Interior and accumulation points Show that every interior point of a set must also be an accumulation point of that set.
Definitions:
Any point $x$ that belongs to $E$ is said to be an interior point of $E$ provided that some interval $(x-c,\ x+c)\subset E$.
Any point $x$ (not necessarily in $E$) is said to be an accumulation point of $E$ provided that  for every $c>0$ the intersection $(x-c,\ x+c)\cap E$ contains infinitely many points.
How do I show that every interior point of a set must also be an accumulation point of that set from these 2 definitions?
 A: There is $\varepsilon > 0$ such that $(x-\varepsilon, x + \varepsilon ) \subset E$. Let $\varepsilon_n = \frac{\varepsilon}{n}$. Take $x_n \in (x-\varepsilon_n, x + \varepsilon_n ) \setminus (x-\varepsilon_{n+1}, x + \varepsilon_{n+1} )$. The set $\left\{ x_i \right\}_{i=1}^\infty \cap (x-c, x + c )$ is infinite for any $c >0 $. 
A: Let $x$ be an interior point of $E$, $(x-c,x+c)\subset E$. Let $d>0$, $(x-inf(c,d),x+inf(c,d))\cap E\subset (x-c,x+c)\subset E$. Since $(x-inf(c,d),x+inf(c,d))\subset (x-d,x+d)$ contains an infinite number of points, the result follows.
A: Let $x$ be an interior point of $E.$ Since $x$ is an interior point of $E$ there exists $c>0$ such that $(x-c,\ x+c)\subset E.$ Now, you have to show that it is an accumulation point. That is, you have to show that for any $d>0$ the set $(x-d,x+d)\cap E$ contains infinitely many points. Of course, it is enough to consider $d\le c.$ But then $(x-d,x+d)\cap E=(x-d,x+d)$ (since $(x-c,\ x+c)\subset E$) which clearly contains infinitely many poitns. (If $d>c$ use that  $(x-c,x+c)\subset (x-d,x+d)\cap E.)$
A: If $(x-c, x+c) \subset E$, then $(x-c,x+c) \cap E = (x-c,x+c)$ which is an interval that contains infinitely many points for any $c > 0$.
From the first definition, we know that for small enough $c_0$, we have $(x-c_0,x+c_0) \subset E$. If $c \le c_0$, then $(x-c,x+c) \cap E = (x-c,x+c)$ contains infinitely many points. Otherwise if $c > c_0$, then $(x-c_0,x+c_0) \subset E$ contains infinitely many points and hence $(x-c_0,x+c_0) \subset (x-c,x+c)\cap E$ implies the superset also contains infinitely many points.
