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spicychicken
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So Tychonoff theorem states products of compact sets are compact in the product topology.
is this true for the box topology? counterexample?
is this true for the box topology? counterexample?
spicychicken said:if S_n is the set with empty sets in each index except n where for index n you have [0,1], then {S_n} is an open cover with no finite subcover...i think
The product of compact sets in box topology is the set of all possible combinations of elements from each of the individual compact sets. It is defined as the Cartesian product of the compact sets, where each element in the product is a tuple consisting of one element from each of the individual sets.
The product of compact sets in box topology is different from other topologies because it uses the box topology, which is defined by taking the product of open sets in each of the individual topologies. This results in a topology that is finer than the product topology and coarser than the Tychonoff topology.
When compact sets are compact in the box topology, it means that the product space is also compact. This is important because compactness implies that the product space is both closed and bounded, which allows for certain theorems and properties to be applied to the product space.
The box topology can affect the convergence of sequences in the product of compact sets by making it more difficult for sequences to converge. This is because the box topology is finer than the product topology, meaning that there are more open sets in the box topology, making it easier for sequences to escape from a particular set.
Yes, the product of compact sets in box topology can be infinite-dimensional. This is because the product of compact sets is defined as the Cartesian product of the individual sets, and there is no limit to the number of sets that can be included in a Cartesian product. Therefore, the product of compact sets in box topology can be infinite-dimensional.