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I am trying to understand a counter-example showing that the box topology and product topology are not equal. Here it is:

Let $\tau$ and $\tau'$ be the product and box topologies respectively. Let $X_i = \mathbb{R}\ \forall i$ and let $U_i = (-1,1)\ \forall i$. Then $U:= \prod_{i=1}^{\infty}U_i$ and $U \in \tau'$ but $U \notin \tau$.

I don't understand why $U \notin \tau$.

Any insight would be appreciated!

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    $\begingroup$ What does a typical open in the product topology look like? Hint: There is some finiteness condition. $\endgroup$ May 20 '16 at 13:13
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    $\begingroup$ Related older question: Product and Box Topologies $\endgroup$ May 20 '16 at 17:13
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The other answers have covered this in a way, but I thought it would be better to make it explicit.

The box topology on $\Pi_i X_i$ is the topology generated by the basis of open sets of the form $\Pi_i U_i$, where $U_i\subset X_i$ is open for each $i$.

The product topology on $\Pi_iX_i$ is the topology generated by the basis of open sets of the form $\Pi_i U_i$, where $U_i\subset X_i$ is open for each $i$ and where $U_i=X_i$ for all but finitely many $i$.

Why do we define the product topology in this strange way, rather than use the more natural seeming box topology? The reason is that we want the continuous maps $Y\to\Pi_i X_i$ to be precisely those maps $f=(f_i\colon Y\to X_i)$ such that each component $f_i$ is continuous. Now in the box topology, we can have a colllection of continuous maps $f_i\colon Y\to X_i$ such that $f_i$ is continuous for each $i$, but the map $$ f=(f_i)_i\colon Y\to\Pi_iX_i $$ is not continuous. For example, let $f$ be the 'identity' map from $\Pi_iX_i$ with the product topology to $\Pi_iX_i$ with the box topology. There are sets that are open in the box topology that are not open in the product topology, so this map is not continuous. But each component of the map $f$ is projection on to one of the factors, so is continuous.

Looking at your example, we may define a function $f$ from $\mathbb R$ to $\Pi_{i=1}^\infty \mathbb R$ by setting $f(x)=(x, 2x, 3x, \dots)$. If we endow $\Pi_{i=1}^\infty\mathbb R$ with the box topology, then this map is not continuous, even though each factor $x\mapsto x,x\mapsto 2x,x\mapsto 3x,\dots$ is a continuous map.

Why is it not continuous? Well, as you said in your question, $\Pi_{i=1}^\infty (-1,1)$ is an open set in $\Pi_{i=1}^\infty$ in the box topology. But the preimage of this set under $f$ is $\{0\}$, since for any $x\ne 0$, there will be some $n$ such that $nx\not\in(-1,1)$.

Now, if we had endowed $\Pi_{i=1}^\infty\mathbb R$ with the product topology instead, we would not have run into this trouble, because $\Pi_{i=1}^\infty(-1,1)$ is not open in the product topology. Indeed, the map $f$ is continuous in the product topology.

Why is that? It suffices to check that the preimage of any basic open subset is open. Let $U=\Pi_{i=1}^\infty U_i$ be open in $\Pi_{i=1}^\infty \mathbb R$. Then $U_i=\mathbb R$ for all but finitely many $i$, so there is some $N$ such that $U_i=\mathbb R$ for all $n> N$. We may therefore write

$$ U=\Pi_{i=1}^N U_i\times\Pi_{i=N+1}^\infty\mathbb R $$

The preimage of this set under $f$ is just the set of all $x\in\mathbb R$ such that $x\in U_1, 2x\in U_2, \dots, Nx\in U_N$. This can be written as the intersection of finitely many open sets, so it is open.


To conclude, even though the definition of the product topology is (slightly) more complicated than that of the box topology, we use it more often because it is much better behaved. For example, we can show that the product of compact spaces - endowed with the product topology - is compact, while the same is not true if we endow the product with the box topology.

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  • $\begingroup$ This is such a brilliant explanation, thanks - really made me understand box and product topologies a lot more in general :) $\endgroup$
    – Joe
    May 20 '16 at 14:48
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    $\begingroup$ Thanks for your kind words. To complete your learning experience, go and prove for yourself the property of the product topology that I mentioned - namely, that continuous functions into a product are precisely the functions that are continuous in each component. At some point, you will find yourself needing to use the finiteness criterion, which means that your proof will not work with the box topology. $\endgroup$ May 20 '16 at 15:02
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Here is one way. Consider the element $x $ with $x_i=0$ for all $i $. The product topology is pointwise convergence. So if we define $x^k $ by $$ x^k_i=\begin {cases} 0,&\ i\leq k\\ i,&\ i>k\end {cases} $$ we have that $x^k\to x $.

If $U$ were open, for $k $ big enough we would have $x^k\in U $, which is not the case. So $U $ is not open.

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(This answer is similar as Martin's, but I post it anyway.)

It may be illuminating to consider an explicit example of a sequence that converges for one of the topologies, but does not for the other.

Let $e_n$ be the standard vector in $\mathbf R^\infty=\mathbf R^{\mathbf N}$ defined by $$ e_n=(0,\ldots,0,1,0,\ldots) $$ where the coordinate equal to $1$ is the $n$-th one. The sequence $\{e_n\}_{n\in\mathbf N}$ does not converge in the box topology. Indeed, if $x\in \mathbf R^\infty$ and the open $\varepsilon$-box around $x$ $$ x+\prod_{i\in\mathbf N}(-\varepsilon,\varepsilon) $$ contains infinitely many elements of the sequence $\{e_n\}$, then it should als contain at least $2$ elements of that sequence. It follows that for all $i\in\mathbf N$, there is an $e_n$ in the open $\varepsilon$-box around $x$, with $n\neq i$, and one has $$ |x_i-0|\lt\varepsilon, $$ for all $\varepsilon>0$. It follows that $x=0$. But now, no open $\tfrac12$-box around the origin contains any element of the sequence $\{e_n\}$. This shows that the sequence $\{e_n\}$ does not converge in the box topology.

However, with respect to the product topology, the sequence converges to the origin! Indeed, the subsets $$ U_j(\varepsilon)=\prod_{i=0}^j(-\varepsilon,\varepsilon)\times\prod_{i=j+1}^\infty\mathbf R $$ constitute a basis for the product topology neighborhoods of the origin. It is clear that $e_n\in U_j(\varepsilon)$ for all $n\geq j+1$, and that statement holds for all $j\in\mathbf N$ and for all $\varepsilon>0$. This proves that, indeed, $e_n\rightarrow 0$ with respect to the product topology.

It becomes even more amusing if you consider the subspace $\mathbf R^{(\mathbf N)}$ of all $\infty$-tuples $(x_i)_{i\in\mathbf N}$ for which only finitely many $x_i$ are nonzero. The induced topologies will again be called the box topology and the product topology. Of course, the sequence $\{e_n\}$ is contained in $\mathbf R^{(\mathbf N)}$. Since it does not converge for the box topology in $\mathbf R^{\mathbf N}$, it cannot converge for the induced box topology on $\mathbf R^{(\mathbf N)}$. Similarly, since the sequence $\{e_n\}$ converges to the origin with respect to the product topology on $\mathbf R^{\mathbf N}$, it also converges to the origin with respect to the induced product topology on $\mathbf R^{(\mathbf N)}$. I find this particularly striking since in $\mathbf R^{(\mathbf N)}$ one has the infinity sphere $$ S^\infty=\left\{(x_i)\mid\sum_{i=0}^\infty x_i^2=1\right\}, $$ which is a well defined subset of $\mathbf R^{(\mathbf N)}$, the increasing union of all finite-dimensional spheres $S^n$. The sequence $\{e_n\}$ is contained in $S^\infty$ and converges to the origin $0$, which does not belong to $S^\infty$ of course! The explanation is simply that $S^\infty$ is not closed with respect to the product topology, but still.

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Let $X=\prod_i X_i$ be a product of topological spaces. The product topology on $X$ is defined as the coarsest topology (i.e. the one with the least open subsets) such that each projection map $\pi_i\colon X\to X_i$ is continuous. This definition makes sense since the intersection of any number of topologies is again a topology.

In your example, assume the product topology $\tau$ on $X$ contains $U$. Then the topology generated by the set $\tau-U$ does still make all the $\pi_i$ continuous, a contradiction to the fact that $\tau$ was supposed to be the coarsest one.

Which sets are open in the product topology? Since the $\pi_i$ are supposed to be continuous, we find that for each $i_0$ and each open set $U_{i_0}$ in $X_{i_0}$, the set $U_{i_0}\times \prod_{i\neq i_0}X_i$ is open in $X$ (I am being liberal in the ordering of the product, you should imagine $U_{i_0}$ being put in the right place). Hence the product topology contains the topology generated by sets of the form $U_{i_0}\times \prod_{i\neq i_0}X_i$. Since this topology also makes the $\pi_i$ continuous, it is equal to the product topology.

Hence we see that the open sets of the product topology are sets of the form $\prod_{j=0}^{k}U_{i_j} \times \prod_{i\neq i_j} X_i$, with $U_{i_j}$ open in $X_{i_j}$. In particular, your set $U$ is not open if the product is taken over infinitely many $X_i$.

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