Week 7

Respond to one of the following two prompts:

i) Define a topological space X that is compact and show that it is compact using the definition of compactness and/or relevant theorems.

ii) Define a topological space that is not compact and show that it is not compact using the definition of compactness and/or relevant theorems. 

In addition look at your classmates’ topological spaces. Verify their compactness or non-compactness.

For example: Every closed and bounded interval [a, b] in ℝ with the standard topology is compact.

Proof: Let Obe an open cover of [a, b].  We need to show that there is a finite subcover of [a, b].  Suppose that there is no finite subcover of [a, b].

Consider dividing the interval [a, b] in half into two half-intervals [a,a+b2] and [a+b2, b].  One of these half-intervals does not have a finite subcover from O; otherwise, [a, b] would have a finite subcover.  Let [a1, b1] be the half-interval that does not have a finite subcover from O.  We can repeat this process of dividing in half, and let [a2, b2] be the half-interval of [a1, b1] that is not finitely-coverable.  Then, we get an infinite collection of half-intervals [an, bn] that are not finitely-coverable.

We have that for each n=1, 2, 3, …:

i) [an, bn] ⊂ [an+1, bn+1]

ii) bn-an = b−a2n

iii) [an, bn] is not finitely-coverable

By Cantor's Nested Intervals Theorem (Theorem 2.11 of Croom), ∩∞n=1[an,bn] is nonempty.  Let x be in this intersection.  Then, x∈[a, b].  Since O is an open cover of [a, b], there is an open set O in O such that x∈O.  Since O is open in ℝ, there is an open interval (c, d)⊂O containing x; in particular, there is an epsilon-neighborhood (x−ε, x+ε)⊂(c, d)⊂O containing x.  

Let N be sufficiently large such that b−a2N < ε.  Since x lies in ∩∞n=1[an,bn], x lies in [aN, bN].  Note that [aN, bN]⊂(x−ε, x+ε) because bN−aN=b−a2N < ε.  So [aN, bN]⊂(x−ε, x+ε)⊂O, which implies that [aN, bN]⊂O.  That is, [aN, bN] is covered by one open set in O; in other words, [aN, bN] is finitely-coverable.  This contradicts the fact that all of the half-intervals [an, bn] are not finitely-coverable.

We can conclude that O has a finite subcover.  Therefore, every open cover of [a, b] has a finite subcover, and [a, b] is compact.