A closed set contains all its limit points

Question

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?

Answer 1

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.