Definition-related question about basis of a topology

Question

$B$ is a basis for a topology if any open set can be written as the union of elements of $B$.

If some set can be written as the union of elements of $B$, is it correct that this set is open?

In other words is $\tau = \cup B$?

Answer 1

Your first statement is not correct: $\mathscr{B}$ is a base for a topology $\tau$ on $X$ if a set $U\subseteq X$ is open if and only if $U$ can be written as a union of elements of $\mathscr{B}$. In other words, both the condition in your first sentence and the condition in your second sentence must be satisfied in order for $\mathscr{B}$ to be a base for $\tau$.

This is not the same as saying that $\tau=\bigcup\mathscr{B}$, however; in fact

$$\bigcup\mathscr{B}=\bigcup\{B:B\in\mathscr{B}\}=X\;.$$

What is true is that

$$\tau=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{B}\right\}\;.$$

Answer 2

If $B$ is a basis, then yes, any set written as a union of elements from $B$ will be open, this follows directly from the definition of a topology, as one must have that any union of open sets is open.

It is not the case that $\tau=\cup B$, rather, $\tau=\{\cup_{i\in I} K_i:K_i\in B\}$, that is, the topology is the collection of all sets which are unions of basis elements (hence why I stressed the word 'from' above; it doesn't matter how many or how few of the basis elements you union together, the resulting set is open, and hence in the topology).

Answer 3

Some authors say basis .Some say base. The English word "bases" is the plural of both words.

If $T$ is a topology on a set $S$ then $B$ is a base for $T$ iff (1) $ B\subset T $ and (2) Every $t\in T $ is equal to $\cup C$ for some $C\subset B.$

A family $B$ of subsets of a set $S$ is a base for a topology on $S$ iff (1) $ \cup B=S $ and (2) whenever $b_1,b_2 \in B$ and $x\in b_1\cap b_2$ there exists $b_3\in B$ with $x\in b_3\subset b_1\cap b_2.$

Observe that a family $B$ of subsets of $S$ which is closed under finite intersections and satisfies $\cup B=S$ is a base for a topology on $S$ but a base for a topology is not necessarily closed under finite intersections. For example if $B=\{(x,x+r)\cup (x+r,x+2r):x\in \mathbb R\land r>0\}$ then $B$ is a base for $\mathbb R,$ but if $b_1,b_2\in B$ then $b_1\cap b_2$ often fails to belong to $B.$

Observe that if $B$ is a base for the topology $T$ and $U\subset B$ then $U\subset T$ so $\cup U\in T.$

Observe that a topology may have many bases. In particular $T$ is a base for $T.$