It has just occurred to me that if we choose $U$ as the basis element then $U\subset U$ and that works to show it is open.
If this were not the case (we demanded a proper subset) then we could always find a smaller basis element, thus we couldn't have a finite basis, so at some point we must accept the relation is not strict.
What you have shown is that the collection of open intervals is a basis for some topology, but not that it is a basis for the standard topology on $\mathbb R$. If $X$ is a topological space, then we say that a collection $\mathcal B$ of open sets of $X$ is a basis (of open sets) for the topology if every open set in $X$ can be written as a union of open sets drawn from $\mathcal B$.
Equivalently, $\mathcal B$ is a basis if the open subsets of $X$ are precisely those sets $U$ such that for every $x\in U$ there exists $V\in\mathcal B$ such that $x\in V\subset U$.
Using the second version, it should be clear that the collection of open intervals in $\mathbb R$ is a basis for the standard topology.
Where do the two conditions you talked about come in? Well, if $X$ is a set, then it can be shown that a collection $\mathcal B\subset\mathcal P(X)$ of subsets of $X$ is a basis of open sets for some topology on $X$ if and only if it satisfies the two conditions you mentioned:
- For every $p\in X$, there is $V\in\mathcal B$ such that $x\in V$.
- For every $U,V\in\mathcal B$, and any $q\in U\cap V$, there is $W\in\mathcal B$ such that $q\in W\subset U\cap V$. Equivalently, $U\cap V$ can be written as the union of sets drawn from $\mathcal B$.
But the topology might not be the one you are interested in. For instance, the singleton collection $\{\mathbb R\}$ satisfies conditions (1) and (2) (for $X=\mathbb R$), and it is indeed a basis of open sets for a topology on $\mathbb R$ (the indiscrete topology), but it is not a basis for the standard topology. Similarly, $\mathbb P(\mathbb R)$ satisfies the conditions for a basis, but it is a basis for the discrete topology on $\mathbb R$, not the standard topology. Finally, the collection $\mathcal B$ of all half-open intervals $[x, y)$ satisfies (1) and (2), but it is a basis for the half-open interval topology, which is not the same as the standard topology (indeed, $[0,1)$ is not open in the standard topology).
It's a good idea to work through the proofs of the claims I have made. Namely, that:
- Show that the two versions of a definition of a basis for a topology given above are equivalent.
- If $X$ is a topological space, and $\mathcal B$ is a basis of open sets for the topology on $X$, show that $\mathcal B$ satisfies (1) and (2).
- If $X$ is a set, and $\mathcal B\subset\mathcal P(X)$ satisfies (1) and (2), show that the collection of all unions of sets contained in $\mathcal B$ is a topology on $X$.
Let us know if you have any trouble with those.
Best Answer
Take an arbitrary $x \in X$. Then $x \in X \subseteq X$ (using $X$ itself as an open subset of $X$). By definition of $\mathscr C$ there is a $C \in \mathscr C$ with $x \in C$. Hence $\mathscr C$ covers $X$.