# Is a topological space isomorphic to some group?

I'm reading topology without tears to develop intuition before attempting Munkres. I noticed how similar a topological space was to a group in Abstract Algebra.

1.1.1 Definitions. Let $$X$$ be a non-empty set. A set $$\tau$$ of subsets of $$X$$ is said to be a topology on $$X$$ if

1. $$X$$ and the empty set, $$\varnothing$$, belong to $$\tau$$,
2. the union of any (finite or infinite) number of sets in $$\tau$$ belongs to $$\tau$$, and
3. the intersection of any two sets in $$\tau$$ belongs to $$\tau$$.

The pair $$(X,\tau)$$ is called a topological space.

Can you see why $$X$$ and $$\varnothing$$ behave like identities? And $$\cup$$ and $$\cap$$ are like binary operations. And this 'group' is closed under both operations.

Is there a group a topological space is isomorphic to? Or are these just coincidences?

• It's a monoid, as others have written. If I'm not mistaken, then this also gives you a contravariant functor from $\mathbf{Top}$ to $\mathbf{Mon}$. May 8, 2015 at 19:42

The system $(\tau, \cap)$ forms a monoid with identity $X$, and $(\tau, \cup)$ forms a monoid with identity $\emptyset$.

A monoid is an associative binary operation with identity; the inverse axiom is excluded. Note, for example, that for $S \in \tau$ nonempty, there can be no $T \in \tau$ such that $S \cup T = \emptyset$, because $S \cup T \supset S \supsetneq \emptyset$.

Not in general. But you're not too far from the truth.1

Generally, if $$X$$ is a set, then $$\mathcal P(X)$$ is a group using symmetric difference as a commutative addition operator. It's even a ring when using $$\cap$$ for multiplication.

But in the case of $$\cap$$ and $$\cup$$ as candidates for addition we run into some problems, since neither is reversible. Namely, $$A\cup B=A\cup C$$ does not imply that $$B=C$$, which means that there is no additive inverse.

We can try and argue with symmetric difference again; but a topology need not be closed under symmetric differences, e.g. $$(-1,1)\mathbin\triangle(0,1)=(-1,0]$$, and the latter is not open in the standard topology of $$\Bbb R$$.

1. Note, by the way, that a group is a set with an operation on that set. Like $$\Bbb Z$$ with $$+$$. Whereas a topological space is a set with a collection of subsets, rather than an operation.

I took the liberty of understanding the question a bit differently. If $$\tau$$, the topology itself, is a group.