# box topology vs product topology

Given a countably infinite family of topological spaces $$(X_1,\tau_1),..,(X_n,\tau_n),...$$, and their product $$X$$, I read that the box topology has as its basis the family:

$$B' = \prod_{i=1}^{\infty}O_i : O_i \in \tau_i$$

... and that the box topology is not compact, but the product topology is compact. The example where box topology is not compact is when each $$(X_i,\tau_i)$$ is a compact finite discrete space - which leads to a non-compact infinite discrete topology over $$X$$.

But under the product topology (where $$O_i\neq X_i$$ for only finitely many $$i$$ and all $$X_i$$ are compact), suppose that we use the basis $$(O_1,...,O_{i-1},U_i,O_{i+1},...)$$, where $$U_i \neq X_i$$, but $$O_j=X_j$$ for all $$j \neq i$$. A countable intersection of these bases is $$(U_1,U_2,U_3,...,U_n,...)$$ which can possibly lead to an infinite set of open covers with no finite subcovers.

But am I right that $$(U_1,U_2,U_3,...,U_n,...)$$ is no longer open in the product topology? --i.e. since the infinite intersection of open sets is not necessarily open. This implies that we won't get the 'non-compactness' of a box topology.

• Where do you get the idea that "the infinite intersection of open sets is closed"?
– bof
Commented Jun 19, 2019 at 3:32
• Thats right, I rephrased the question Commented Jun 19, 2019 at 6:17
• The (Tychonoff) product topology is compact IF every $O_i$ is compact. This is called The Tychonoff Theorem. Commented Jun 19, 2019 at 6:18
• The family of sets of the form $(O_1,...,O_{i-1},U_i,...)$ that you describe, is not a base (basis) but is a sub-base (sub-basis). Commented Jun 19, 2019 at 6:24
• Ok that came to mind, but shouldnt those sets meet the conditions for a base under product topology? (p.s. i also rephrased the question to make each $X_i$ compact Commented Jun 19, 2019 at 6:28

$$(1).$$ Unless some $$U_i$$ is empty, the set $$U=\prod_{i\in \Bbb N}U_i$$ is not open in the product topology:

Suppose $$U$$ is open and $$p\in U.$$ Then $$p\in C\subset U$$ for some basic open set $$C=\prod_{i\in \Bbb N}C_i$$ where each $$C_i$$ is non-empty & open in $$X_i$$ and the set $$D=\{i:C_i\ne X_i\}$$ is finite.

The key is that D is not empty. Consider, for some (any) $$j\in D,$$ the projection $$p_j:\prod_{i\in \Bbb N} X_i\to X_j$$ where $$p_j(x_1, x_2, x_3,...)=x_j.$$ The image $$p_j(C)$$ of $$C$$ under $$p_j$$ is a subset of the image $$p_j(U)$$ because $$C\subset U.$$ But then $$X_j=p_j(C)\subset p_j(U)=U_j,$$ which is false.

In other words: If $$U$$ is not empty then for any $$j,$$ the set of the $$j$$-th co-ordinates of the members of $$U$$ is $$U_j,$$ but if $$U$$ is open and non-empty then for any $$j\in D,$$ the set of the $$j$$-th co-ordinates of the members of $$U$$ is $$X_j.$$

$$(2).$$ It is possible that the box topology is compact. For example if $$X_i$$ is the only non-empty open subset of each $$X_i$$ then the only non-empty open subset of $$X=\prod_{i\in \Bbb N}X_i$$ in the box topology is $$X$$ itself.

On the other hand, suppose each $$X_i$$ is a two-point discrete space (which is obviously compact). Then the box topology on $$\prod_{i\in \Bbb N}X_i$$ is an infinite discrete space (which is obviously not compact).

• Just a question, i read that a topology can include any intersection of its elements, which include the basic open sets. In what you wrote the contradiction happened because $p_j(C) \not \subseteq p_j(U)$ - and $C = X_j$ for some $j$.... But if any intersection is allowed, does it not include infinite intersections s.t. we obtain an infinite set s.t. $C \neq X_j$ for any $j$? Commented Jun 20, 2019 at 5:31
• Or does the rule that any intersection of basic open sets is a union of basic open sets take precedence... such that there is only a finite number of $C_j \neq X_j$ for any possible intersection? (which in turn limits the number of possible intersections to only finite intersections in the product topology) Commented Jun 20, 2019 at 5:34
• An intersection of FINITELY MANY open sets is open. An intersection of infinitely many open sets might or might not be open.... The $union$ of any collection of open sets is open. Commented Jun 20, 2019 at 6:04
• Ok thanks I guess that goes along with my 2nd comment Commented Jun 20, 2019 at 7:47

Assume $$U=(U_1,U_2,U_3,...,U_n,...)$$ is open , then there must be an topology basis $$O \subset U$$ . This leads to a contradiction since only finte coordinate of $$O$$ can be subset of $$U_i$$ .