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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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    Never mind, I see now. Thanks all!2012-10-18

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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!2012-10-18
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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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    If $U \notin \mathscr{T}$ then U is closed or neither closed nor open, but in the example I assume the set closed... And $\mathscr{T}$ is fixed.2012-10-19