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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

  1. $\mathcal{T}$ makes all $f_i$ continuous.
  2. If $\mathcal{T}'$ is any other topology on $X$ that makes all $f_i$ continuous, then $\mathcal{T} \subseteq \mathcal{T}'$.
user264745
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2 Answers2

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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.

Alessandro
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  • Unfortunately I have spoiled the solution by watching Henno Brandsma post on initial topology. You have given the best hint I have ever seen. Since I already know the answer($\mathcal{T}$ is constructed by subbasis $S$ set of all inverse image of open sets of codomain). To be honest, I may not have found the subbasis without looking at the solution. Anyway your hint was just perfect to solve the problem. – user264745 Jan 24 '22 at 16:59
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    I'm glad. I don't see how the first statement is circular though. Also, maybe I'm biased because I already know the solution, but if a certain topology $\tau$ makes the functions $f_\alpha$ continuous, then $\tau$ must contain every subset of the form $f_\alpha^{-1}(U)$, for every $\alpha$ and every $U\subseteq X_\alpha$ open (this is the very definition of continuous function), so $\tau$ must contain the family $$S={f^{-1}\alpha(U)|\alpha\in J,U\text{ open in }X\alpha}$$Anyway, glad you found the solution at last – Alessandro Jan 24 '22 at 17:04
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    My bad, First statement is not circular, I interpreted differently(incorrect). – user264745 Jan 24 '22 at 17:21
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This is the so-called initial topology on $A$ wrt the maps $f_\alpha, \alpha \in J$. See my post here for an existence proof and all details and applications of this notion.

Short answer: it's the topology on $A$ with subbase $$\{f_\alpha^{-1}[O]\mid j \in J ; O \subseteq X_\alpha \text{ open }\}$$

Henno Brandsma
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  • Thank you for the answer. That post is really helpful. I don’t understand the following step in the proof of existence part: as $\mathcal{T}^\prime$ is closed under finite intersection and unions, we have that $\mathcal{T} \subseteq \mathcal{T}^\prime$. And I have one more question, we constructed $\mathcal{T}$ by subbasis but in part(a) of this exercise we’re not given any subbasis, in part(b) we have to show $\mathcal{T}$ have a subbasis $S$, is that mean we have to prove existence of initial topology, without any info about subbasis? – user264745 Jan 24 '22 at 16:47
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    @user264745 well, you can consider the set of all topologies on $A$ such that all $f_\alpha$ are continuous. This is always anon-empty set of topologies on $A$ as the discrete topology makes any function continuous. The intersection of that set of topologies is itself a topology and almost by definition minimal such topology. The subbase part is just to make the abstract definition more concrete. – Henno Brandsma Jan 24 '22 at 16:56
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    @user264745 the subbase elements are in $\mathcal{T}$ and in $\mathcal{T’}$ as well. The base elements are finite intersections so are also in both topologies. All open sets are unions of base sets so also in both. – Henno Brandsma Jan 24 '22 at 17:01
  • So you’re saying since $S\subseteq \mathcal{T}^\prime$ and $\mathcal{T}$ is topology generated by $S$, $\mathcal{T} \subseteq \mathcal{T}^\prime$. – user264745 Jan 24 '22 at 17:12
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    @user264745 yes that’s why subbases exist. They’re generating sets for their topology in that sense. – Henno Brandsma Jan 24 '22 at 17:13