Does a subbasis always contains an open cover?

Question

Why does a subbasis $\gamma \subset \mathcal{T}$ of a topological space $(X,\mathcal{T})$ always have an open cover of $X$, i.e. why does there exist $U_i \in \gamma$ with $X=\bigcup\limits_{i \in I} U_i$ ?

Answer 1

In your other question Alexander's subbase theorem you point out that there are two definitions of subbasis, one not requiring the $X$ belongs to the subbasis.

If $X$ is required to belong to the subbasis (or more generally, if the union of the elements of the subbasis is required to be $X$), then the subbasis itself is an open cover. (So take your subsover $\mathcal U$ to be the same as $\gamma$.)

If the union of the elements of a subbasis $\gamma$ is NOT required to be $X$, so if $\cup\gamma\neq X$, then there is no way to find an open cover $\mathcal U\subseteq\gamma$. For any such $\mathcal U\subseteq\gamma$ we have $\cup\mathcal U\subseteq\cup\gamma\neq X$. For a specific example, take any non-empty set $X$ and let $\gamma=\emptyset$. Then $\gamma$ is a subbasis for the antidiscrete (i.e. trivial topology $\{\varnothing,X\}$) but $\cup\gamma=\varnothing\not=X$.

Answer 2

Let $B$ be the set of all finite intersections of elements of $\gamma$, so $B$ is a basis for $\mathcal{T}$. Since $\mathcal{T}$ contains $X$, and every set in $B$ is a subset of $X$, we have $\bigcup B = X$. Then $\bigcup \gamma \supseteq \bigcup B = X$.

Answer 3

In some books (e.g. Topology by James R. Munkres), the definiton of subbasis $S$ for a topology on $X$ is the collection of subsets of $X$ whose union equals $X$. From this definition we can prove that "the collection of all finite intersections elements of $S$ is a basis", which is your definition of subbasis. I mean to say that both definitions are, in fact, equivalent.