Let $x \in X$. Show that the subset $\{x\} \subset X$ is closed with respect to $\tau(d)$

general-topology

Let $x \in X$. Show that the subset $\{x\} \subset X$ is closed with respect to $\tau(d)$

To solve this problem I have looked at properties of closed sets:

$a)$ The empty set $\emptyset$ and the whole space $X$ are closed.

$b)$ If $A \subset X$ and $B \subset X$ are closed, so is their union $A
\cup B$.

$c)$ Let $I$ be a set and let $(A_i)_{i \in I}$ be a family of closed subsets of $X$. Then their intersection $\bigcap_{i \in I} A_i$ is closed as well.

To show that the empty set is closed, we have to see that $X$ \ $\emptyset$ is open, and $X = X$ \ $\emptyset$ is open in any topology by definition. Likewise, $X$ \ $X = \emptyset$ is open, hence $X$ is closed.

Not sure how to answer $a)$ and $b)$

Thanks in advance.

Best Answer

For now let $(X,\tau)$ be a general topological space. You have already shown that $X$ and $\varnothing$ are closed. For (b), let $A$ and $B$ be closed subsets of $X$. To prove that $A \cup B$ is closed, we need only check that $(A \cup B)^\complement = A^\complement \cap B^\complement$ is open. But this is immediate since $A^\complement$ and $B^\complement$ are open and finite intersections of open sets are open.

Treating (c), let $\{A_i\}_{i \in I}$ be a family of closed subsets of $X$. Then, \begin{align*} \left( \bigcap_{i \in I} A_i \right)^\complement = \bigcup_{i \in I} A_i^\complement \end{align*} is open because arbitrary unions of open sets are open (and every $A_i^\complement$ is open by definition). This proves (c).

Now, given an arbitrary topological space $(X,\tau)$, it is not true in general that every singleton $\{x\}$ is closed in $X$. As I pointed out in the comments, if we give $\mathbb{R}$ the indiscrete topology $\{\mathbb{R}, \varnothing\}$, then every singleton will not be closed in $\mathbb{R}$.

Let us also recall the following definition:

Definition. A space $(X,\tau)$ is said to be $T_1$ if for any $x,y \in X$ with $x \neq y$, there exists an open set $U \ni x$ that does not contain $y$.

Without assuming that $(X,\tau)$ is either Hausdorff or $T_1$, it will be impossible to show that singletons are closed. In fact, the following equivalence holds:

Theorem. Let $(X,\tau)$ be a topological space. Then, $(X,\tau)$ is $T_1$ if and only if every singleton $\{x\}$ is closed in $X$.

In light of this theorem, it's clear that one cannot hope to prove what you want without any additional assumptions on the space $(X,\tau)$.

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