This is another post StoneCech compactification. The links for other posts on StoneCech compactification can be found below. In this post, we prove a few basic facts about , the StoneCech compactification of the discrete space of the nonnegative integers, . We use several characterizations of StoneCech compactification to find out what is like. These characterizations are proved in the blog posts listed below. Let denote the cardinality of the real line . We prove the following facts.
 The cardinality of is .
 The weight of is .
 The space is zerodimensional.
 Every infinite closed subset of contains a topological copy of .
 The space contains no nontrivial convergent sequence.
 No point of is an isolated point.
 The space fails to have many properties involving the existence of nontrivial convergent sequence. For example:
 The space is not first countable at each point of the remainder .
 The space is not a Frechet space.
 The space is not a sequential space.
 The space is not sequentially compact.
 No point of the remainder is a point.
 The remainder does not have the countable chain condition. In fact, it has a disjoint open collection of cardinality .
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Characterization Theorems
For any completely regular space , let be the set of all continuous functions from into . The StoneCech compactification is the subspace of the product space which is the closure of the image of under the evaluation map (for the details, see Embedding Completely Regular Spaces into a Cube).
The brief sketch of we present here is not based on the definition using the evaluation map. Instead we reply on some characterization theorems that are stated here (especially Theorem U3.1). These theorems uniquely describe the StoneCech compactification of a given completely regular space . For example, satisfies the function extension property in Theorem C3 below. Furthermore any compactification of that satisfies the same property must be (Theorem U3.1). So a “C” theorem tells us a property possessed by . The corresponding “U” theorem tells us that there is only one compactification (up to equivalence) that has this property.

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Theorem C1
Let be a completely regular space. Let be a continuous function from into a compact Hausdorff space . Then there is a continuous such that .

Theorem U1
If is any compactification of that satisfies condition in Theorem C1, then must be equivalent to .
See Two Characterizations of StoneCech Compactification.
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Theorem C2
Let be a completely regular space. Among all compactifications of the space , the StoneCech compactification of the space is maximal with respect to the partial order .

Theorem U2
The property in Theorem C2 is unique to . That is, if, among all compactifications of the space , is maximal with respect to the partial order , then .
See Two Characterizations of StoneCech Compactification.
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Theorem C3
Let be a completely regular space. The space is embedded in its StoneCech compactification .

Theorem U3.1
Let be a completely regular space. Let . Let be a compactification of such that each continuous can be extended to a continuous . Then must be .

Theorem U3.2
If is any compactification of that satisfies the property in Theorem C3 (i.e., is embedded in ), then must be .
See C*Embedding Property and StoneCech Compactification.
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The following discussion illustrates how we can use some of these characterizations theorem to obtain information about and in particular.
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Result 1 and Result 2
According to the previous post (StoneCech Compactification is Maximal), we have for any completely regular space , where is the density (the smallest cardinality of a dense set in ). With being a countable space, .
Result 1 is established if we have . Consider the cube where is the unit interval . Since the product space of many separable space is separable (see Product of Separable Spaces), is separable. Let be a countable dense set. Let be a bijection. Clearly is a continuous function from the discrete space into . By Theorem C1, is extended by a continuous . Note that the image is dense in since contains the dense set . On the other hand, is compact. So . Thus is a surjection. The cardinality of is . Thus we have .
From the same previous post (StoneCech Compactification is Maximal), it is shown that . Thus . The same function in the above paragraph shows that (see Lemma 2 in StoneCech Compactification is Maximal). Thus we have
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Result 3
A space is said to be zerodimensional whenever it has a base consisting of open and closed sets. The proof that is zerodimensional comes after the following lemmas and theorems.

Theorem 1
Let be a normal space. If and are disjoint closed subsets of , then and have disjoint closures in .
Proof of Theorem 1
Let and be disjoint closed subsets of . By the normality of and by the Urysohn’s lemma, there is a continuous function such that and . By Theorem C3.1, can be extended by . Note that and . Thus .

Theorem 2
Let be a completely regular space. Let be a closed and open subset of . Then (the closure of in ) is also a closed and open set in .
Proof of Theorem 2
Let be a closed and open subset of . Let . Define by letting for all and for all . Since both and are closed and open, the map is continuous. By Theorem C3, is extended by some continuous . Note that and . Thus and have disjoint closures in , i.e. . Both and are closed and open in since .

Lemma 3
For every , (the closure of in ) is both closed and open in .
Note that Lemma 3 is a corollary of Theorem 2.

Lemma 4
Let be a set that is both closed and open in . Then where .
Proof of Lemma 4
Let . Either or . Thus . We claim that . Since , it follows that . To show , pick . If , then . So focus on the case that . It is clear that where . But every open set containing must contain some points of . These points of must be points of . Thus we have .
Proof of Result 3
Let be the set of all closed and open sets in . Let . Lemma 3 shows that . Lemma 4 shows that . Thus . We claim that is a base for . To this end, we show that for each open and for each , we can find with . Let be open and let . Since is a regular space, we can find open set with . Let .
We claim that . Suppose . There exists open such that and misses . But must meets some points of , say, . Then , which is a contradiction. So we have .
It is now clear that . Thus is zerodimensional since is a base consisting of closed and open sets.
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Result 4 and Result 5
Result 5 is a corollary of Result 4. We first prove two lemmas before proving Result 4.

Lemma 5
For each infinite , (the closure of in ) is a homeomorphic copy of and thus has cardinality .
Proof of Lemma 5
Let . Let be any function (necessarily continuous). Let be defined by for all and for all . By Theorem C3, can be extended by . Let .
Note that the function extends . Thus by Theorem U3.1, must be . Since is a countably infinite discrete space, must be equivalent to .

Lemma 6
For each countably infinite such that is relatively discrete, (the closure of in ) is a homeomorphic copy of and thus has cardinality .
Proof of Lemma 6
Let such that is discrete in the relative topology inherited from . There exist disjoint open sets (open in ) such that for each , . Since is zerodimensional (Result 3), can be made closed and open.
Let be a continuous function. We show that can be extended by . Once this is shown, by Theorem U3.1, must be . Since is a countable discrete space, must be equivalent to .
We first define by:
The function is well defined since each is in at most one . By Theorem C3, the function is extended by some continuous . By Lemma 4, for each , . Thus, for each , . Note that is a constant function on the set (mapping to the constant value of ). Thus for each . So let . Thus is the desired function that extends .
Proof of Result 4
Let be an infinite closed set. Either is infinite or is infinite. If is infinite, then by Lemma 5, is a homeomorphic copy of . Now focus on the case that is infinite. We can choose inductively a countably infinite set such that is relatively discrete. Then by Lemma 6 is a copy of that is a subset of .
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Result 6
We prove that no point in the remainder is an isolated point. To see this, pick and pick an arbitrary closed and open set with . Let (thus an arbitrary open set in the remainder containing ). By Lemma 4, where . According to Lemma 5, is a copy of and thus has cardinality . The set is minus a subset of . Thus must contains many points. This means that can never be open in the remainder . In fact, we just prove that any open and closed subset of (thus any open subset) must have cardinality at least .
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Result 7
The results under Result 7 are corollary of Result 5 (there is no nontrivial convergent sequence in ). To see Result 7.1, note that every point in the remainder is not an isolated point and hence cannot have a countable local base (otherwise there would be a nontrivial convergent sequence converging to ).
A space is said to be a Frechet space if and for each , there is a sequence of points of such that . A set is said to be sequentially closed in if for any sequence of points of , implies . A space is said to be a sequential space if is a closed set if and only if is a sequentially closed set. If a space is Frechet, then it is sequential. It is clear that is not a sequential space.
A space is said to be sequentially compact if every sequence of points in this space has a convergent subsequence. Even though is compact, it cannot be sequentially compact.
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Result 8
Result 7.1 indicates that no point of remainder can have a countable local base. In fact, no point of the remainder can be a point (a point that is the intersection of countably many open sets). The remainder is a compact space (being a closed subset of ). In a compact space, if a point is a point, then there is a countable local base at that point (see 3.1.F (a) on page 135 of [1] or 17F.7 on page 125 of [2]).
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Result 9
The space is a separable space since is a dense set. Thus has the countable chain condition. However, the remainder does not have the countable chain condition. We show that there is a disjoint collection of many open sets in .
There is a family of infinite subsets of such that for every with , is finite. Such a collection of sets is said to be an almost disjoint family. There is even an almost disjoint family of cardinality (see A Space with Gdelta Diagonal that is not Submetrizable). Let be such a almost disjoint family.
For each , let and . By Lemma 3, each is a closed and open set in . Thus each is a closed and open set in the remainder . Note that is a disjoint collection of open sets in .
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Blog Posts on StoneCech Compactification

Post #0: Embedding Completely Regular Spaces into a Cube
Post #1: A Beginning Look at StoneCech Compactification
Post #2: Two Characterizations of StoneCech Compactification
Post #3: C*Embedding Property and StoneCech Compactification
Post #4: StoneCech Compactification is Maximal
Post #5 (this post): StoneCech Compactification of the Integers â€“ Basic Facts
Post #6: StoneCech Compactifications â€“ Another Two Characterizations
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Reference
 Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
 Willard, S., General Topology, AddisonWesley Publishing Company, 1970.
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