Let $A$ be a set; let $\{X_\alpha \}_{\alpha \in J}$ be an indexed family of spaces; and let $\{ f_\alpha \}_{\alpha \in J}$ be an indexed family of functions $f_\alpha \colon A \to X_\alpha$.
(a) Show there is a unique coarsest topology $\mathscr{T}$ on $A$ relative to which each of the functions $f_\alpha$ is continuous.
I can’t solve this problem. Maybe I don’t understand “exactly” what to prove. For instance, closure $\overline{A}$ is the smallest closed subset of $X$ containing $A$ means (1) $\overline{A}$ is closed in $X$ and it contains $A$, and (2) if $A\subseteq V$ where $V$ is closed in $X$ then $\overline{A} \subseteq V$. You can also give hint(s) to solve this problem.
Edit: The following definition I sto.. ahaa aaa took from Henno Brandsma answer
Let $(X,\mathcal{T})$ be a topological space.
Let $I$ be an index set, and let $Y_i (i \in I)$ be topological spaces
and let $f_i: X \rightarrow Y_i$ be a family of functions.
Then $\mathcal{T}$ is called the initial topology with respect to the maps $f_i$
iff
- $\mathcal{T}$ makes all $f_i$ continuous.
- If $\mathcal{T}'$ is any other topology on $X$ that makes all $f_i$ continuous, then $\mathcal{T} \subseteq \mathcal{T}'$.
Best Answer
You need to show that there is a unique topology $\tau$ on $A$ such that every map $f_\alpha: A\rightarrow X_\alpha$ is continuous and $\tau$ is contained in every other topology making the maps $f_\alpha$ continuous (all at once).
You can construct $\tau$ by noticing that a topology makes each $f_\alpha$ continuous if and only if it contains a certain family $S$ of subsets of $A$, can you see what family?
Once you've proved the previous result, you're also done: This family $S$ will not be a topology in general, so you take the topology $\tau$ generated by $S$. This $\tau$ contains $S$, so it makes every $f_\alpha$ continuous and, by construction, is contained in every other topology making $f_\alpha$ continuous.