Are the axioms of a topological space superfluous?

Question

A topology on a set $X$ is a family $\mathcal{T}$ of subsets of $X$, which are open sets and satisfy:

(1) $\emptyset, X \in \mathcal{T}$.

(2) Any union of elements of $\mathcal{T}$ belongs to $\mathcal{T}$.

(3) Any finite intersection of elements of $\mathcal{T}$ belongs to $\mathcal{T}$.

An open set $\mathcal{T}$ is defined as a set such that for every $x \in \mathcal{T}$ there is some open interval $(x-\epsilon,x+\epsilon) \subseteq \mathcal{T}$, with $\epsilon > 0$.

Then by the definition of an open set it is not difficult to argue that:

(4) The union of any family of open sets is open.

(5) Any finite intersection of open sets is open.

Does this mean that conditions (2) and (3) just follows from the fact $\mathcal{T}$ is an open set?

Furthermore, isn't it an assumption of set theory that $\emptyset$ is a subset of any set and thus automatically $\emptyset \in X$?

Answer 1

Your logic is backwards here. The three axioms you give are the axioms defining a general topological space. When you start talking about open intervals, you're talking about the set $\mathbf{R}$ (open intervals don't make sense in a general set). You're specifying a collection of sets of $\mathbf{R}$ that you want to be the opens of a topology, and because, as you observe, the union of any family of such sets, and any finite intersection of such sets, is again such a set, the collection is indeed a topology (well, once you also verify that $\mathbf{R}$ itself and $\emptyset$ satisfy the condition you give involving open intervals).

Answer 2

Your definition is wrong. The elements of $\mathcal{T}$ are the open sets by definition. It makes no sense to speak of the interval $(x-\epsilon,x+\epsilon)$ in an arbitrary set.

Answer 3

1. Not every topological space has a notion where $(x-\epsilon,x+\epsilon)$ makes sense.

2. $\cal T$ is a collection of subsets of $X$, whereas $(x-\epsilon,x+\epsilon)$ is a subset of $X$. Membership is not inclusion, usually.

3. $\varnothing$ is a subset of every set, but it's not necessarily a member of every set.

4. By definition, the (4) and (2); and (5) and (3) are equivalent. You just reformulated the statement. The reason is that the elements of $\cal T$ are defined as open sets.

Answer 4

Without a topology, an open set cannot be defined. It is important that this notion is well understood. You were talking about $\mathbb{R}^1$ - so define the usual topology $T_\mathbb{R}$ on $\mathbb{R}^1$. Then, $(x+\epsilon,x-\epsilon)$ can be define as an open set, i.e., an element of $T_\mathbb{R}$.

The elements of $\mathcal{T}$ are open sets, however, it does not make sense to say that "conditions (2) and (3) just follows from the fact $\mathcal{T}$ is an open set?", the reason being that an open set cannot be defined without first defining a topology.

Also, the empty set is not an element of any set - however, by the definition of a topology, it has to be an element of $\mathcal{T}$.