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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 ?

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  • $\begingroup$ If by "continuous set" you mean an interval, the only possibility is a singleton $\{ a \}$. $\endgroup$
    – Crostul
    Jul 18 '18 at 20:19
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    $\begingroup$ 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 $[0,1]$ $\endgroup$
    – Peter
    Jul 18 '18 at 20:21
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    $\begingroup$ 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. $\endgroup$ Jul 18 '18 at 21:33
  • $\begingroup$ /= {$\emptyset$,x} is a misunderstanding $\endgroup$ Jul 19 '18 at 2:46
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    $\begingroup$ 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 $\endgroup$ Jul 20 '18 at 8:09
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By "continuous", I think you mean connected. But a connected set might not have all of its limit points.

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.

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  • $\begingroup$ 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. $\endgroup$
    – User8128
    Jul 18 '18 at 20:57
  • $\begingroup$ @User8128 Thank you for correcting my mistake. $\endgroup$
    – NicNic8
    Jul 18 '18 at 20:58

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