# Definition of topological space: Is Ω equal to the powerset of X?

A topological space is a set $X$ and a collection $\Omega$ of subsets of $X$ such that:

• $\emptyset \in \Omega$ and $X \in \Omega$
• The union of any collection of $\Omega$ is in $\Omega$
• The intersection of any finite collection of sets in $\Omega$ is again in $\Omega$

My question is, is the set $\Omega$ essentially the powerset of $X$? Or is the powerset of $X$ just a special case of a suitable collection of subsets of $X$?

• It can be the power set of $X$, but it need not be. There are a lot of different topologies on a set $X$. It is always a subset of the power set, but most topologies are not closed under taking the complement. – Thomas Andrews Jan 17 '16 at 19:43
• Why does this question have so many similar answers? – Unit Jan 17 '16 at 20:03
• @Unit All the answers appeared right after the question was posted. I suspect they were being composed in parallel. I know mine was. The answers are similar because the question has essentially one answer. – Ethan Bolker Jan 18 '16 at 13:39

The power set of $X$ is a special case of such a collection $\Omega,$ called the discrete topology on $X$. All topologies on $X$ will be subcollections of the power set of $X$ (by definition).

For a wildly different example (at least, if $X$ has more than one point), consider the indiscrete topology on $X$: $\Omega=\{\emptyset,X\}.$

• I'd remark that the discrete topology is basically the degenerate special case where the space has no topological properties anymore, but only behaves as a plain old set. In that sense, if the set of open sets were always the powerset, then we wouldn't need topology... – leftaroundabout Jan 17 '16 at 20:08
• @leftaroundabout The indiscrete/trivial topology also is in some sense "behaving as a plain old set". Both are adjoints to the forgetful functor that forgets about topological structure: ncatlab.org/nlab/show/discrete+and+codiscrete+topology – Mark S. Jan 18 '16 at 16:24

The power set is a good example that defines a topological space. This is usually called the discrete topology. This is definitely not the only way to define a topology though!

On the other extreme, for any set $X$ $\tau=\{\emptyset, X\}$ is a topology on $X$.(do you see how this satisfies all your axioms?)

In general though, we don't care much about these topologies since their structures aren't difficult to understand. A large chunk of more interesting topologies are those that are induced by a collection of subsets. For example, the usual topology on $\mathbb{R}$ is the smallest topology that contains all the open intervals. It may not be obvious that this doesn't include the entire power set, so a good way to convince yourself of that would be to consider a singleton, for simplicity say $\{0\}$. Since topologies are closed under arbitrary unions and finite intersections, the question is if we can express $\{0\}$ with a combination of unions and only finitely many intersections of open intervals. Any two non-disjoint open intervals intersect in another open interval though and any arbitrary union of open intervals can always be written as a disjoint union of open intervals so it ends up we find that the only types of objects that we get from doing any combination of finitely many intersections and arbitrary unions of open intervals is a disjoint union of open intervals. Since $\{0\}$ is not a disjoint union of open intervals, $\{0\}$ is not a member of our topology.

I would suggest writing out a proof of this(in actual rigor) if you have time as I think that would be a good exercise to help you get an idea of why topologies aren't always just power sets. Hope this helps!

• It most certainly does! Thank you for your extensive answer! – Iordan Iordanov Jan 17 '16 at 19:57
• No problem. One thing I should have mentioned: the reason that I chose to look at a singleton is because if a topology contains all the singletons, then it is the discrete topology, which means that if a topology isn't the whole power set, then there must be singletons that are not in the topology. – Sean English Jan 17 '16 at 20:01
• Why the downvote on this and all the other answers but one (I suspect it's the same person)? – Ethan Bolker Jan 18 '16 at 0:51

The power set of $X$ is a topology for $X$, but it certainly is not the only one. Consider $\mathbb R$ with $\mathscr B=\{(a,b):a,b\in\mathbb R, a<b\}$ and $\Omega=\{\bigcup U_i: U_i\in\mathscr B\}$. You are probably already aware that this is a topology, but $\Omega\ne\mathscr P(\mathbb R)$. In general, a topology of $X$ is a subset of its power set with the stated properties.

• That's not a topology by itself. We probably know what you meant, though. – Ian Jan 17 '16 at 19:43
• Yes, you are confusing the basis of the topology on the reals with the topology on the reals. – Thomas Andrews Jan 17 '16 at 19:44
• Yes, I've fixed my answer. – Tim Raczkowski Jan 17 '16 at 19:45

The power set is an example of a topology, but a topology can be much smaller - for example, there's always the trivial topology $\Omega = \{\emptyset, X\}$.

The powerset of $X$ is a special case of a topology - the special case that makes every function to another topological space continuous. It's the largest possible topology.

But a topology need not be all the subsets of $X$. Many other subsets of the power set may satisfy the axioms. The smallest is the subset containing just $X$ itself and the empty set.

• Why did you delete your previous answer and re-post it? – Thomas Andrews Jan 17 '16 at 19:46
• @ThomasAndrews By accident, while I was making a trivial edit. – Ethan Bolker Jan 17 '16 at 19:48

Ω is a subset of the powerset of X. It can be the entire powerset (at most), the set {∅,X} (at least), or another subset that meets the properties.

The power set can be a lot bigger than the topology. For instance, in the real numbers under the usual topology, there is a bijection between $\mathbb{R}$ and the topology $\mathcal{T}$, but due to Cantor's famous diagonalization argument, the power set of the reals has strictly greater cardinality than the reals.

The fact that the usual topology is in bijection with $\mathbb{R}$ comes from the fact that each open set is a union of rational intervals, and there are only countable many of these.

• Could I get an explanation from the downvoter? This is the most important topological space and I have quantified the distinction between its powerset and its topology. – Noah Olander Jan 17 '16 at 21:52
• Every answer in the thread excepting the first was downvoted. Likely someone unhappy about the numerous similar answers. – user296602 Jan 18 '16 at 6:34

Let X = {a,b} be a set with 2 elements. There are four distinct topologies on X:

{∅, {a,b}} (the trivial topology)
{∅, {a}, {a,b}}
{∅, {b}, {a,b}}
{∅, {a}, {b}, {a,b}} (the discrete topology)


(wikipedia: Finite topological space)

The second and third of these are topologies, but are not power sets of X.

Afaict if all the elements of X are elements of the topology then there is only one topology, the discreet, identical to the powerset of X.

• Note: It's the discrete topology. Discreet is if you don't talk about sensitive things. – celtschk Jun 25 at 4:15