# Limit point of the closed set $[0,1]$

I am trying to understand the concept of limit point in general.

For the open set $(0,1)$, earlier discussions on stackexchange showed that all point in the open set $(0,1)$ is a limit point. Similarly, we can say the same about the closed set $[0,1]$, right ?. For every point $x$ in the set $[0,1]$, we can find $\epsilon$ neighborhood ($V_{\epsilon}$) such that $V_{\epsilon}(x) \cap A \neq \{\emptyset,x\}$.

If the above comment is true, anyone can give an example of a continuous set containing isolated points ?

• If by "continuous set" you mean an interval, the only possibility is a singleton $\{ a \}$. Jul 18 '18 at 20:19
• Yes it's true. Continuous set is unfortunately a little bit ambiguous. But if I understood well, indeed, set as $[a,b]$ has no isolated point. But if you take the discrete topology and $x\in [0,1]$, then $[0,1]\cap\{x, -3\}=\{x\}$ and thus $x$ is an isolated point of $[0,1]$ Jul 18 '18 at 20:21
• You need to show that for every $\varepsilon$-neighbourhood of $x$, $V_\varepsilon(x) \cap A$ is contains a point in $A \setminus \{x\}$, not that there is one. But indeed all $x\in [0,1]$ are limit points. Jul 18 '18 at 21:33
• /= {$\emptyset$,x} is a misunderstanding Jul 19 '18 at 2:46
• As an intersection of A and an open set will never have the empty set as a member, your statement is always true. You are misusing set notation. Not subset of {x} is correct. @Shew Jul 20 '18 at 8:09

Here's an example. Consider the set $(0,1)$. It is true that every point in the set is a limit point. For example, let's look at the point $1/2$. Is this a limit point? To be a limit point, we need to create a sequence of points in the set that approach this point. The Sequence $1/2, 1/2, 1/2, ...$ is such a set of points.
What about the point $0$? Is $0$ a limit point of this set? It is! Consider the sequence $1/2, 1/4, 1/8, 1/16, 1/32, ...$ This sequence is getting closer and closer to the point $0$, and no point in this sequence will ever leave the set $(0,1)$. Therefore, the set $(0,1)$ does not contain all its limit points.
Any set that contains all of its limit points is called a "closed" set. So $(0,1)$ is not a compact set. But $[0,1]$ is.
• A set that contains all it's limit points is closed, but not necessarily compact; for example, $[0,\infty)$ contains all it's limit points but is not compact. Jul 18 '18 at 20:57