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 ?

  • $\begingroup$ If by "continuous set" you mean an interval, the only possibility is a singleton $\{ a \}$. $\endgroup$
    – Crostul
    Jul 18 '18 at 20:19
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.