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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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What about the family $\mathcal{F}=\{(-\infty, a), (b, +\infty)\mid a, b \in \mathbb{R}\}$? If you take finite intersections first and arbitrary unions second, then you get the usual topology on the line. If you do that in reverse you get a non-Hausdorff topology. Please check this last assertion as I'm not too sure about it. –  Giuseppe Negro May 1 '12 at 9:35
    
Sorry, my answer was misguided; I deleted it. –  joriki May 1 '12 at 10:47
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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. –  joriki May 1 '12 at 13:38

2 Answers 2

up vote 2 down vote accepted

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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In your first example I do not see anything really different from Giuseppe's example. If I understand, the sets one can obtain are of the form $(-\infty,a)\cup (b,+\infty)$. So, which is a set that is an arbitrary union of the sets obtained but that is not in the collection? –  Oo3 May 1 '12 at 19:34
    
@Oo3: There's no difference in the sense of different open sets; I was merely pointing out how this family of open sets could occur in your context of rendering functions continuous. Of course one can trivially get any family of open sets in that context by using one two-element $Y_i$ with topology $\{\{\},\{a\},\{a,b\}\}$ per open set, so in a sense the whole function aspect of the question is redundant and you could have just asked about subbases directly; but it seemed worthwhile to point out a slightly more natural example that leads to Guiseppe's family of open sets. –  joriki May 1 '12 at 19:55
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@Oo3: If you take unions first, you get almost nothing new; all the unions are of the forms $(-\infty,a)=\mathbb R\setminus[a,\infty)$ or $(b,\infty)=\mathbb R\setminus(-\infty,b]$ or $(-\infty,a)\cup(b,\infty)$, which is either all of $\mathbb R$ or $\mathbb R\setminus[a,b]$. Then if you form finite intersections of these, you can only remove a finite number of closed intervals, so the result is a finite union of open intervals, whereas if you do it the right way around you get arbitrary unions of open intervals. –  joriki May 1 '12 at 19:59
    
Yes, now I see it. Imho it was not so easy to get - at first glance - without any additional comment. –  Oo3 May 1 '12 at 20:24

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: $$O_r:=\bigcup_{n\in\Bbb N}\{x\mid x_{2n}<r\},\quad O'_r:=\bigcup_{n\in\Bbb N}\{x\mid x_{2n+1}<r\}.$$ 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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Maybe I'm wrong and certainly I don't know the structure of sequences in $\ell^{\infty}$ very well, but it seems possible to have some sequence having $r_ {2j}$ (a fixed negative value) with index $2j$ and another one which has the same value with index $2j-1$ or $2j+1$. So $r_{2j}$ would be surely in $\mathcal{O}_r \cap \mathcal{O}_r^\prime$ for some $r<0$. What do you think? –  Oo3 May 2 '12 at 8:32
    
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. –  Davide Giraudo May 2 '12 at 8:37

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