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I understand the product topology in finite dimensions but I'm having trouble visualizing it in infinite dimension. I think that the best way to understand would probably be to first do some examples. Take the countable as well as uncountable product of $\mathbb{R}$ with itself, the space of sequences and functions of $\mathbb{R}$. What sets are open in each?

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The basic open sets are sets that place "open" restrictions in only finitely many coordinates, and leave the rest off the coordinates "free". This is true whether the index is uncountable or countable. In other words, you have something which is "open" in the product topology on some finite subcollection of components, but is "everything" in the complement of those components. –  Arturo Magidin Jul 6 '12 at 22:03
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Alternately, the closed sets in the product topology are generated by products of the closed sets in each factor. Most textbooks don't mention this for some reason. –  Qiaochu Yuan Jul 6 '12 at 23:26
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The following diagram of $\Bbb R^{\Bbb N}$ may be helpful: enter image description here Each of the vertical lines represents one factor of the product and is therefore a copy of $\Bbb R$. A point $x$ in the product is a sequence $\langle x_n:n\in\Bbb N\rangle$ of real numbers, one from each factor; I’ve represented that by the jagged line that crosses the diagram from left to right. Its intersection with the $n$-th factor (vertical line) represents a single real number, $x_n$.

A basic open set in the product is determined by picking a finite set $F$ of factors, picking an open set in each, and requiring the value of $x_n$ for $n\in F$ to be in the open set chosen for that factor. For instance, in the diagram I’ve shown an open set that restricts $x_2,x_5$, and $x_8$ to the heavy black open intervals in those factors; my point $x$ is in that open set, because $x_2,x_5$, and $x_8$ do lie in the heavy black intervals. I’ve shown another basic open as well: this one restricts the possible values of the third, sixth, and seventh coordinates of a point in the product to the open intervals shown in red. My point $x$ is not in this basic open set, because $x_7$ is not in the red open interval of the seventh factor, even though it does hit the other two restrictions.

You can think of the points in the first basic open set as those whose graphs lie between the blue lines in the picture below: enter image description here

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For example:

For $R^N$, where $N$ is a countable set, Suppose $\{U_\alpha :\alpha \in F, F \subset N; F \text{ is finite}\}$ are open sets in $R$, then the set $\Pi_{\alpha\in F}U_\alpha \times R^{N-F}$ is open in $R^N$. The basic open set in $R^N$ is always of this form.

For $R^C$, where $C$ is an uncountable set, Suppose $\{U_\alpha :\alpha \in F, F \subset C; F \text{ is finite}\}$ are open sets in $R$, then the set $\Pi_{\alpha\in F}U_\alpha \times R^{C-F}$ is open in $R^C$. The basic open set in $R^C$ is always of this form.

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What is $\mathbb{R}^{N-F}$? I know what $\mathbb{R}^n$ is when n is an cardinal, but not when its a set. –  Parakee Jul 7 '12 at 0:42
    
@Parakee: $R^{N-F}=\Pi_{i \in N-F} R_i$ –  Paul Jul 7 '12 at 0:49
    
what is $\mathbb{R}_i$? –  Parakee Jul 7 '12 at 2:15
    
@Parakee: $R_i$ is $R$; however, we let $R_i$ denote the "ith" coordinate. –  Paul Jul 7 '12 at 2:39
    
So how is $\mathbb{R}^{N-F}$ not the same as $\mathbb{R}^N$? –  Parakee Jul 7 '12 at 2:56
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I think a simpler example is $\displaystyle\prod_\mathbb{N} \mathbb{F}_2$, i.e. the space of all 0-1 sequences. You can visualize this as the set of infinite paths in a rooted tree:

enter image description here

Then the basic open sets in the product topology are those paths which eventually lie in a given full subtree, i.e. in the set of all "children" of a given node.

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Wouldn't each sequence be a open set itself? –  Parakee Jul 7 '12 at 14:23
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