I am reading Topology by Munkres. The following lemma is proven:
A topology $\mathcal{T}$ equals all unions of elements of the basis $\mathcal{B}$.
The reasonign is like this: let $U \in \mathcal{T}$, choose for each $x \in U$ a basis element $B_x \in \mathcal{B}$. Then:
$$U = \bigcup_{x \in U} B_x$$
I do not understand the final step. Shouldn't it be:
$$U \subseteq \bigcup_{x \in U} B_x$$
Consider a topology on $X$ with a single basis element $B = X$. Then $U \subset B$, if $U \neq X$. Alternatively, pick an open subset of a basis: $U \subset B_i$. Then $U \subset \bigcup_{x \in U} B_x = B_i$.
Munkres defines a basis on $X$ as a collection $\mathcal{B}$ of subsets $B_i$ of $X$ satisfying:
$$\forall x \in X \Rightarrow \exists B_i: x \in B_i$$
$$x \in B_1 \cap B_2 \Rightarrow \exists B_3: x \in B_3 \subset B_1 \cap B_2$$
Best Answer
A basis $\cal B$ for a particular topology $\cal T$ is a basis that further has the property that for each $U \in \cal T$ and for every $x \in U$ there exists $B \in \cal B$ with $x \in B \subset U$. Thus you would have $B_x \subset U$ for all $X$ and consequently $$\bigcup_{x \in U} B_x \subset U$$too.