A set is Closed iff it contains all its Limit Points

Last updated Nov 1, 2022

Suppose $(X, \tau)$ is a Topological Space. Then $K \subset X$ is Closed If and Only If $\bar{K} \subset K$. That is, $K$ is Closed If and Only If for all $x \in X$ Limit Point of $K$, $x \in K$.

$(\Rightarrow)$ Since $K$ is Closed and A Set is Closed iff it is its own Closure, $\text{cl} K = K$. We also know that $\bar{K} = \text{cl} K$. So $\bar{K} = K$ and thus $\bar{K} \subset K$. $\checkmark$

$(\Leftarrow)$ Suppose $\bar{K} \subset K$. Recall All points in a Set are Limit Points, so $K \subset \bar{K}$. Thus $\bar{K} = K$. Since $\bar{K} = \text{cl} K$ and $\text{cl} K$ is Closed, we have that $K = \text{cl} K$ is Closed. $\checkmark$ $\blacksquare$

$\blacksquare$

1. This seems like a trivial result, but I anticipate using it occasionally. It’s easier to reference this result than repeat the line of reasoning in the proof everytime

• Closure
• Limit Point
• A Set is Closed iff it is its own Closure
• Closure of a Set is all its Limit Points