Is every point contained in a proper open set of a topological space?

Question

Let $X$ be a topological space. Let $a\in X$. Is it always true that $a$ is contained in a proper open set of $X$? I don't know how to derive it directly by the axioms of a topological space.

Answer 1

No. Indeed, the "trivial topology" in which only the empty set and the full space are open, is a valid topology. It's not very interesting, but it does satisfy the axioms.

Answer 2

No, this need not be the case: if $X$ is a set and $p \in X$ then the following defines a topology on $X$ (th excluded point topology w.r.t. $p$):

$$\mathcal{T}= \{A \subseteq X: p \notin A\} \cup \{X\}$$

It's easy to check this satisfies the axioms of a topology. And it's also clear that the only open set that contains $p$ is $X$ itself.

But very often additional assumptions on $X$ exist, e.g. $T_1$ (for every pair $x \neq y$ of points in $X$ there is an open set $O$ such that $x \in O$, and $y \notin O$. This guarantees that there are enough open sets so that every point is contained in a proper open set. The above example is merely $T_0$, not $T_1$. Where $T_0$ means that for every $x \neq y$ we have an open set $O$ such that ($x \in O$ and $y \notin O$) or ($x \notin O$ and $y \in O$), which does hold as one of the $x$ or $y$ is unequal to $p$ (say $x$) and we then use $O=\{x\}$. So $T_0$ does not guarantee this property, and the stronger $T_1$ does. So if you desire such a property for a space $X$, assume such "separation axioms" on it.