Notes on Products

Definition:

Given topological spaces $\QTR{Large}{S}$ and $\QTR{Large}{T}$, a basis for the product topology on MATH is given by all sets of the form MATH, where MATH, and MATH are open sets.

Note that the sets MATH form a basis since MATH so the sets are in fact closed under finite intersections.

Remarks:

Given topological spaces $\QTR{Large}{S}$ and $\QTR{Large}{T}$, , let MATH and MATH be the projection functions (i.e. MATH ) The product topology on MATHcan also be described as the topology generated by all sets of the form MATH, or MATH where MATH, and MATH are open sets. By generated we mean "take arbitrary unions of finite intersections of such sets."

Note that:

  1. MATH MATH MATH

  2. MATH and MATH are continuous with respect to this topology, essentially by definition.

  3. MATH is continuous if and only if MATH and MATH are continuous.

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Homework:

Due: Thursday 20

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Lemma:

Let MATH then MATH is continuous, where MATH. Similarly for MATH

Proof:

We need only show that the inverse image of basis elements are open. But MATH or MATH depending on whether or not MATH

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Theorem:

If $\QTR{Large}{S}$ and $\QTR{Large}{T\ }$be two compact topological spaces then MATH is compact.

Proof:

In general, since any open set is the union of basis elements it suffices to show that any open covering by basis elements admits a finite subcovering. (WHY?)

Let MATH be an open covering of MATH Since $\QTR{Large}{S}$ is compact and MATH is continuous, for each MATH, we can select a finite subset MATH such that MATH covers $\QTR{Large}{S}$.

Let MATH MATH.

Note:MATH MATH

Next, since $\QTR{Large}{T}$ is compact, there is a finite subset MATH MATH MATHof the MATH that covers $\QTR{Large}{T}$. Thus

MATH covers MATH, as does MATH since MATH MATHfor each $\QTR{Large}{i}$.

MATH MATH