0
$\begingroup$

I'm constructing a proof regarding a closed set $Y$. In it, I make the assumption that $Y$ is not closed. This means that $Y$ does not contain all its limit points $k_0$. Does this then mean, that one of these points $k_0$ must instead be a isolated point? Since, If I have understood this correctly, the opposite of a limit point is an isolated point. In general, what does it mean when a set does not contain all its limit points?

$\endgroup$
2
  • $\begingroup$ Not at all. Consider for example $(0,1)$. Both $0$ and $1$ are limit points of this set. But they are not isolated points. $\endgroup$
    – MJD
    Feb 6, 2020 at 14:40
  • $\begingroup$ Indeed every limit point is a point of closure but the converse does not hold always, a point of closure which is not a limit point is called an isolated point, and the set containing all of such isolated points is called derived set which is closed itself $\endgroup$
    – user715522
    Feb 6, 2020 at 14:50

2 Answers 2

2
$\begingroup$

You idea of “opposite” here is a bit confused.

Say we have a set $S$. A point $p$ could be

  1. In $S$, or not in $S$
  2. A limit point of $S$, or not a limit point of $S$

All four combinations are possible. Consider the set $$S = [0, 1) \cup \{2\}.$$ Then examples of each of the four combinations of properties are:

$$ \begin{array}{c|cc} & \text{in $S$} & \text{not in $S$} \\ \hline \text{limit point} & 0 & 1 \\ \text{not limit point} & 2 & 3 \\ \end{array} $$

Every point in the line is one of these four types. The points $0$ and $2$ are in $S$, but $0$ is a limit point and $2$ is not. The points $1$ and $3$ are not in $S$, but $1$ is a limit point and $3$ is not.

Points of type $2$ are called isolated points. They are exactly those points that are in $S$ but are not limit points of $S$.

You can see in the chart that "limit point" is not really the "opposite" of "isolated point". That is true for points of $S$ (types $0$ and $2$). But a point of type $3$ is neither a limit point of $S$ nor an isolated point of $S$.

When there are no points of type $1$, we say that $S$ is closed. If $S$ is not closed, then that means there is a point of type $1$. But that has no bearing on whether there is an isolated point, which is about whether there are any points of type $2$. There could be points of both types, or neither, or points of one type but not the other.

$\endgroup$
1
$\begingroup$

No, it does not imply there exists at least one isolated point.

The assumption that a set does not contain all its limit points implies there exist a limit point of the set not in the set, not "not all points in the set are limit points".

$\endgroup$

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.