Notes on the Definition of Compact Spaces

Definition:

In a topological space $\QTR{Large}{T}$, a collection $\QTR{Large}{B}$ of open subsets of $\QTR{Large}{T}$ is said to form a basis for the topology on $\QTR{Large}{T}$ if every open subset of $\QTR{Large}{T\ }$can be written as a union of sets in $\QTR{Large}{B}$.

Definitions:

That is, given any open covering of $\QTR{Large}{T}$, it is possible select a finite subset of the open covering, that still forms an open covering of $\QTR{Large}{T}$.

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Remarks:

  1. Any basis is an open covering since $\QTR{Large}{T}$ itself is an open set and hence must be the union of basis elements

  2. An open covering may not be a basis . For example, $\QTR{Large}{\{T\}}$ is always an open covering of $\QTR{Large}{T.}$

  3. MATHis not compact. MATH is an open covering without a finite refinement.

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Theorem:

Every closed subspace $\QTR{Large}{S}$ of a compact space $\QTR{Large}{T\ }$is also compact.

Proof:

Let MATH be an open covering of $\QTR{Large}{S}$. Let MATH be a collection of open sets in $\QTR{Large}{T}$ such that MATH.

Since $\QTR{Large}{S}$ is closed MATH is a covering of $\ \QTR{Large}{T}$.

Since $\QTR{Large}{T}$ is compact, let MATH and possibly $\QTR{Large}{T-S}$ is a finite subcovering of $\QTR{Large}{T}$. Hence MATHis a finite subcovering of $\QTR{Large}{S}$.

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Definition:

A subset MATHis bounded if it is contained in some closed $\QTR{Large}{n}$-cube. That is MATH

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Theorem(Heine-Borel):

A subset MATHis compact if and only if is closed and bounded.

Proof:

We will show that MATH is compact. Thus, since $\QTR{Large}{S}$ is a closed subset of a compact space it is compact.

Now suppose MATHis compact, we need to show it is closed and bounded.

To show it is closed, let MATH for each point MATH we can find MATH and MATH such

that MATH.

Since MATH is an open cover of $\QTR{Large}{S}$ we can find a finite subcover MATH.

And since MATH and MATH, MATH is a open set containing $\QTR{Large}{x}$ in MATH. Finally, since is MATH is arbitrary, MATH is open so is $\QTR{Large}{S}$ closed.

To show $\QTR{Large}{S}$ is bounded, fix MATH and consider the cover MATH. Choose a finite subcover MATH. Let MATH

Homework due March 16:

Show that if MATH MATH and hence $\QTR{Large}{S}$ is bounded.

Proof:

Assume MATH and MATH.

Then MATH