TITLE: Interior and accumulation points
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.