Criterion for a Set of a Topological Space to be Closed

# Criterion for a Set of a Topological Space to be Closed

Recall from the Accumulation Points of a Set in a Topological Space page that if $(X, \tau)$ is a topological space and $A \subseteq X$ then a point $x \in X$ is said to be an accumulation point of $A$ if every open neighbourhood of $x$ contains points of $A$ different from $A$.

Also recall that $A$ is said to be closed if $A^c = X \setminus A$ is open.

We will now look at an important theorem relating these two concepts in that a set $A$ is closed if and only if $A$ contains all of its accumulation points.

Theorem 1: Let $(X, \tau)$ be a topological space and $A \subseteq X$. Then $A$ is closed if and only if $A$ contains all of its accumulation points. |

**Proof:**$\Rightarrow$ Suppose that $A \subseteq X$ is closed and that $A$ does not contain all of its accumulation points. Then there exists an accumulation point $x$ of $A$ such that $x \not \in A$. Since $x \not \in A$, we must have that $x \in A^c = M \setminus A$.

- Since $x$ is an accumulation point of $A$, by definition, every open neighbourhood $U \in \tau$, $x \in U$, of $x$ contains elements of $A$ different from $x$. So every open neighbourhood of $x$ contains points of $A$.

- So there exists NO open set such that $x \in U \subseteq A^c = X \setminus A$ (since if $x \in U$, then there exists an $a \in A$, $a \neq x$, such that $a \in A$ and clearly $a \not \in A^c = X \setminus A$).

- Therefore $x \not \in \mathrm{int} (A^c)$. Then $\mathrm{int}(A^c) \neq A^c$. So $A^c$ is not open, which implies that $A$ is not closed - a contradiction.

- Hence, the assumption that $A$ does not contain all of its accumulation points is false. Therefore $A$ must contain all of its accumulation points.

- $\Leftarrow$ Now suppose that $A$ contains all of its accumulation points. We want to show that then $A$ is closed. Consider the complement $A^c = X \setminus A$. If we can show that $A^c$ is open, then it follows that $A$ is closed. Note that $A^c$ is open if and only if $\mathrm{int}(A^c) = A^c$. Let $x \in A^c$. Since $A$ contains all of its accumulation points, it follows that $x \in A^c$ is NOT an accumulation point of $A$. By the definition of $x$
*not*being an accumulation point of $A$, we have that there exists an open neighbourhood $U$ ($U \in \tau$) of $x$ such that $x \in U$ and such that $U$ does not contain any points of $A$ different from $x$ (i.e., if $a \in A$ then $a \not \in U \setminus \{ x \}$. But $x \not \in A$ too since $x \in A^c$. Therefore $U \in A^c$.

- So, for all $x \in A^c$ there exists an open neighbourhood $U \in \tau$ such that:

\begin{align} \quad x \in U \subseteq A^c \end{align}

- By definition we have that $x \in A^c$ is an interior point of $A^c$, so $\mathrm{int} (A^c) = A^c$. So $A^c = X \setminus A$ is open. Therefore $A$ is closed. $\blacksquare$