# Can we define topology recursively?

Let $X$ be a topological space with a fixed topology $\mathscr{T}$. We know that the following are equivalent for all $U \subseteq X$.

1. $U \in \mathscr{T}$.
2. For all $x \in U$ there is $U_{x} \in \mathscr{T}$ such that $x \in U_{x} \subseteq U$.

My question is, is it okay to define topology as follows?

Definition. Let $X$ be a set. A subset of power set $\mathscr{T} \subseteq \mathcal{P}(X)$ is called a topology on $X$ if for every $U \in \mathscr{T}$ the following is true: for all $x \in U$, there exists $U_{x} \in \mathscr{T}$ such that $x \in U_{x} \subseteq U$.

If this is okay, I don't know what made most of people introduce topology with three axioms. Is it because this definition is self-referencing? Before I sat in my introductiory topolgy course, I always thought that this was going to be introduced in the beginning of the class rather than more axiomatic definition.

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What is $U$ in your definition? – Chris Eagle Oct 18 '12 at 22:57
Your definition does not require the union of open sets to be open. Isn't this something desirable? – Andrés E. Caicedo Oct 18 '12 at 22:57
The role of axioms in mathematics is to give a rigorous framework to describe properties of objects. Some long time ago, people figured what are the properties we expect open sets to have, and ended up with a very concise way of saying that. Not everything needs to be recursive, and not everything can be recursive. – Asaf Karagila Oct 18 '12 at 23:00
@Chris Thanks, I meant to say for all $U \in\mathscr{T}$. Let me fix it right away. – user123454321 Oct 18 '12 at 23:02
@AndresCaicedo - the definition had a typo. Sorry for the confusion. – user123454321 Oct 18 '12 at 23:04

## 2 Answers

No. For one, any covering of $X$ would be a topology. Alternatively, the set of singletons (not the discrete topology) would be a topology on every space by your logic.

You need to be able to take unions and stay inside the topology to do anything remotely interesting.

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That makes sense. So recursive definition has more sets in $\mathscr{T}$. What a stupid question. I totally forgot the fact that we are already given more information that $\mathscr{T}$ is a topology in the very first logical equivalence. Thanks! – user123454321 Oct 18 '12 at 23:15

This definition only guarantee you that all $U \subset \mathscr{T}$ are open, this is because in the definition of topology is important:

1. Both the empty set and $X$ are elements of $\mathscr{T}$
2. Any union of elements of $\mathscr T$ is an element of $\mathscr{T}$
3. Any intersection of finitely many elements of $\mathscr{T}$ is an element of $\mathscr{T}$

This is no recursively. And the first part tell you that if you have any set, maybe a closed set, you can find a open set of any point of that set that belong to the fixed topology.

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I don't think this answer is relevant to what I asked. See anon's answer. – user123454321 Oct 18 '12 at 23:19
"And the first part tell you that if you have any set, maybe a closed set, you can find a open set of any point of that set that belong to the fixed topology." - What do you mean? – user123454321 Oct 19 '12 at 0:17
If U is open then $U \in \mathscr{T}$, but if U is closed for all $x \in U$, you can find $U_{x} \in \mathscr{T}$ which is open by definition. – José Ramírez Oct 19 '12 at 0:26
Note that I did not use the definition of the word "closed" as one should not use it as we are discussing the definition of the word "open." – user123454321 Oct 19 '12 at 3:17
I used the set closed as a example and it's evident the definition. – José Ramírez Oct 19 '12 at 3:56