The basic axioms of a topological space in terms of open sets says:
A topological space is an ordered pair (X, τ), where X is a set and τ is a
collection of subsets of X, satisfying the following axioms:
a) The empty set and X itself belong to τ.
b) Any (finite or infinite) union of members of τ still belongs to τ.
c) The intersection of any finite number of members of τ still belongs to τ.
The elements of τ are called open sets and the collection τ is called a topology
on X.
No problem.
Now…The interval $[0,1)$ is neither open nor closed.
Problem: The simple proof of this shows how the open interval $(0-\epsilon,0+\epsilon)$ does not lie entirely within $[0,1)$. Now this makes sense. But this "lies entirely within" condition is not listed above. So why is it enough to state it as a proof?
Best Answer
There is another (equivalent) definition for a topology : in terms of neighborhoods (see e.g. Bourbaki General Topology Ch I). For the usual topology of $\mathbb{R}$, the neighborhoods of $x_0$ are precisely the sets which contain some $(x_0-\epsilon,x_0+\epsilon)$ and the open sets are precisely those sets $U\subset \mathbb{R}$ such that $$ (\forall x_0\in U)(\exists \epsilon>0)((x_0-\epsilon,x_0+\epsilon)\subset U) $$
the last inclusion gives a meaning to lies entirely. So, there is no contradiction.