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Consider Theorem 19.1 from Munkres' topology:

The box topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$. The product topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$ and $U_\alpha$ equals $X_\alpha$ except for finitely many values of $\alpha$.

I am having a hard time seeing how the last sentence follows from the subbasis-definition of the product topology. I don't find Munkres' preceding explanation to be helpful.

Andrew Thompson
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  • What is the defining subbasis for the product topology, in this context? – Ben Grossmann Sep 15 '14 at 16:22
  • $\mathcal{S} = \cap_{\beta \in J}S_\beta$ where $S_\beta = {\pi^{-1}\beta(U\beta): | : U_\beta \text{ open in } X_\beta}$. – Andrew Thompson Sep 15 '14 at 16:24
  • Read the last paragraph of Page 116. Sometimes you just need to finish the whole chapter before you ask. – Troy Woo Sep 15 '14 at 16:24
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    The reason we require finitely many $U_\alpha$ to be proper subsets only becomes relevant when considering the product of infinitely many topological spaces. I can't recall the pathologies that occur if you allow an number of $U_\alpha$ to be proper subsets, but I would be surprised if Munkres doesn't mention it somewhere. – Benjamin Sep 15 '14 at 16:25
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    @AndrewThompson consider the fact that to generate open sets, we may take at most finite intersections of the open sets in our subbasis. – Ben Grossmann Sep 15 '14 at 16:26
  • If I recall correctly is to have Tychonoff theorem which states that the product space of compact spaces is also compact. – Julio Cáceres Sep 15 '14 at 16:30
  • @Omnom, I am aware of that, but I am still having a hard time. The subbasis is a collection of elements whose union equals all of $X$, and the topology generated by it is the collection of all unions and finite intersections of elements of the subbasis. As such, to generate open sets in the product topology, we may take arbitrary unions and finite intersections of elements of the subbasis. This is clear, and I have to admit that the desired understanding is somewhat closer now, but I'd be lying if I said that its crystal. – Andrew Thompson Sep 15 '14 at 16:35

2 Answers2

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The product topology can be described in a few equivalent ways. One of these ways is "the weakest topology on the set $X=\prod X_\alpha$, that makes all the projections $X\to X_\alpha$ continuous.

Since all the projections need to be continuous, it follows immediately that any subset of the form $\prod U_\alpha$, where $U_\alpha=X_\alpha$ for all $\alpha$ except for one $\alpha$ for which $U_\alpha\subset X_\alpha$ is any open subset, needs to be open. Taking finite intersections of all the subsets of this form yields the basis mentioned in the question, and the topology it generates has the desired property of the product.

Another way to describe the product topology is by the universal property, which is: For any topological space $A$ and continuous maps $f_\alpha:A\to X_\alpha$, there is a unique continuous map $F:A\to \prod X_\alpha$, such that for every $\alpha$ we have $f_\alpha=\pi_\alpha\circ F$.

Now take for example $\mathbb{R}^\mathbb{N}$, the product of countably many copies of the real line. Let $A=\mathbb{R}$, and for every $n\in\mathbb{N}$ let $f_n:A\to\mathbb{R}$ be given by $t\mapsto n\cdot t$. Obviously, there is exactly one way to define $F:A\to\mathbb{R}^\mathbb{N}$ that will satisfy the above property. The topology on the product needs to be such that this map will be continuous. Consider now $U=\prod_{n\in\mathbb{N}}U_n$, where for every $n$, $U_n=(-1,1).$ Obviously, $$F^{-1}(U)=\bigcap_{n\in\mathbb{N}}\left(-\frac{1}{n},\frac{1}{n}\right)=\{0\},$$which is not open. This shows that the above $U$ shouldn't be open in the product topology.

Amitai Yuval
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  • Thank you for the edit. One last bit, it may very well be identical to what you added after the edit, but can you show a case where continuity fails for some projections if $U_\alpha \ne X_\alpha$ for infinitely many? – Andrew Thompson Sep 15 '14 at 17:00
  • The projections will never fail to be continuous, since every projection ignores all the coordinates except for one. The defect that does occur is exactly the part that was added at the edit. – Amitai Yuval Sep 15 '14 at 17:06
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    This is a good way to understand "the weakest topology such that...". If you have too many open subsets in the product topology, it will not interfere with the projections being continuous, but it will not be the smallest topology such that... – Amitai Yuval Sep 15 '14 at 17:08
  • Perfectly explained, thank you. – Meitar Apr 30 '22 at 14:11
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There are two ways to view the generation of a topology from a subbasis $\mathcal S$.

  1. The topology generated by $\mathcal S$ is the smallest (coarsest) topology in which all the sets in $\mathcal S$ are open.

  2. We first transform $\mathcal S$ into a basis $\mathcal B$ consisting of all (nonempty) finite intersections of sets in $\mathcal S$, and then take the topology generated by this basis.

These two ways can be shown to be equivalent, though the second method is somewhat more illustrative in the current situation, since we outright say that the basis open sets are finite intersections of subbasic open sets. This means that $U$ is open in the generated topology if and only if for each $x \in U$ there are $W_1 , \ldots , W_n$ (for some $n \geq 1$) in $\mathcal S$ such that $x \in W_1 \cap \cdots \cap W_n \subseteq U$.

Since the subbasis for the product topology is the family of all sets $\pi_\beta^{-1} ( U )$ where $\beta$ an index, and $U$ is open in $X_\beta$, then the basic open sets are of the form $$\pi_{\beta_1}^{-1} ( U_{\beta_1} ) \cap \cdots \cap \pi_{\beta_n}^{-1} ( U_{\beta_n} ),$$ where $\beta_1 , \ldots , \beta_n$ are indexes, and $U_{\beta_i}$ is open in $X_{\beta_i}$ for all $i$. We may then note that if $\beta_i = \beta_j = \beta$, then $\pi_{\beta_i}^{-1} ( U_{\beta_i} ) \cap \pi_{\beta_j}^{-1} ( U_{\beta_j} ) = \pi_{\beta}^{-1} ( U_{\beta_i} \cap U_{\beta_j} )$, and $U_{\beta_i} \cap U_{\beta_j}$ is open in $X_\beta$, and so each index need only appear (at most) once.

Lääne-Viru
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