How does this definition of a topology on $X$ not cause $X \in X$? From Wikipedia:
Formally, let X be a set and let $\tau$ be a family of subsets of X. Then $\tau$ is called a topology on X if:

*

*Both the empty set and X are elements of $\tau$.


*Any union of elements of τ is an element of $\tau$.


*Any intersection of finitely many elements of τ is an element of $\tau$.
I'm confused about (1). If $\tau$ is a family of subsets of X, and X is a member/element of $\tau$  then doesn't that suggest that X is a member of itself, which generally speaking isn't allowed?
 A: Other answers already describe the general situation well, but a concrete example may also help.
Think about the real numbers, $\newcommand{\R}{\mathbb{R}}\R$.  Certainly this isn’t an element of itself — $\R$ is not a real number.  The standard topology on it is given by $\tau_\R := \{ U \subseteq \R \mid \forall x \in U,\, \exists \epsilon > 0,\ B_\epsilon(x) \subseteq U\}$ — the collection of all open sets of reals (in the standard metric definition of “open”).
Now the biggest subset possible subset of $\R$ is all the reals, i.e. $\R$ itself, and it’s certainly open.  So $\R \in \tau_\R$.  And generally this is what the axiom says: In any topological space, the whole space (viewed as a subset of itself) is open.
A: "then doesn't that suggest that X is a member of itself” No, it suggests $X$ is a subset of itself, because $\tau$ is a family of subsets of $X$.
A: Careful. A topology is a subcollection of $2^X$, the power set of $X$. So if say $X = \{a,b\}$, maybe $\tau = \{\varnothing, \{a\}, \{b\}, X\}$.  Thus $X \in \tau$, but this says nothing about $X \in X$, which would, you're right, be problematic.  In other words, a collection of subsets of $X$, such as $\tau$, is a different 'kind' of object than $X$ itself.
