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I am reading about metrization theorems and can't figure out What is the difference between a topological space which has a $\sigma$-locally finite base to a paracompact topological space.

A topological space is called paracompact, if every cover has an open refinement which is locally finite.

A topological space has a $\sigma$-locally finite base if it has a base which is a countable union of locally finite families.

Any help? Thank you!

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The following is either quoted or paraphrased from "Counterexamples in Topology" by Lynn Arthur Steen and J. Arthur Seebach, Jr. I will skip some of the later proofs.

"A topological space is metrizable iff it is regular (T4) and has a $\sigma$-locally finite base, that is, a base which is the countable union of locally finite families. Although this requirement is very close to paracompactness, and though every metric space is paracompact, there exist regular paracompact spaces which are not metrizable." (p.37)

Thus, we need to find a space that is regular but not second countable, meaning that it does not contain a countable basis. This, in turn means the space is still paracompact, but not metrizable, and thus the space is not $\sigma$-locally finite.

The "Right Half-Open Interval Topology" on the set X of all real numbers is defined by open sets of the form $[a,b)$ with $a,b \in \Bbb R \subset X$.

The space is not second countable: Define $S:=\lbrace\left[x_i,y_i \right)| i \in \Bbb Z^+ \rbrace$. If it is a countable basis, we can find an $a \in X$ with $a \neq x_i$ for any $i \in \Bbb Z^+$. However, $[a,b)$ cannot be formed by any union of any elements of $S$. Thus, the space is not metrizable. (75-76)

The space is completely normal, which implies it is normal, which implies it is regular. Because it is Lindelöf, which means every open cover has a countable subcover, we conclude that is is paracompact. (75-76)

In sum, paracompact but does not have a $\sigma$-locally finite base, since no metric can be defined that generates the desired topology.

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  • $\begingroup$ Thank you. I am familier with the book and the example but understood your version better. The only thing I can't figure yet is how to prove that regularity implies paracompactness. If I understand correctly, you used it in line 8 of your argument. Thanks! $\endgroup$
    – user83081
    Commented Jan 19, 2014 at 13:19
  • $\begingroup$ @user83081 He is not saying that regularity implies paracompactness (it doesn't, they are like apples and elephants). That third paragraph is in fact quite clumsy. I'd rewrite it as follows: "Thus, we need to find a space that is regular but not metrisable but is paracompact. This, in turn means the space is still paracompact, but not metrizable, and thus the space is not σ-locally finite." This is achieved by the example of Right Half Open Interval Topology he mentions, not just because it is not second countable (metric spaces need not be second countable)... $\endgroup$ Commented Jan 4, 2016 at 13:09
  • $\begingroup$ ...but because it is separable (Q is dense in this topology as well) AND not second countable, while we know that in metric spaces, these two properties coincide. Hope it is clear now. $\endgroup$ Commented Jan 4, 2016 at 13:09

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