Certain covering properties and separation properties allow open covers to shrink, e.g. paracompact spaces, normal spaces, and countably paracompact spaces. The shrinking property is also interesting on its own. This post gives a more indepth discussion than the one in the previous post on countably paracompact spaces. After discussing shrinking spaces, we introduce three shrinking related properties. These properties show that there is a deep and delicate connection among shrinking properties and normality in products. This post is also a preparation for the next post on Dowker space and Morita’s first conjecture.
All spaces under consideration are Hausdorff and normal or Hausdorff and regular (if not normal).
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Shrinking Spaces
Let be a space. Let be an open cover of . The open cover of is said to be shrinkable if there is an open cover of such that for each . When this is the case, the open cover is said to be a shrinking of . If an open cover is shrinkable, we also say that the open cover can be shrunk (or has a shrinking). Whenever an open cover has a shrinking, the shrinking is indexed by the open cover that is being shrunk. Thus if the original cover is indexed in a certain way, e.g. , then a shrinking has the same indexing, e.g. .
A space is a shrinking space if every open cover of is shrinkable. The property can also be broken up according to the cardinality of the open cover. Let be a cardinal. A space is shrinking if every open cover of cardinality for is shrinkable. A space is countably shrinking if it is shrinking.
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Examples of Shrinking
Let’s look at a few situations where open covers can be shrunk either all the time or on a limited basis. For a normal space, certain covers can be shrunk as indicated by the following theorem.
Theorem 1
The following conditions are equivalent.
 The space is normal.
 Every pointfinite open cover of is shrinkable.
 Every locally finite open cover of is shrinkable.
 Every finite open cover of is shrinkable.
 Every twoelement open cover of is shrinkable.
The hardest direction in the proof is , which is established in this previous post. The directions are immediate. To see , let and be two disjoint closed subsets of . By condition 5, the twoelement open cover has a shrinking . Then and . As a result, and . Since the open sets and cover the whole space, and are disjoint open sets. Thus is normal.
In a normal space, all finite open covers are shrinkable. In general, an infinite open cover of a normal space does not have to be shrinkable unless it is a pointfinite or locally finite open cover.
The theorem of C. H. Dowker states that a normal space is countably paracompact if and only every countable open cover of is shrinkable if and only if the product space is normal for every compact metric space if and only if the product space is normal. The theorem is discussed here. A Dowker space is a normal space that violates the theorem. Thus any Dowker space has a countably infinite open cover that cannot be shrunk, or equivalently a normal space that forms a nonnormal product with a compact metric space. Thus the notion of shrinking has a connection with normality in the product spaces. A Dowker space space was constructed by M. E. Rudin in ZFC [2]. So far Rudin’s example is essentially the only ZFC Dowker space. This goes to show that finding a normal space that is not countably shrinking is not a trivial matter.
Several facts can be derived easily from Theorem 1 and Dowker’s theorem. For clarity, they are called out as corollaries.
Corollary 2
 All shrinking spaces are normal.
 All shrinking spaces are normal and countably paracompact.
 Any normal and metacompact space is a shrinking space.
For the first corollary, if every open cover of a space can be shrunk, then all finite open covers can be shrunk and thus the space must be normal. As indicated above, Dowker’s theorem states that in a normal space, countably paracompactness is equivalent to countably shrinking. Thus any shrinking space is normal and countably paracompact.
Though an infinite open cover of a normal space may not be shrinkable, adding an appropriate covering property to any normal space will make it into a shrinking space. An easy way is through pointfinite open covers. If every open cover has a pointfinite open refinement (i.e. a metacompact space), then the pointfinite open refinement can be shrunk (if the space is also normal). Thus the third corollary is established. Note that the metacompact is not the best possible result. For example, it is known that any normal and submetacompact space is a shrinking space – see Theorem 6.2 of [1].
In paracompact spaces, all open covers can be shrunk. One way to see this is through Corollary 2. Any paracompact space is normal and metacompact. It is also informative to look at the following characterization of paracompact spaces.
Theorem 3
A space is paracompact if and only if every open cover of has a locally finite open refinement such that for each .
A proof can be found here. Thus every open cover of a paracompact space can be shrunk by a locally finite shrinking. To summarize, we have discussed the following implications.
Diagram 1
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Three Shrinking Related Properties
None of the implications in Diagram 1 can be reversed. The last implication in the diagram cannot be reversed due to Rudin’es Dowker space. One natural example to look for would be spaces that are normal and countably paracompact but fail in shrinking at some uncountable cardinal. As indicated by the the theorem of C. H, Dowker, the notion of shrinking is intimately connected to normality in product spaces . To further investigate, consider the following three properties.
Let be a space. Let be an infinite cardinal. Consider the following three properties.
The space is shrinking if and only if any open cover of cardinality for the space is shrinkable, i.e. the following condition holds.

For each open cover of , there exists an open cover such that for each .
The space has Property if and only if every increasing open cover of cardinality for the space is shrinkable, i.e. the following holds.

For each increasing open cover of , there exists an open cover such that for each .
The space has Property if and only if the following holds.

For each increasing open cover of , there exists an increasing open cover such that for each .
A family is increasing if for any . It is decreasing if for any .
In general, any space that is shrinking for all cardinals is a shrinking space as defined earlier. Any space that has property for all cardinals is said to have property . Any space that has property for all cardinals is said to have property .
The first property shrinking is simply the shrinking property for open covers of cardinality . The property is shrinking with the additional requirement that the open covers to be shrunk must be increasing. It is clear that shrinking implies property . The property appears to be similar to except that has the additional requirement that the shrinking is also increasing. As a result implies . The following diagram shows the implications.

Diagram 2
The implications in Diagram 2 are immediate. An example is given below showing that shrinking does not imply property . If , then all three properties are equivalent in normal spaces, as displayed in the following diagram. The proof is in Theorem 5.

Diagram 3
The property has a dual statement in terms of decreasing closed sets. The following theorem gives the dual statement.
Theorem 4
Let be a normal space. Let be an infinite cardinal. The following two properties are equivalent.
 The space has property .
 For each decreasing family of closed subsets of such that , there exists a family of open subsets of such that and for each .
First bullet implies second bullet
Let be a decreasing family of closed subsets of with empty intersection. Then is an increasing family of open subsets of where . Let be an open cover of such that for each . Then where is the needed open expansion.
Second bullet implies first bullet
Let be an increasing open cover of . Then is a decreasing family of closed subsets of where . Note that . Let be a family of open subsets of such that and for each . For each , there is open set such that since is normal. For each , let . Then is a family of open subsets of required by the first bullet. It is a cover because . To show , let such that . Then . Since and is open, . Let . Since , , which means , a contradiction. Thus .
Now we show that the three properties in Diagram 3 are equivalent.
Theorem 5
Let be a normal space. Then the following implications hold.
shrinking Property Property shrinking
Proof of Theorem 5
shrinking Property
Suppose that is shrinking. By Dowker’s theorem, is a normal space. We can think of as a convergent sequence with as the limit point. Let be an increasing open cover of . Define and as follows:
It is straightforward to verify that and are disjoint closed subsets of . By normality, let and be disjoint open subsets of such that and . For each integer , define as follows:
The set consists of all integers and the limit point . From the way the sets are defined, is an increasing open cover of . The remaining thing to show is that for each . Suppose that and . Then by definition of . There exists an open set such that and . Since is an open set containing , . Let . By definition of , there is some open set such that and , a contradiction since is supposed to miss . Thus for all integers .
The direction Property Property is immediate.
Property shrinking
Consider the dual condition of in Theorem 4, which is equivalent to shrinking according to Dowker’s theorem.
Remarks
The direction shrinking Property is true because shrinking is equivalent to the normality in the product . The same is not true when becomes an uncountable cardinal. We now show that shrinking does not imply in general.
Example 1
The space is the set of all ordinals less than with the ordered topology. Since it is a linearly ordered space, it is a shrinking space. Thus in particular it is shrinking. To show that does not have property , consider the increasing open cover where for each . Here consists of all ordinals less than . Suppose has property . Then let be an increasing open cover of such that for each .
Let be the set of all limit ordinals in . For each , and thus . Thus there exists a countable ordinal such that misses points in . Thus the map is a pressing down map. By the pressing down lemma, there exists some such that is a stationary set in , which means that intersects with every closed and unbounded subset of . This means that for each , would miss . This means that for each , . As a result would not be a cover of , a contradiction. So does not have property .
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Property
Of the three properties discussed in the above section, we would like to single out property . This property has a connection with normality in the product (see Theorem 7). First, we prove a lemma that is used in proving Theorem 7.
Lemma 6
Show that the property is hereditary with respect to closed subsets.
Proof of Lemma 6
Let be a space with property . Let be a closed subspace of . Let be an increasing open cover of . For each , let be an open subset of such that . Since the open sets are increasing, the open sets can be chosen inductively such that for all . This will ensure that will form an increasing cover.
Then is an increasing open cover of where . By property , let be an increasing open cover of such that . For each , let . It can be readily verified that is an increasing open cover of . Furthermore, for each (closure taken in ).
Let be an infinite cardinal. Let be a discrete space of cardinality . Let be a point not in . Let . Define a topology on by letting be discrete and by letting open neighborhood of be of the form where and has cardinality less than . Note the similarity between and the convergent sequence in the proof of Theorem 5.
Theorem 7
Let be a normal space. Then the product space is normal if and only if has property .
Remarks
The property involves the shrinking of any increasing open cover with the added property that the shrinking is also increasing. The increasing shrinking is just what is needed to show that disjoint closed subsets of the product space can be separated.
Notations
Let’s set some notations that are useful in proving Theorem 7.
 The set is an open set in containing the point and is defined as follows.
 .
 For any two disjoint closed subsets and of the product space , define the following sets.
 For each , let and .
 Let and .
 For each , choose open such that , and (due to normality of ).
 Choose open such that , and (due to normality of ).
Proof of Theorem 7
Suppose that has property . Let and be two disjoint closed sets of . Consider the following cases based on the locations of the closed sets and .

Case 1. and .
Case 2a.
Case 2b. Exactly one of and intersect the set .
Case 3. Both and intersect the set .
Remarks
Case 1 is easy. Case 2a is the pivotal case. Case 2b and Case 3 use a similar idea. The result in Theorem 7 is found in [1] (Theorem 6.9 in p. 189) and [4]. The authors in these two sources claimed that Case 2a is the only case that matters, citing a lemma in another source. The lemma was not stated in these two sources and the source for the lemma is a PhD dissertation that is not readily available. Case 3 essentially uses the same idea but it has enough differences. For the sake of completeness, we work out all the cases. Case 3 applies property twice. Despite the complicated notations, the essential idea is quite simple. If any reader finds the proof too long, just understand Case 2a and then get the gist of how the idea is applied in Case 2b and Case 3.
Case 1.
and .
Let . It is clear that and .
Case 2a.
Assume that . We now proceed to separate and with disjoint open sets. For each , define as follows:
Then is an increasing open cover of . By property , there is an increasing open cover of such that for each . The shrinking allows us to define an open set such that and .
Let . It is clear that . Next, we show that . Suppose that . Then . As a result, . Let be open such that and . Since for all , it follows that for all . It is clear that for all . What has been shown is that there is an open set containing the point that contains no point of . This means that . We have established that .
Case 2b.
Exactly one of and intersect the set . We assume that is the set that intersects the set . The only difference between Case 2b and Case 2a is that there can be points of outside of in Case 2b.
Now proceed as in Case 2a. Obtain the open cover , the open cover and the open set as in Case 2a. Let . It is clear that . We claim that . Suppose that . Since (as in Case 2a), there exists open set such that and . There also exists open such that and . It is clear that for all . This means that is an open set containing the point such that misses the open set . Thus .
Case 3.
Both and intersect the set .
Now project and onto the space .
Note that is simply the copy of and is the copy of in . Since is normal, choose disjoint open sets and such that and .
Let and . Let and . Note that is closed in , is open in and . Similarly is closed in , is open in and .
We now define two increasing open covers using property . Define and and and as follows:
The open cover is an increasing open cover of . The open cover is an increasing open cover of .By property of and , both covers have the following as shrinking (by Lemma 6). The two shrinkings are:
such that
for each and such that both and are increasing open covers. Note that the closure is taken in and the closure is taken in .
For each , let be the interior of and be the interior of (with respect to ). Note that is meaningful since is a subset of the closure of the open set . Similar observation for . To make the rest of the argument easier to see, note the following fact about and .

(closure with respect to )
(closure with respect to )
For each , choose open set such that
The last point is possible because misses and . Define the open sets and as follows:
It is clear that . We claim that . To this end, we show that if , then . If , then either for some or .
Let . Note that . Since , . Choose an open set such that and misses . Note that misses for all since for all . It is clear that misses for all .
We can also choose open such that and misses . It is clear that misses for all . Thus there is an open set containing the point such that contains no point of .
Let . First we find an open set containing such that misses . From the way the open sets are defined, it follows that for all . Furthermore . Thus is the desired open set. On the other hand, there exists such that . Note that are chosen so that for all . Since for all , for all . Thus the open set contains no points of for any . Then the open set contains no point of . This means that . Thus .
In each of the four cases (1, 2a, 2b and 3), there exists an open set such that and . This completes the proof that is normal assuming that has property .
Now the other direction. Suppose that is normal. Then it can be shown that has property . The proof is similar to the proof for shrinking Property in Theorem 5.
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Reference
 Morita K., Nagata J.,Topics in General Topology, Elsevier Science Publishers, B. V., The Netherlands, 1989.
 Rudin M. E., A Normal Space for which is not Normal, Fund. Math., 73, 179486, 1971. (link)
 Rudin M. E., Dowker Spaces, Handbook of SetTheoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 761780.
 Yasui Y., On the Characterization of the Property by the Normality of Product Spaces, Topology and its Applications, 15, 323326, 1983. (abstract and paper)
 Yasui Y., Some Characterization of a Property, TSUKUBA J. MATH., 10, No. 2, 243247, 1986.
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