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

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*$X$ and the empty set, $\varnothing$, belong to $\tau$,

*the union of any (finite or infinite) number of sets in $\tau$ belongs to $\tau$, and

*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?
 A: 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 $S \cup T \supset S \supsetneq \emptyset$.
A: 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 $\Bbb R$.


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*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.
