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:

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

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

  3. 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?

  • $\begingroup$ $X$ is a subset of itself, so can be a element of a family (i.e. set) of subsets of $X$ $\endgroup$
    – Henry
    Aug 19, 2022 at 13:52

3 Answers 3


"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 $X$.


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.

  • $\begingroup$ I'd not seen the notation $2^X$ for the power set of $X$ before; especially if the OP is relatively new to thinking about set theory and topologies it might be a bit obstructive compared to something like $\mathcal{P}(X)$. If one is already learning to keep track of a set, its subsets and its elements, putting a notation based around functions into the mix is quite a lot. $\endgroup$
    – dbmag9
    Aug 19, 2022 at 13:19

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.


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