The previous post discusses several definitions of the tightness of a topological space. In this post, we discuss another way of characterizing tightness using the notion of free sequences.
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The main theorem
Let be a space. The tightness of , denoted by , is the least infinite cardinal number such that for each and for each , there is a set with cardinality such that . There are various different statements that can be used to define (discussed in this previous post).
A sequence of points of a space is called a free sequence if for each , . When the free sequence is indexed by a cardinal number , the free sequence is said to be of length .
The cardinal function is the least infinite cardinal number such that if is a free sequence of length , then . Thus is the least upper bound of all the free sequences of points of the space . The cardinal function is another way to characterize tightness of a space. We prove the following theorem.
Theorem 1
Let be a compact space. Then .
All spaces considered in this post are regular spaces.
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One direction of the proof
We first show that . Suppose that . We show that . Suppose not. Then there is a free sequence of points of of length greater than , say where . For each , let and .
let . By , there is some such that . Furthermore, since is a free sequence. Then choose some open such that and . Note that contains at most many points of the free sequence .
Let . The collection is an open cover of the compact space . Thus some finite is a cover of . Then all the open sets are supposed to cover all the elements of the free sequence . But each is supposed to cover at most many elements of and there are only finitely many in , a contradiction. Thus .
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Some lemmas
To show , we need some basic results technical lemmas. Throughout the discussion below, is an infinite cardinal number.
A subset of the space is a set if is the intersection of many open subsets of . Clearly, the intersection of many sets is a set.
Lemma 2
Let be any space. Let be a subset of . Then for each , there is a subset of such that is closed and .
Proof of Lemma 2
Let where each is open and is an infnite cardinal number . Note that for each , . We assume that the space is regular. We can choose open sets such that for each integer , and . Consider the following set .
The set is a subset of and . To see that is closed, note that can be rearranged as follows:
The right hand side is the intersection of closed sets, showing that a closed set. This concludes the proof of Lemma 2.
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For any set , define as follows:
In general is the part of that can be “reached” by the closure of a “small enough” subset of . Note that if and only if for each , .
For any set , define the set as follows:
A point is an accumulation point of the set if for all open set with . As a contrast, is the set of all accumulation points of . Any point is like an accumulation point of except that sets are used instead of open sets. It is clear that .
Lemma 3
Let be a compact space as before. Let be any infinite cardinal number. Let . Then .
Proof of Lemma 3
It is clear that . We only need to show . Suppose that we have and . This means there exists a subset of such that and . By Lemma 2, there is a closed subset of such that .
Since is a closed subset of a compact space and is a subset, there is a base for the set such that has cardinality (see the exercise below). For each , since is an open set containing . Choose . Let . Note that and . Thus . On the other hand, . Thus .
Let’s look at what we have. The sets and are disjoint closed sets. We also know that is a base for . There exists such that and . But and , a contradiction. Thus . This concludes the proof of Lemma 3.
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Let is an infinite cardinal number as before. Recall the concept of a closed set from this previous post. A set is a closed set if for each with , we have . Theorem 2 in the previous post states that

if and only if every closed set is closed.
This means that

if , then there is some closed set that is not closed.
The above observation will be used in the proof below. Another observation that if is a closed set, we have .
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The other direction of the proof
We now show that . First we show the following:

If , then there exists a free sequence of length where is the next cardinal number larger than .
Suppose . According to the observation on closed set indicated above, there exists a set such that is a closed set but is not closed. By another observation on closed set indicated above, we have . By Lemma 3, .
Since is not closed, choose . Then . This means the following:

For each subset of with ,
The fact indicated in (1) will make the construction of the free sequence feasible. To start the construction of the free sequence, choose . Let . Suppose that for , we have obtained and with the following properties:
 For each , is a closed subset of with ,
 For each , ,
 For all , ,
 For all , .
We now proceed to choose define and choose . Consider the set . Note that and . Thus . Since , and . By Lemma 2, there exists some closed subset of such that and . Let , which is still a closed and subset of the space . By the observation (1), . Then choose .
The construction we describe can be done for any as long as . Thus the construction yields the sequence . We now show that is a free sequence. Let . From the construction step for , we see that . From how is defined in step , we see that for any . This means that . Since is closed, . This shows that . We have shown that is a free sequence of points of .
As a result of assuming , a free sequence of length is obtained. Thus if , then . Then it must be the case that . This concludes the proof of Theorem 1.
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Remarks
The easier direction of Theorem 1, the direction for showing , does not require that the space is compact. The proof will work as long as the Lindelof degree of .
The harder direction, the direction for showing , does need the fact the compactness of the space (see the exercise below). Proving for a wider class of spaces than the compact spaces will probably require a different proof than the one given here. One generalization is found in [1]. It obtained theorem in the form of for pseudoradial regular spaces as well as other theorems with various sufficient conditions that lead to .
Theorem 1 has been applied in this blog post to characterize the normality of for any compact space .
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Exercise
Let be a compact space. Let be a closed subset of such that is the intersection of many open subsets of . Show that there exists a base for the closed set such that . To say that is a base for , we mean that is a collection of open subsets of such that each element of contains and that if with open, then for some .
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Reference
 Bella A., Free sequences in pseudoradial spaces, Commentationes Mathematicae Universitatis Carolinae, Vol 27, No 1 (1986), 163170
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