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COMPACT, a tutorial by Scott W. Williams 

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Disjoint or nonintersecting closed sets one of which is compact in a (Hausdorff) topological space can be expanded to disjoint open sets  this is called separating them.
The distance, in the plane, between the disjoint closed sets, graphs of y=0 and xy=1 is 0. Neither of these sets is compact and is necessary for this process. Suppose H and K are disjoint closed sets of a space X. If we can not expand these two to disjoint open sets then there must be some kind of "hole" present: Consider the Tychonov Plank, the space X is the product with the upper right hand corner removed; i.e., . The closed sets are the top A= and the right hand side B=.
{Key to the proof is that the hole at the end of A (see 1.4.5) is so far away that any open set containing B contains countable sequences converging to A; i.e., limit points belonging to A. Thus, there can be no disjoint open sets containing A and B.}
Even stronger (superficially) than "expansion of disjoint closed sets to disjoint open sets" is "separation of closed sets by a continuous function."
3.1. THEOREM. If A and B are disjoint closed subsets of a compact space X there is a continuous f:X[0,1] such that f(A)=0 and f(B)=1.
Let us return to products: Fréchet was the first to define a finite product of topological spaces [Fréchet1910]. That the product of two compact spaces are compact is intuitively clear: A "hole" in the product of X and Y ought to imply it in a factor. Induction shows "two" can be replaced by "finitely many." But what about "infinitely many?"
In 1923 Tietze first gave the definition of a topology, called the box topology, on a product of infinitely many spaces to be that which is generated by the product of open sets [Tietze1923]:
This definition is the "right" way to define product in many areas of mathematics (example, in Algebra). However, even the product of countably many two element sets is not compact.
Further, an important old (at least the 1960's and prehaps the 1920's) and major unsolved problem [Williams1984] in topology asks,
Does the product (with the box topology) of countably many copies of [0,1] satisfy the conclusion of Theorem 3.1 ?
The answer to large products in general is "NO!"; i.e., there are A and B are disjoint closed subsets of the product of uncountably many copies of [0,1] for which no continuous function f:X[0,1] satisfies both f(A)=0 and f(B)=1 [Lawrence1994].
Clearly, Tietze's topology is not good for proving theorems in infinite products (e.g., the preservation of compactness, connectedness, metric etc.), and thus we use a product topology [Tychonov1930] which, like compactness, extends finite delicately  the topology is generated by a product of open sets which, only finitely often, may be different from the entire factor:
This guarantees that a hole in the product must come from a hole in at least one factor and the standard:
3.2. THEOREM. The product of arbitrarily many compact spaces are compact.
page 1: the beginnings 
page 2: a special example 
page3: functional separation 

page4: the universe in a box 
page5: open covers 