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Let $\{(X_\alpha,\mathscr{T}_\alpha):\alpha\in\Lambda\}$ be an indexed family of topological spaces, and for each $\alpha\in\Lambda$ let $f_\alpha:X\to X_\alpha$ be a function. Furthermore, let $\mathscr{T}$ be the weak topology on X induced by $\{f_\alpha:\alpha\in\Lambda\}$. Then $\mathscr{L}=\{f_\alpha^{-1}(U_\alpha):\alpha\in\Lambda; U_\alpha\in\mathscr{T}_\alpha\}$ is a subbasis for $\mathscr{T}$.

I have tried this: Since $f_\alpha$ is continuous therefore $f_\alpha^{-1}(U_\alpha)$ is open then, intersections of elements of $\mathscr{L}$ are open. Let $\mathscr{B}$ be the set of all of those finite intersections then if $B_1,B_2\in\mathscr{B}$ and $x\in B_1\cap B_2$, ($x\in X$), then there exist $B\in\mathscr{B}$ such that $x\in B$ and $B\subseteq B_1\cap B_2$. I need to prove that $X=\cup\{B:B\in\mathscr{B}\}$.

  • First I would like to know some important properties of a weak topology.
  • And of course I would like to see a proof of this theorem please.
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2 Answers

It looks to me as if what you are calling "weak topology" in your question is commonly referred to as initial topology.

As for the question why the set you describe is a subbase for the weak topology on $X$: I am not sure there is anything to show. That it is a subbase means that it generates the (weak) topology that is, that the weak topology is the smallest topology containing $\mathscr L$ as a subset. But if every map $f_\alpha$ is to be continuous, the topology must contain at least all sets in $\mathscr L$. But the weak topology is defined to be the smallest topology such that all $f_\alpha$ are continuous. Hence they have to be equal.

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Hi, thanks for your answer. When you say "they have to be equal" who do you mean? –  Fernando Valle Jan 1 '13 at 13:39
@FernandoValle I mean the weak topology and the topology generated by $\mathscr L$. –  Matt N. Jan 1 '13 at 13:41
@Fernando: Certainly: by definition a topology is closed under taking finite intersections and arbitrary unions. –  Brian M. Scott Jan 1 '13 at 14:33
That’s the usage in the topological vector space setting, but the term is also used in general topology. The terms initial topology and final topology weren’t in common use when I learned my topology; I suspect that they come from category theory, or at least from people with a category-theoretic background. –  Brian M. Scott Jan 1 '13 at 16:16
Good! finally I understanding, I thank both for your answer. –  Fernando Valle Jan 2 '13 at 2:31
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For a set $X$ with maps $f_\alpha: X \rightarrow X_\alpha$, where $X_\alpha$ are topological spaces, the weak topology is the smallest topology on $X$ such that the $f_\alpha$ are continuous (explicitly, one can take it to be the intersection of all topologies on $X$ for which $f_\alpha$ are continuous).

To prove your claim, just show that $\mathcal{L}$ have to be open in any topology for which $f_\alpha$ are continuous.

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Hi thanks for you answer, I know that all the elements of $\mathscr{L}$ are open, that is a result of the continuity of $f_\alpha$ –  Fernando Valle Jan 1 '13 at 14:02
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