Quotient map is continuous

Question

In the above proof, I do not get how the union of equivalence classes in $U$ is open in $X$. Can someone explain this to me? Thank you

Answer 1

the quotient topology construction may be easier to follow if we introduce an explicit indexing set $A$ to label the equivalence classes as disjoint subsets of $X$. of course $A$ is equipollent to the quotient set, but it plays a distinct role in transmitting as much as possible of the topology on $X$ to its quotient $\bar X$. we "list" as marked subsets of $A$ those collections of equivalence classes which happen to form open subsets of $X$. then, having formed the quotient set, we use the indexing by $A$ to translate these open collections in $X$ into the open subsets of a topology on $\bar X$.

the equivalence relation allows us to write $X$ as a disjoint union of equivalence classes: $$ X = \bigcup_{\alpha \in A} X_{\alpha} $$ where $A$ is any indexing set of suitable cardinality.

for any subset $B \subset A$ we may say $B$ is $\tau$-compliant iff $$ \bigcup_{\alpha \in B} X_{\alpha} \in \tau $$ where $\tau$ is the collection of open sets defining the topology on $X$.

let $\bar X$ be the quotient set, with elements $\bar x_{\alpha}$ corresponding to the equivalence classes $X_{\alpha}$. we may define any subset of $\bar X$ in terms of a corresponding subset of the index set $A$.

the open sets of the quotient topology of $\bar X$ are just those subsets of $\bar X$ which are of the form: $$ \bigcup_{\alpha \in B} \{\bar x_{\alpha}\} $$ for some $\tau$-compliant subset of $A$.