq (p-\epsilon,p+\epsilon)$ whenever $0 < \delta \leq \epsilon$, if we're given a value of $\epsilon$ that is too large, we can just construct $c$ using a smaller value of $\epsilon$.
In this case, $\epsilon>0$ is too large iff $\epsilon \geq 2(p-a)$ (we can get this by manipulating the inequality $c =  \frac{p + p - \epsilon}{2} \leq a$). So when we're given an $\epsilon>0$, we can take $\epsilon_{0} = \min\{\epsilon, (p-a)\}$. And so then $\epsilon_{0}\leq \epsilon$, and $\epsilon_{0} \leq (p-a) < 2(p-a)$. Then we can set $c = \frac{p + p - \epsilon_{0}}{2}$ which would give us $c\in (p-\epsilon_{0},p+\epsilon_{0})\subseteq (p-\epsilon,p+\epsilon)$, and $c\in (a,b)$ with $c\neq p$.
