$[0,1]$ defined in $\mathbb{R}$ is a closed set. A closed set contains all its limit points. The points $0$ and $1$ do not have all their neighbourhoods within $[0,1]$.
So they are not limit points ? But $0$ is a limit point of set 1/n which should be a subset of $[0,1]$ ?
Definition of limit point: A real number x is a limit point of a set S if every neighborhood of x contains an infinite number of elements of S.
Wikipedia definition: A point x in X is a limit point of S if every neighbourhood of x contains at least one point of S different from x itself. This suggests $0$ is a limit. So, the first definition is wrong ??
Then what are the limit points of $[0,1]$ ?
How do we "find" limit points of "simple sets"(if there is such a thing) like $[0,1]$?
I am aware of the fact that openness, closedness are not mutually exclusive. I am struggling with visualising the concept of closure of set.
Best Answer
Hint
In $\mathbb R$, all neighborhoods of a point $x$ contain a neighborhood of the form $B_\epsilon(x) = (x-\epsilon, x+\epsilon)$. Thus a point is a limit point of $S$ iff $(x-\epsilon, x+\epsilon) \cap S$ contains infinitely many points for all $\epsilon > 0$. Now note that for $\epsilon < 1$ $$(-\epsilon, \epsilon) \cap [0,1] = [0, \epsilon)$$ wich contains infinitely many points and for $\epsilon \ge 1$, $(-\epsilon, \epsilon) \cap [0,1] = [0,1]$ wich contains infinitely many points so $0$ is indeed a limit point of $[0,1]$.
Proceed in the same way with $1$.