1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Where do you get the idea that "the infinite intersection of open sets is closed"? $\endgroup$
    – bof
    Commented Jun 19, 2019 at 3:32
  • $\begingroup$ Thats right, I rephrased the question $\endgroup$
    – Link L
    Commented Jun 19, 2019 at 6:17
  • $\begingroup$ The (Tychonoff) product topology is compact IF every $O_i$ is compact. This is called The Tychonoff Theorem. $\endgroup$ Commented Jun 19, 2019 at 6:18
  • $\begingroup$ 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). $\endgroup$ Commented Jun 19, 2019 at 6:24
  • $\begingroup$ 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 $\endgroup$
    – Link L
    Commented Jun 19, 2019 at 6:28

2 Answers 2

3
$\begingroup$

$(1).$ 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).

$\endgroup$
4
  • $\begingroup$ 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 $j$? $\endgroup$
    – Link L
    Commented Jun 20, 2019 at 5:31
  • $\begingroup$ 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) $\endgroup$
    – Link L
    Commented Jun 20, 2019 at 5:34
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jun 20, 2019 at 6:04
  • $\begingroup$ Ok thanks I guess that goes along with my 2nd comment $\endgroup$
    – Link L
    Commented Jun 20, 2019 at 7:47
0
$\begingroup$

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 $U_i$ .

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .