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let $\{X_{\alpha}\} $ be an indexed family of spaces.Let $\mathbb{T}$ be a coarsest topology such that $f_{\alpha}:A \to X_{\alpha}$ be a continuous function. Let $S_{\beta}=f^{-1}(U_{\beta})$ where $U_{\beta}$ is open in $X_{\beta}$ then show that $S=\cup S_{\beta}$ is a subbasis of the topology $\mathbb{T}$.

How do I show that $S$ satisfies the subbasis condition? If $S$ satisfies the subbasis condition of any topology $\mathbb{T'}$ then it will be easy to conclude that $\mathbb{T'} \subset \mathbb{T}$ How to proceed after this?

Arctic Char
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Guria Sona
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1 Answers1

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Your question is sloppy, notationwise. You should rather state that you have a family of maps $\{f_\alpha: A \to X_\alpha\mid \alpha \in \Lambda\}$ where all the $X_\alpha$ are topological spaces and $A$ has no topology yet.

Then $$\mathcal{S}=\{f_\alpha^{-1}[U]\mid \alpha \in \Lambda; U \subseteq X_\alpha \text{ open }\}$$

is the subbase you mean. If you need to check a condition, it's only that it covers $A$ (in my book any family of subsets of $A$ is a subbase for a topology on $A$, so there is nothing to check) and that is trivial here as for any $\alpha \in \Lambda$ we can see that $A=f^{-1}[X_\alpha] \in \mathcal{S}$ (so we do need a non-empty set of maps/spaces).

The subbase $\mathcal{S}$ generates a topology $\Bbb T$ and if $\Bbb T'$ is any topology on $A$ that makes all $f_\alpha$ continuous, then it's clear that by definition of continuity $\mathcal{S} \subseteq \Bbb T'$ and so $\Bbb T \subseteq \Bbb T'$ and indeed $\Bbb T$ is the coarsest topology with that property.

Henno Brandsma
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