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Let $X$ be a generic set and let $(Y_i)_i$ be a family of topological spaces. Let $(\varphi_i)_i$ be a collection of functions of the kind $X \to Y_i$. It is possible to determine a topology (that would be the coarser) in which all those functions are continuous, as follows:

1) if $\omega_i$ is an open set from some of the $Y_i$, then $\varphi_i^{-1}(\omega_i)$ is necessarily an open set (because we are supposing the functions to be continuous). So we can obtain a family $U$ of subsets in $X$ given by these pre-images.

2) We can consider finite intersections of members from $U$ obtaining a space $\phi$ that includes $U$ and that is stable under finite intersections. $\phi$ may not be stable for arbitrary unions.

3) Then we can consider the family $\mathcal{F}$ obtained by forming arbitrary unions of elements from $\phi$. It can be proven that $\mathcal{F}$ is stable under arbitrary unions and finite intersections.

The process of taking finite intersections first and then arbitrary unions cannot be reversed because we can obtain a family of subsets that is not stable under arbitrary unions.

Do you know some concrete examples showing why the "reverse" construction fails?

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    @Giuseppe: Your example (in the comment, not in the deleted answer) is right, since arbitrary unions of these sets don't produce anything new beyond unions of two of them. However, the result of first taking arbitrary unions and then finite intersections isn't a non-Hausdorff topology; it isn't a topology at all, since it's not closed under arbitary unions.2012-05-01

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There have to be at least two $Y_i$, since the preimages of open sets of a single $Y$ are already closed under arbitrary unions and finite intersections. We can construct an example using the family from Guiseppe's comment by taking $Y_1$ to be $\mathbb R$ endowed with the open sets $(-\infty,a)$ for $a\in[-\infty,\infty]$ and $Y_2$ to be $\mathbb R$ endowed with the open sets $(b,\infty)$ for $b\in[-\infty,\infty]$, with $\phi_i(x)=x$.

An example with standard topologies is $\mathbb R^2$ with $Y_1=Y_2=\mathbb R$, with $\phi_i$ the projection onto the $i$-th component. Then the standard topologies on $Y_1$ and $Y_2$ induce the standard topology on $\mathbb R$, since the subbase of preimages of open sets of $Y_1$ and $Y_2$ consists of the sets $U\times\mathbb R$ and $\mathbb R\times V$, with $U,V\subseteq\mathbb R$ an open set. Finite intersections yield sets of the form $U\times V$, and then arbitrary unions lead e.g. to

$\bigcup_{t\in\mathbb R}\left((t,\infty)\times(-\infty,t)\right)\;,$

the open set of points strictly under the main diagonal. This cannot be formed using a finite intersection of arbitrary unions, since each term of the intersection would have to cover the entire set, which implies that it must contain a set of the form $\mathbb R^2\setminus\left([-\infty,t]\times[t,\infty]\right)$, and each such set can only exclude one point of the diagonal from the intersection, so it would require (uncountably) infinitely many of them to exclude the entire diagonal.

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    Yes, now I see it. Imho it was not so easy to get - at first glance - without any additional comment.2012-05-01
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Since it's tagged functional analysis and is linked with the weak topology, I will give an example in this spirit (hoping it's correct), although I won't consider all the linear maps. Take $X=\ell^{\infty}(\Bbb R)$, the space of bounded sequence of real numbers endowed with the uniform norm. Consider $f_n$ the (continuous) linear functional defined by $f_n(\{x_n\}_k)=x_n$, and put for $r$ real number: $O_r:=\bigcup_{n\in\Bbb N}\{x\mid x_{2n} Assume that we can write $\bigcup_{r<0}(O_r\cap O'_r)$ as an union of the form $\bigcup_{j\in J}f_j^{-1}(O_j)$, where $O_j$ is an non-empty open subset of the real line and $J\subset \Bbb N$. Assume that $2j\in J$ for some $j$ (otherwise take $2k+1$ for some $k$), and $r_{2j}\in O_{2j}$. The sequence $x=r_{2j}e_{2j}$ is in $f_{2j}^{-1}(O_{2j})$ but not in $\bigcup_{r<0}(O_r\cap O'_r)$.

So after taking the arbitrary unions and finite intersections, we have to take again the arbitrary unions of such sets, which finally gives the same result.

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    In fact, what I meant in this example is that the fact to we in $\bigcup_{j\in J}f_j^{-1}(O_j)$ is determined only by the fact to be in one $f_j^{-1}(O_j)$; the choose for the other coordinates are free.2012-05-02