Can every basis act as a subbasis?

Question

Given a basis $B$ for a topology, is $B$ also a subbasis as its finite intersections of single elements will produce basis $B$ and all other finite intersections are in $B$ because $B$ is a generating topology using only arbitary unions?

Answer 1

Yes, a base $B$ is a subbase with the special property that all finite intersections of elements of $B$ already are unions of elements of $B$. This means that we don't have to do finite intersections of $B$ first to generate the topology (like with a subbase), but just use unions only.

Answer 2

Any family $F$ of subsets of a set $X$ such that $\cup F=X$ is a sub-base for a topology on $X.$ If $B$ is a base for a topology on $X$ then $\cup B=X$ so $B$ is also a sub-base for the same topology.

BTW. If $B$ is a family of subsets of $X$ then $B$ is a base for a topology on $X$ iff

(i)... $\;\cup B=X,\;$ and

(ii)... whenever $b_1, b_2\in B$ and $p\in b_1\cap b_2,$ there exists $b_3\in B$ with $p\in b_3\subset b_1\cap b_2.$

Note that condition (ii) is satisfied if $b_1\cap b_2\in B$ whenever $b_2,b_2 \in B$ and $\emptyset \ne b_1\cap b_2.$

Many authors use "base" instead of "basis".