Given a set $X$ and an indexed family $(Y_{i}, \tau_i)_{i \in I}$ of topological spaces with functions $f_{i}: X \rightarrow Y_{i}$. The initial topology $\tau$ on $X$ induced by the collection $(f_i)_{i\in I}$ is the coarsest topology on $X$ such that each $$ f_{i}:(X, \tau) \to (Y_{i}, \tau_i) $$ is continuous. Such $\tau$ does exist and $\{f_i^{-1} (O_i) \mid i\in I, O_i \in \tau_i\}$ is its subbase. I would like to prove the following characterization, i.e.,
Theorem Let $(x_d)_{d\in D}$ be a net in $X$ and $x\in X$. Then $x_d \to x$ in $\tau$ if and only if $f_i(x_d) \to f_i(x)$ for all $i \in I$.
Could you have a check on my attempt?
My attempt One direction is obvious. Let's prove the reverse. Assume $f_i(x_d) \to f_i(x)$ for all $i \in I$. Let $U$ be an open neighborhood (nbh) of $x$ in $\tau$. There exist a finite subset $J$ of $I$ and open nbh $O_i$ of $f_i (x)$ in $\tau_i$ such that $$ \bigcap_{i\in J} f_i^{-1} (O_i) \subset U. $$
Because $f_i (x_d) \to f_i (x)$, there is $d_i \in D$ such that $f(x_d) \in O_i$ for all $d \in D$ such that $d_i \le d$. We pick $\hat d \in D$ such that $d_i \le \hat d$ for all $i \in J$. Such $\hat d$ exists because $J$ is finite. Then $$ x_d \in \bigcap_{i\in J} f_i^{-1} (O_i) \subset U \quad \forall d\in D \text{ such that } \hat d \le d. $$
It follows that $x_d \to x$. This completes the proof.