Figure 1 below is the boundary of the unit square (just the edges and not the interior), the starting point of the construction.

We draw a vertical line segment at the middle of the square in Figure 1. The middle line segment splits the square into two halves. Next we draw a vertical line segment in the middle of left half of the square, which is then split into two halves. Next we draw a vertical line segment in the middle of the resulting left half. The same process is repeated successively, each time a vertical line segment is drawn in the left half created as a result of the previous vertical line segment. Figure 2 below shows the resulting vertical line segments of the first several iterations.

We start with the corner and erase the upper horizontal edge to the next vertical line (from the point to the point ). Next, we start with the point and erase the lower horizontal edge to the next vertical line (from the point to the point ). The same process is repeated successively by alternating between the upper edge and the lower edge. Figure 3 below shows the resulting topologist’s sine curve.

The left vertical line segment in Figure 3 consists of the points where . We call the vetical line segment . We call the “sine curve” , which converges to . The topologist’s sine curve is the set , which has the topology inherited from the Euclidean plane. The space is compacted, connected. On the other hand, is both not pathwise connected and locally connected.

To see that it is connected, note that the curve is connected. Also note that the closure of is the entire topologist’s sine curve (). Since the closure of a connected set is connected, is also connected.

However, is not pathwise connected. For example, there is no connected path linking the point and . In general, there is no connected path linking any point in to any point in .

A space is locally connected at a point if there is a neighborhood base consisting of open connected sets at that point. The topologist’s sine curve is not locally connected at any point on the vertical line secgment .

An interesting subspace of the topologist’s sine curve is . This is the subset with only one point of the vertical line segment . The space is not locally compact at the point . However, is the continuous image of , demonstrating that continuous image of a locally compact space needs not be locally compact.

Figure 4 below is the extended topologist’s sine curve, which is obtained by the same process as indicated above except that no lower horizonatal edges are removed. What is interesting about the extended sine curve is that it is pathwise connected but not locally connected.

Some general observations. In general, a space is disconnected if there are two disjoint nonempty open sets and such that . We say is disconnected by and . A space is connected if no disconnection exists. A space is pathwise connected if for any two points with , there exists a continuous function such that and . The function is called a path from to .

It is well known that the Cartesian product of connected spaces is also a connected space (see [1] or [3]).

Pathwise connectedness is also preserved by taking Cartesian product. Suppose that is a pathwise connected space for each . Then is pathwise connected. To see this, suppose that with .

For each , if , choose a continuous such that and . For each , if , let be defined by for all .

We now define . For each , let be the element of the product space such that for each . It is clear that is continuous and and . In other words, the function is a path from to .

Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Steen, L. A., Seebach, J. A.,Counterexamples in Topology, 1995, Dover Edition, Dover Publications, New York.
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Let be the real line and be the set of all rational numbers. Then is the set of all irrational numbers. The Michael Line is set topologized by making be discrete and by having points in retaining the usual open intervals. The Michael Line is hereditarily paracompact. The set is not a set in both the usual topology of the real line as well as the Michael Line. Note that is a closed subset of the Michael Line. Thus the Michael Line is not perfectly normal.

The Michael Line Proof
Let and . Suppose that is not an -set in the usual topology of . Then is not normal where is considered as a subspace of the Michael Line and has the usual topology.

Let and . Both sets are closed in . We show that and cannot be separated by disjoint open sets. Let and be open sets in such that and .

For each , choose an open interval such that and . Assume that is at the center of . For each positive integer , let . We have . Since is not an -set in the usual topology of , there is some and there is a rational such that is in the Euclidean closure of .

Choose an integer such that . Choose such that . This is a crucial point that will be needed in the generalization of the proof, the point being that the second factor in the product is dense in real line.

Note that . Choose an integer such that and . Since is in the Euclidean closure of , choose such that . The point belongs to both and .

It is clear that . To see that , note that the following:

The above derivation shows that . We have since .

A Corollary to the Proof
The key characteristics of the above argument are that all isolated points in the Michael line subspace cannot form an -set in usual topology and that the complement of the limit points in the Michael Line is dense in the Euclidean topology. The following is the summarization.

Let and . Furthermore, let and . Suppose that is not an -set in the usual topology of and that is dense in the usual topology of the real line. Consider as a modified Michael Line (i.e. is discrete and points in have the usual open intervals) and having the usual Euclidean topology. Then is not normal.

The proof of this statement goes just like the argument indicated above, almost word for word. The following example is an application of this result, an example of a product of a Lindelof space and a separable metric space whose product is not Lindelof.

Example
A subset of the real line is a Bernstein set if both and have nonempty intersection with every uncountable closed subset of the real line. Clearly if is a Bernstein set, so is its complement. Such sets are constructed by transfinite induction, which goes like this. Let be all the uncountable closed subsets of the real line where each . To start, pick and . Suppose that and have been chosen for all . Then pick and , taking care to ensure that and cannot be any of the previously chosen points. The induction can go forward at each step since any uncountable closed subset of the real line has cardinality continuum. The set is a Bernstein set, as is .

Now let be a Berstein set and let . Let . Consider as a modified Michael Line where is discrete and points in have the Euclidean open interval. Then is a Lindelof space. Note that any open set (in the modified Michael Line ) containing contains all but countably many points in . Otherwise there would be a Euclidean uncountable closed set that misses .

On the other hand, any Bernstein set is dense in the Euclidean real line. It is also the case that any Bernstein set cannot be an -set in the Euclidean real line. If a Bernstein set is an in , then its complement is a dense set that would contain an uncountable closed subset of the real line.

By the above general statement, is not normal where the second factor has the usual Euclidean topology. Thus is not Lindelof.

Recall that a space is productively Lindelof if its product with any Lindelof space is Lindelof (see [1] and [3]). Any Bernstein set as a subspace of the Euclidean real line is an example of a separable metrizable space that is not a productively Lindelof space. A nice property such as being a separable metric space (or even a subset of the real line) is no guarantee of the productively Lindelof property. The Bernstein example shows that the productively Lindelof property is a very interesting property.

Reference

1. Alster, K., On spaces whose product with every Lindelof space is Lindelof, Colloq. Math. 54 (1987), 171–178.
2. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
3. Tall, F., Productively Lindelof spaces may all be D, Canad. Math. Bull. to apear.
4. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

The fact that the product of any -compact space with any Lindelof space is Lindelof is due to the Tube Lemma.

We now define a productively Lindelof space that is not -compact. Let where is any uncountable set and . The set is discrete in and open neighborhoods at have the form where is countable. The only compact subsets of this space are finite sets. Thus is not -compact.

To see that is productively Lindelof, let be any Lindelof space. Let be any open cover of . Assume that consists of open sets of the form where is open in and is open in .

There exists a countable such that covers . Suppose that . Also assume that for each , where is countable.

Note that each is a Lindelof space since it is the product of a countable space (thus -compact) with a Lindelof space. It is also clear that each point either belongs to a set in or to for some .

For each , choose countable such that covers . Then is a countable subcover of .

Reference

1. Alster, K., On spaces whose product with every Lindelof space is Lindelof, Colloq. Math. 54 (1987), 171–178.
2. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
3. Tall, F., Productively Lindelof spaces may all be D, Canad. Math. Bull. to apear.
4. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Let and be topological spaces. A slice in the Cartesian product is a subspace of the form or where and . A tube is an open subset of the Cartesian product that is of the form or where is open in and is open in . In the Euclidean plane, a slice would be either a vertical line or a horizontal line and open strips (vertical or horizontal) are examples of tubes.

Tubes are one type of open subsets of the Cartesian product . The Tube Lemma is applicable when one of the factors is compact. Let be the factor that is compact. A good way of thinking about the lemma is that when you consider the slices as “points”, the tubes , where , behave like a base. The following is a statement of the lemma.

The Tube Lemma
Let be a space and let be a compact space. For each , and for each open set of such that , there is an open set such that .

Proof. Let and let be open in the product space such that . For each , choose open sets and such that . Since is compact, we can find finitely many whose union is , say, . Let . It follows that .

Remarks
The lemma is not true when none of the factors is compact. Let and with the usual topology. Let be defined by:

The open set contains the slice . But no tube can be situated between this slice and .

The Tube Lemma can be used in proving that the product of two compact spaces is compact. By induction, it follows that the product of finitely many compact spaces is compact. However, the lemma cannot be used in proving the compactness of product space with infinitely many compact factors (the Tychonoff Theorem).

The Tube Lemma also shows that both the Lindelof property and paracompactness are preserved in taking two-factor Cartesian product as long as one of the factors is compact. As a corollary, the product of two Lindelof spaces is Lindelof if one of the factors is -compact. We have the following theorems.

Theorem 1
Let and be compact spaces. Then is compact.

Proof. Let be an open cover of . For each , let be a finite subcollection of such that is a cover of . By the Tube Lemma, there is an open set such that . Since is compact, there are finitely many such that . It follows that covers .

Theorem 2
Let be a Lindelof space and be compact spaces. Then is Lindelof.

Proof. The same proof in Theorem 1 applies except that there are countably many that cover , leading to a countable subcover of the original open cover.

Corollary 3
Let be a Lindelof space and be -compact spaces. Then is Lindelof.

Theorem 4
Let be a paracompact space and be compact spaces. Then is paracompact.

Proof. The proof begins just as in Theorem 1. Let be an open cover of . For each , let be a finite subcollection of such that is a cover of . By the Tube Lemma, there is an open set such that . Since is paracompact, let be a locally finite open refinement of such that for each .

Let . It can be shown that is a cover of , is a refinement of , and is locally finite. The first two points are clear. To show that it is a locally finite collection of sets, let . There is some open such that and meets only finitely many . Then meets only finitely many sets in .

• Footnote 2: The reader should recall that paracompact spaces are normal, and that regular Lindelof spaces are paracompact. A Lindelof space with all open subsets is hereditarily Lindelof, and conversely if is regular.
• Footnote 4: This example is new for . In contrast to this example, S. Willard has shown that, if is paracompact with Lindelof and separable, then must be Lindelof.

The title of the paper is Paracompactness and the Lindelof Property in Finite and Countable Cartesian Products. The example of the Sorgenfrey Line shows that the product of two Lindelof spaces needs not be normal. The goal of [2] is to present several examples demonstrating that higher powers of paracompact or Lindelof spaces can behave unpredictably too.

Footnote 2 gives background information about paracompact, Lindelof spaces, and hereditarily Lindelof spaces. The last sentence in the footnote is that any Lindelof space is hereditarily Lindelof if and only if it is perfectly normal. Footnote 4 provides some contrasting information to Example 1.4 in [2]. In the following discussion, all spaces are assumed to be Hausdorff and regular.

Discussion of Footnote 2
The last sentence in the footnote is essentially the following theorem:

Theorem 1
For any Lindelof space , the space is hereditarily Lindelof property if and only if is perfectly normal.

A space is perfectly normal if it is normal and that every open subspace is an set. Thus, an alternative way to check whether a Lindelof space is hereditarily Lindelof is to check whether every open subset is (or every closed subset is ). In particular, any compact space with a closed subset (or even a singleton set) that is not cannot be hereditarily Lindelof. Some examples: the unit square with the lexicographic order, the ordinal , and where is any uncountable cardinal.

To prove Theorem 1, we need the following proposition.

Proposition 1
Any space is hereditarily Lindelof if and only if every open subspace of is Lindelof.

Proof. The direction is clear. To see , let and let be an open cover of . Let be a collection of open subsets of such that for each , for some . Then is Lindelof. We can find countably many sets in whose union equals . It follows that we can find a countable subcollection of that covers .

Proof of Theorem 1. Suppose is hereditarily Lindelof. The normality of comes from the fact that is regular and Lindelof. Let be an open subset. For each , let be open such that (this comes from the fact that is a regular space). Since is Lindelof, we can find countably many such that the union of these countably many equals . This shows that every open subset of is an set.

Suppose the Lindelof space is perfectly normal. To show that is hereditarily Lindelof, it suffices to show that every open subset of is Lindelof. This follows from that fact that every open subset of is an set and that the Lindelof property is hereditary with respect to subsets.

Discussion of Footnote 4
Example 1.4 in [2] provides, under the Continuum Hypothesis, for each positive integer , a regular space such that is Lindelof and is paracompact, but is not Lindelof. For , Example 1.4 is essentially a negative answer to the question: if is paracompact and each of the factors is Lindelof, must be Lindelof? Footnote 4 in [2] says that if one of the factors is separable, then must be Lindelof. We have the following theorem.

Theorem 2
If is paracompact such that is Lindelof anf is separable, then is Lindelof.

To prove Theorem 2, we need the following two results.

Theorem 3
If is paracompact and has a dense Lindelof subspace, then must be Lindelof.

Proof. Suppose that is paracompact. Let be a dense Lindelof subspace. To show that is Lindelof, it suffices to show that every locally finite open cover of has a countable subcover.

Let be a locally finite open cover of . For each , choose open such that and only meets finitely many sets in . For each such , let be the union of the finitely many that intersect .

Since is Lindelof, we can find countably many whose union contains . Consider the countably many corresponding . We claim that the countably many that are associated with these countably many form a cover of . Let . Choose some such that . Since is dense in , choose some . Then belongs to one of the countably many that cover . Thus, is associated with one of the corresponding . This completes the proof of Theorem 3.

Theorem 4
If is Lindelof and is a compact space, then is Lindelof.

Proof. It is known that is Lindelof if one factor is Lindelof and the other factor is compact (see this previous post). As a corollary, if one of the factor is the union of countably many compact spaces, is Lindelof.

Proof of Theorem 2. Suppose that is paracompact and that is Lindelof and is separable. Let be a countable dense subset of . Then is Lindelof by Theorem 4. Furthermore, is a dense Lindelof subspace of , By Theorem 3, must be Lindelof.

Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Michael, E., Paracompactness and the Lindelof property in Finite and Countable Cartesian Products, Compositio Math. 23 (1971) 199-214.
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Theorem
Let be a completely regular space. Then the following conditions are equivalent:

1. is hereditarily separable.
2. is hereditarily Lindelof for all positive integer .
3. is hereditarily Lindelof.

As an introduction to this theorem, we present the proof to one direction of this theorem for .

Observation
Let be any completely regular space. We have the following obervation:

If is hereditarily separable, then is hereditarily Lindelof.

Suppose is not hereditarily Lindelof. We aim to show that is not hereditarily separable by producing a non-separable subspace of .

Let be a subspace that is not Lindelof. Let be a collection of open subsets of such that covers and no countable subcollection of covers .

For each , choose such that . By the completely regularity of , choose a continuous such that maps to and . Let . It can be shown that is a non-separable subspace of . That is, no countable subset of can be dense in .

For any completely regular space , it is also known (see [2]) that is separable if and only if has a weaker separable metrizable topology (i.e. has a weaker topology such that with this weaker topology is a separable metrizable space). The result in [2] combined with the observation presented here provides a way to obtain sepearable that is not hereditarily separable. Look for any that is not hereditarily Lindelof but has a weaker separable metrizable topology. One such example is the Michael Line.

The observation we make here is a rather weak result. The double arrow space is hereditarily Lindelof. Yet is not even separable since is compact space that is not metrizable. Note that is not hereditarily Lindelof since it contains a copy of the Sorgenfrey plane (see the previous post on double arrow space).

Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Noble, N., The density character of function spaces, Proc. Amer. Math. Soc. 42:1 (1974) 228-233.
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.
4. Zenor, P., Hereditarily m-separability and the hereditarily m-lindelof property in product spaces and function spaces, Fund. Math. 106 (1980), 175-180

Lemma
Any countably compact space that is countable in size is metrizable and thus first countable.

Proof. Let be countably compact such that . Then is compact (any Lindelof countably compact space is compact). In any countable space, the set of all singleton sets is a countable network. Any compact Hausdorff space with a countable network is metrizable and thus first countable. See Spaces With Countable Network.

Theorem
Let be a sequential space. Then is countably compact if and only if is sequentially compact.

Proof. The direction always holds without the space being sequential.

Suppose is countably compact. Suppose that is not sequentially compact. Then there is a sequence of points of with no convergent subsequence. Let be the set of all terms in this sequence, i.e. . Note that is sequentially closed. Since is sequential, is closed in . As a closed subset of a countably compact space, is countably compact. By the lemma, is first countable. Since is an infinite compact space, has a non-isolated point . This means some sequence of points of converges to , contradicting the assumption that has no convergent subsequence. Therefore must be sequentially compact.

Previous posts on sequential spaces and k-spaces:
Sequential spaces, I
Sequential spaces, II
Sequential spaces, III
Sequential spaces, IV
Sequential spaces, V
k-spaces, I
k-spaces, II
A note about the Arens’ space
An observation about sequential spaces

Reference

1. Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Henkel, D. Solution to Monthly Problem 5698, American Mathematical Monthly 77, p. 896, 1970
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

The first example is from the problem section of Mathematical Monthly in 1970 (see [2]). Let be the real line and let be the set of all irrational numbers. Let . Let and define a new topology on by calling a subset open if and only if where is a usual open subset of the real line and is a subset of that is at most countable. This new topology on the real line is finer than the Euclidean topology. Thus is a Hausdorff space. Every point of is a non-isolated point and is the the sequential limit of a sequence of rational numbers, satisfying the condition that every non-isolated point is the sequential limit of a non-trivial convergent sequence.

In the topology for , every countably infinite subset of the set is closed in . Thus no sequence of points of can converge to a point not in . Therefore is sequentially closed and non-closed in , making not a sequential space.

Not only that every countably infinite subset of is closed in , every countably infinite subset of is relatively discrete. Then it follows that for every compact , is finite (and is thus closed in ). Thus is also not a k-space.

Another example is that of a product space. Any uncountable product where each factor has at least two points is not sequential. This follows from the fact that is not sequential (see Sequential spaces, IV). Furthermore, in any product space with infinitely many factors each of which has at least two points, every point is the sequential limit of a non-trivial convergent sequence. Thus any product space with uncountably many factors, each of which has at least two points, is another example of a non-sequential space where there are non-trivial convergent sequences at every point.

Previous posts on sequential spaces and k-spaces:
Sequential spaces, I
Sequential spaces, II
Sequential spaces, III
Sequential spaces, IV
Sequential spaces, V
k-spaces, I
k-spaces, II
A note about the Arens’ space

Reference

1. Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Henkel, D. Solution to Monthly Problem 5698, American Mathematical Monthly 77, p. 896, 1970
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Let be the set of all nonnegative integers. Let be the set of all positive integers. In one formulation, the Arens’ space is the set with the open neighborhoods defined by:

• The points in are isolated;
• The neighborhoods at each are of the form for some ;
• The neighborhoods at are obtained by removing from finitely many and by removing finitely many isolated points in each of the remaining .

Another formulation is that of a quotient space. For each , let be a convergent sequence such that is the limit. Let be a topological sum of the convergent sequences . We then identify for each . The Arens’ space is the resulting quotient space and let denote this space (in the literature is used). Note that the Arens’ space has been previously defined in this blog (see An example of a quotient space, II). Note that the quotient space is topologically identical to . In the remainder of this note, we work with in discussing the Arens’ space.

The Arens’ space is sequential since it is a quotient space of a first countable space. The subspace is not sequential, proving that the Arens’ space is not a Frechet space.

We now show that any sequential space that is not Frechet contains a copy of the Arens’ space. We have the following theorem.

Theorem
Let be a sequential space. Then is Frechet if and only does not contain a copy of the Arens’ space.

Proof
This direction is clear since the Frechet property is hereditary.

For any , let be the set of limits of sequences of points of . Suppose is not Frechet. Then for some , there exists such that . Since is non-closed in and since is sequential, there is a sequence of points of converging to . We can assume that for all but finitely many (otherwise ). Thus without loss of generality, assume for all .

For each , there is a sequence of points of converging to . It is OK to assume that all are distinct and all are distinct across the two indexes. Let where and . Then is a homeomorphic copy of the Arens’ space.

Remark
The above theorem is not valid outside of sequential spaces. Let be a countable space with only one non-isolated point where is not sequential (for example, the subspace of the Arens’ space). Clearly contains no copy of the Arens’ space. Yet is not Frechet (it is not even sequential).

Previous posts on sequential spaces and Frechet spaces:
Sequential spaces, I
Sequential spaces, II
Sequential spaces, III
Sequential spaces, IV
Sequential spaces, V
k-spaces, I
k-spaces, II

Reference

1. Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Sequential spaces, I
Sequential spaces, II
Sequential spaces, III
Sequential spaces, IV
Sequential spaces, V
k-spaces, I

In a given space , is a -set in if where each is open in , i.e. is the intersection of countably many open sets. A point is a -set in if the singleton is the intersection of countably many open subsets of . It is a well known fact in general topology that in a compact Hausdorff space , if is a -set in , then there is a countable local base at . It follows that if every point of a compact Hausdorff space is a -set in , then is first countable (see The cardinality of compact first countable spaces, II).

Theorem
Let be a space in which every point is a -set in . Then if is a k-space then is a sequential space.

Proof. Suppose is not closed in . We show that is not sequentially closed in , i.e. there is a sequence such that and .

Since is a k-space and is not closed, there is a compact such that is not closed in . Every point of is a -set in and thus a -set in . It follows that is first countable.

Let such that (the closure is taken in ). Since is first countable, there is a sequence such that . This means is not sequentially closed in .