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