k-spaces, I

There are many examples in general topology of defining a new topology on a set based on a given topology already defined on the set. For example, given a topology \tau on X, one can define a finer topology \tau_s consisting of all sequentially open subsets of X (based on the original topology \tau). Sequential spaces are precisely those spaces for which the original topology coincides with \tau_s (see the post Sequential spaces, II). A related concept is the notion of k-spaces. We show that the compactly generated open sets form a finer topology and that k-spaces are precisely those spaces for which the compactly generated topology coincides with the original topology. We also give an external characterization of k-spaces, namely, those spaces that are quotient images of locally compact spaces. All spaces under consideration are Hausdorff.

Let X be a space. We say A \subset X is a compactly generated closed set in X if A \cap K is closed in K for any compact K \subset X. We say B \subset X is a compactly generated open set in X if X-B is a compactly generated closed set in X. The space X is said to be a k-space if

A \subset X is a compactly generated closed set in X if and only if A is a closed set.

The direction \Leftarrow of the above statement always holds. So a space is a k-space if it satisfies the direction \Rightarrow in the above statement. We also want to mention that in the above definition, “closed” can be replaced by “open”.

Suppose \tau is the topology of the space X. Then define \tau_k as the set of all compactly generated open sets in X. It can be easily verified that \tau_k is a topology defined on the set X and that \tau_k is a finer topology than the original topology \tau, i.e. \tau \subset \tau_k. It follows that X is a k-space if and only if \tau=\tau_k.

Much of the discussion here mirrors the one in Sequential spaces, II. In a sequential space, the topology coincides with \tau_s, the open sets generated by convergent sequences (a particular type of compact sets). In a k-space, the topology conincides with \tau_k, the open sets generated by the compact sets. A sequential space is the quotient space of a topological sum of disjoint convergent sequences. Any k-space is the quotient space of a topological sum of disjoint compact sets.

We have the following theorem.

Theorem
For any space X, the following conditions are equivalent:

  1. X is a k-space.
  2. X is a quotient space of a locally compact space.

Proof. 1 \Rightarrow 2 Let X be a k-space. Let \mathcal{K} be the set of all compact subsets of X. Let Y=\oplus_{K \in \mathcal{K}}K be the topological sum of all K \in \mathcal{K} where each K \in \mathcal{K} has the relative topology inherited from the space X. Then Y is a locally compact space. There is a natural mapping we can define on Y onto X. The space Y is a disjoint union of all compact subsets of X. We can map each compact set K \in \mathcal{K} onto the corresponding compact subset K of X by the identity map. Call this mapping f where f:Y \mapsto X. We claim that the quotient topology generated by this mapping coincides with the original topology on X.

Let \tau be the given topology on X and let \tau_f be the quotient topology generated on the set X. Clearly, \tau \subset \tau_f. We need to show that \tau_f \subset \tau. Let O \in \tau_f. Since X is a k-space, if we can show that O is a compactly generated open set, then O \in \tau.

Let K \subset X be compact. We need to show that O \cap K is open in K. Since O \in \tau_f, f^{-1}(O) is open in Y. Since K is open in Y, f^{-1}(O) \cap K is open in Y. It is also the case that f^{-1}(O) \cap K is open in K. We can consider f^{-1}(O) \cap K as a subset of K \subset Y and as a subset of K \subset X. As a subset of K \subset X, we have f^{-1}(O) \cap K=O \cap K. Thus O \cap K is open in K and O \in \tau.

2 \Rightarrow 1 Let Y be locally compact and let f:Y \mapsto X be a quotient map. We show that X is a k-space. To this end, we show that if A \subset X is a compactly generated closed set in X, A is closed in X. Or equivalently, if A is not closed in X, then A is not a compactly generated closed set in X. Under the quotient map f, A is closed in X if and only if f^{-1}(A) is closed in Y.

Suppose A is not closed in X. Then f^{-1}(A) is not closed in Y. Then there is y \in \overline{f^{-1}(A)}-f^{-1}(A). Let U \subset Y be open in Y such that y \in U and \overline{U} is compact. Then f(\overline{U}) is compact. It follows that A \cap f(\overline{U}) is not closed in f(\overline{U}). Note that f(y) \in f(\overline{U}) and f(y) \notin A \cap f(\overline{U}). However, f(y) \in \overline{A \cap f(\overline{U})}. Thus A is not a compactly generated closed set in X. \blacksquare

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, II

Sequential spaces are topological spaces in which convergent sequences are sufficient to define the topology. In any topological space (X,\tau), the convergent sequences can generate a finer topology \tau_s. Sequential spaces are precisely those spaces whose topology coincide with \tau_s. We also present an external characterization of sequential spaces, namely those spaces that are quotient images of first countable spaces. This post is a continuation of Sequential spaces, I. All spaces under consideration are Hausdorff.

For any space X, A \subset X is sequentially closed in X if whenever \left\{x_n\right\}_{n=1}^{\infty} is sequence of points of A and x_n \mapsto x, then x \in A. In other words, A contains all sequential limits of convergent sequences of points of A. A space X is sequential if this condition holds: if A \subset X is sequentially closed in X, then A is closed in X.

Given a topological space (X,\tau) where \tau is the topology. Consider the following topology:

\tau_s=\left\{U \subset X: X-U \text{ is sequentially closed in }X\right\}

It is straightforward to verify that \tau_s is a topology on the set X and that it is finer than the original topology \tau, i.e. \tau \subset \tau_s. It can also be verified that X is a sequential space if and only if \tau=\tau_s. In the previous post Sequential spaces, I, all the sequential examples are quotient images of first countable spaces. It turns out that sequential spaces are precisely those spaces that are quotient images of first countable spaces. We have the following theorem.

Theorem
For any space X, the following conditions are equivalent:

  1. X is a sequential space.
  2. X is a quotient space of a metric space.
  3. X is a quotient space of a first countable space.

Proof. We prove 1 \Rightarrow 2. The direction of 2 \Rightarrow 3 is clear. The direction 3 \Rightarrow 1 following from Theorem 2 in Sequential spaces, I.

1 \Rightarrow 2 Let X be a sequential space. Let \mathcal{K} be the set of all compact subsets of X that are of the form:

K=\left\{x\right\} \cup \left\{x_n:n=1,2,3,\cdots\right\}

where x_n \mapsto x. We call such sets compact sequences. We define a topology on each K \in \mathcal{K}. We let each point x_n in the convergent sequence be isolated. The open neighborhood at the limit x is made up of x together with all but finitely many x_n. Note that the topology just described for each K is the relative topology inherited from X.

Let Y=\oplus_{K \in \mathcal{K}}K be the topological sum of all compact sequences in \mathcal{K}. This is simply the disjoint union of all the compact sequences K with each K having the relative topology inherited from X. The resulting space Y is first countable as well as a metric space. Since each point in Y belongs to some compact sequence K \subset X, we can define f:Y \mapsto X such that f maps each point in Y to the corresponding point in the sequence in X. We claim that the quotient topology generated by this map coincides with the original topology on X.

Let \tau be the given topology on X and let \tau_f be the quotient topology generated on the set X. Clearly, \tau \subset \tau_f. We need to show that \tau_f \subset \tau. Let O \in \tau_f. Since X is sequential, if we can show that for any convergent sequence that converges to a point in O, all but finitely many terms of the sequence must be in O, then O \in \tau.

Let x_n \in X such that x_n \mapsto x \in O. Let K=\left\{x\right\} \cup \left\{x_1,x_2,\cdots\right\}. We know that f^{-1}(O) is open in Y since O \in \tau_f. We know that with x as a point in K \subset Y, x \in f^{-1}(O) \cap K. Since f^{-1}(O) \cap K is open in Y, f^{-1}(O) contains all but finitely many x_n \in K. Thus f(x_n)=x_n \in O for all but finitely many n. This means O \in \tau. We just show that the quotient topology on X coincides with the given topology. \blacksquare

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

Any topological space where there is a countable base at every point is said to satisfy the first axiom of countability or to be first countable. In this post we discuss several properties weaker than the first axiom of countability. All spaces under consideration are Hausdorff. Any first countable space X satisfies each of the following conditions:

  1. If A \subset X and x \in \overline{A} then there is a sequence \left\{x_n\right\} of points in A such that the sequence converges to x.
  2. The set A \subset X is closed in X if A is sequentially closed in X, which means that: if \left\{x_n\right\} is a sequence of points in A such that \left\{x_n\right\} converges to x \in X, then x \in A.
  3. The set A \subset X is closed in X if this condition holds: if K \subset X is compact, then A \cap K is closed in K.

Spaces satisfying:

  • condition 1 are called Frechet spaces,
  • condition 2 are called sequential spaces,
  • condition 3 are called k-spaces.

All three of these conditions hold in first countable spaces. We have the following implications:

First countable \Rightarrow \ Frechet \Rightarrow \ Sequential \Rightarrow \ k-space

This post is an introductory discussion of these notions. Each of the above implications is not reversible (see the section below on examples). After we discuss sequential spaces, we take a look at the behavior of these four classes of spaces in terms of whether the property can be passed onto subspaces (the property being hereditary) and in terms of images under quotient maps. For previous posts in this blog on first countable spaces and quotient spaces, see the links at the end of the post. Excellent texts on general topology are [1] and [2].

For a subsequent discussion on sequential space, see Sequential spaces, II.

The Forward Implications

First countable \Rightarrow \ Frechet
Suppose x \in \overline{A} and A \subset X. Let U_1,U_2,\cdots be a local base at x. Then choose x_n \in A \cap U_n and we have x_n \mapsto x.

Frechet \Rightarrow \ Sequential
Let A \subset X be sequentially closed in X. Suppose A is not closed in X. Then there is x \in \overline{A} such that x \notin A. By Frechet, there is a sequence x_n \in A such that x_n \mapsto x. Since A is sequentially closed, x \in A, a contradiction. So any sequentially closed set in a Frechet space must be a closed set.

Sequential \Rightarrow \ k-space
Suppose A \subset X is not closed in X. Since X is sequential, there is a sequence x_n \in A such that x_n \mapsto x and x \notin A. The set K=\left\{x_n:n=1,2,3,\cdots\right\} \cup \left\{x\right\} is a compact set. Note that A \cap K is not closed in K. This shows that X is a k-space.

Sequential Spaces

Sequential spaces are ones in which the topology can be completely described by convergent sequences. Let X be a space. Let A \subset X. The set A is said to be sequentially closed in X if whenever we have a convergent sequence of points in A, the sequential limit must be in A. In other words, A contains all the limits of the convergent sequences of points in A. For U \subset X, U is sequentially open in X if this condition holds: if \left\{x_n \in X: n=1,2,3,\cdots\right\} is a sequence of points converging to some x \in U, then x_n \in U for all but finitely many n. It can be verified that:

U is sequentially open in X if and only if X-U is sequentially closed in X.

Theorem 1
A space X is Frechet if and only if every subspace of X is sequential.

Proof
\Rightarrow Suppose W \subset X is not a sequential space. Then there is A \subset W such that A is sequentially closed in W but A is not closed in W. We show that X is not Frechet. There is a point w \in W such that w is a limit point of A (in the subspace W) and w \notin A. Since A is sequentially closed, no sequence of points in A can converge to w (otherwise w \in A).

Clearly, the point w is also a limit point of A with respect to the toplology of X, i.e. w \in \overline{A} with respect to X. Since no sequence of points in A can converge to w, X is not Frechet.

\Leftarrow Suppose X is not Frechet. Then there is x \in \overline{A} such that A \subset X and no sequence of points in A can converge to x. Consider the subspace Y=A \cup \left\{x\right\}. The set Y-\left\{x\right\} is sequentially closed in Y but is not closed in Y. \blacksquare

Theorem 2
Every quotient space of a sequential space is always a sequential space.

Proof. Let X be a sequential space. Let f:X \mapsto Y be a quotient map. We show that Y is a sequential space. Suppose that B \subset Y is sequentially closed. We need to show that B is closed in Y. Because f is a quotient map, B is closed in Y if and only if f^{-1}(B) is closed in X. So we need to show f^{-1}(B) is closed in X.

Suppose x_n \in f^{-1}(B) for each n=1,2,3,\cdots and the sequence x_n converges to x \in X. Then f(x_n) \in B. Since the map f is continuous, f(x_n) \mapsto f(x). Since B is sequentially closed, f(x) \in B. This means x \in B. Thus f^{-1}(B) is sequentially closed in X. Since X is sequential, f^{-1}(B) is closed in X. \blacksquare

Corollary 3
Every quotient space of a first countable space is sequential. Every quotient space of a Frechet space is sequential.

Some Examples

Example 1. First countable \nLeftarrow \ Frechet
The example is defined in the post An example of a quotient space, I. This is a non-first countable example. There is only one non-isolated point p in the space. It is easy to verify it is a Frechet space.

Example 2. Frechet \nLeftarrow \ Sequential
We consider the space Y defined in the post An example of a quotient space, II. Note that the space Y is the quotient image of a first countable space. Thus Y is sequential by Corollary 3. Consider the subspace Z=\left\{(0,0)\right\} \cup V. Within Z, no sequence of points in V can converge to the point (0,0). However, (0,0) is a limit point of V. Thus Z is not sequential. By Theorem 1, Y is not Frechet.

Example 3. Sequential \nLeftarrow \ k-space
Any compact space is a k-space. Let \omega_1 be the first uncountable ordinal. Then \omega_1+1=[0,\omega_1] with the ordered topology is compact. Note that [0,\omega_1) is sequentially closed but not closed. Thus [0,\omega_1] is not sequential.

Example 4. A space that is not a k-space
This example is also defined in the post An example of a quotient space, II. Consider the subspace Z=\left\{(0,0)\right\} \cup V. Every compact subset of Z is finite. So V \cap K is closed in K for every compact K \subset Z. But V is not closed.

Comments About Subspaces

Which of the four properties discussed here are preserved in subspaces? Or which of them are hereditary? It is fairly straightforward to verify that first countability is hereditary and so is the property of being Frechet. By Theorem 1, for any sequential space that is not Frechet has a subspace that is not sequential. Thus the property of being a sequential space is not hereditary. However, closed subspaces and open subspaces of a sequential space are sequential.

The property of being a k-space is also not hereditary. The space Y defined in An example of a quotient space, II is a sequential space (thus a k-space). Yet the subspace Z=\left\{(0,0)\right\} \cup V is not a k-space.

Comments About Quotient Mappings

Continuous image of a first countable space needs not be first countable. The other three properties (Frechet, sequential and k-space) are also not necessarily preserved by continuous mappings. A quick example is to consider any space X that does not have any one of the four properties. Then consider D=X with the discrete topology. Then the indentity map from D onto X is continuous.

Example 1 shows that the property of being first countable is not preserved by quotient map. Example 2 shows that the Frechet property is not preserved by quotient map. Theorem 2 shows that the property of being sequential space is preserved by quotient map. We have the following theorem about k-spaces under quotient map.

k-spaces

The spaces that are k-spaces are called compactly generated spaces. In a k-space, the closed sets and open sets are generated by compact sets. For example, for a k-space X, A \subset X is closed in X if and only if A \cap K is closed in K for every compact K \subset X. Let’s take another look at sequential spaces. The following definition is equivalent to the definition of sequential space given above:

A \subset X is closed in X if and only if A \cap K is closed in K for every compact K \subset X of the form \left\{x\right\} \cup \left\{x_1,x_2,x_3,\cdots\right\} where the x_n are a convergent sequence and x is the sequential limit.

Thus the sequential spaces are compactly generated by a special type of compact sets, namely the convergent sequences.

Theorem 4
Quotient images of k-spaces are always k-spaces.

Proof. Let X be a k-space. Let f:X \mapsto Y be a quotient map. We wish to show that Y is a k-space. Suppose B \subset Y is not closed in Y. Since f is a quotient mapping, f^{-1}(B) is not closed in X. Since X is a k-space, there is a compact K \subset X such that f^{-1}(B) \cap K is not closed in K. Let x \in K such that x \in \overline{f^{-1}(B) \cap K}-(f^{-1}(B) \cap K). We have just produced a compact set f(K) in Y such that B \cap f(K) is not closed in f(K). Note that f(x) \in f(K) and f(x) is a limit point of B \cap f(K). This implies that if B \cap C is closed in C for every compact C \subset Y, then B must be closed in Y (i.e. Y is a k-space). \blacksquare

The discussion on sequential space continues with the post Sequential spaces, II.

Links to Previous Posts

An example of a quotient space, II
An example of a quotient space, I
The cardinality of compact first countable spaces, III
The cardinality of compact first countable spaces, II
The cardinality of compact first countable spaces, I

Reference

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

The product of first countable spaces

All spaces under consideration are Hausdorff. First countable spaces are those spaces where there is a countable local base at every point in the space. This is quite a strong property. For example, every first countable space that is also compact has a cap on its cardinality and the cap is the cardinality of the real line (the continuum). See The cardinality of compact first countable spaces, I in this blog. In fact, if the compact and first countable space is uncountable, it has cardinality continuum (see The cardinality of compact first countable spaces, III). Any metric space (or metrizable space) is first countable. In this post, we discuss the product of first countable spaces. In this regard, first countable spaces and metrizable spaces behave similarly. We show that the product of countably many first countable spaces is first countable while the product of uncountably many first countable is not first countable. For more information on the product topology, see [2].

The Product Space
Consider a collection of sets A_\alpha where \alpha \in S. Let W=\bigcup \limits_{\alpha \in S} A_\alpha. The product \prod \limits_{\alpha \in S} A_\alpha is the set of all functions f:S \mapsto W such that for each \alpha \in S, f(\alpha) \in A_\alpha. If the index set S=\left\{1,2,\cdots,n\right\} is finite, the functions f can be regarded as sequences (f_1,f_2,\cdots,f_n) where each f_i \in A_i. If the index set S=\mathbb{N}, we can think of elements f of the product as the sequence (f_1,f_2,\cdots) where each f_i \in A_i. In general we can regard f \in \prod \limits_{\alpha \in S} A_\alpha as functions f:S \mapsto W or as sequences f=(f_\alpha)_{\alpha \in S}.

Consider the topological spaces X_\alpha where \alpha \in S. Let X=\prod \limits_{\alpha \in S} X_\alpha be the product as defined above. The product space of the spaces X_\alpha is X with the topology defined in the following paragraph.

Let \tau_\alpha be the topology of each space X_\alpha, \alpha \in S. Consider Y=\prod \limits_{\alpha \in S} O_\alpha where for each \alpha \in S, O_\alpha \in \tau_\alpha (i.e. O_\alpha is open in X_\alpha) and O_\alpha=X_\alpha for all but finitely many \alpha \in S. The set of all such sets Y is a base for a topology on the product X=\prod \limits_{\alpha \in S} X_\alpha. This topology is called the product topology of the spaces X_\alpha, \alpha \in S.

To more effectively work with product spaces, we consider a couple of equivalent bases that we can define for the product topology. Let \mathcal{B}_\alpha be a base for the space X_\alpha. Consider B=\prod \limits_{\alpha \in S} B_\alpha such that there is a finite set F \subset S where B_\alpha \in \mathcal{B}_\alpha for each \alpha \in F and B_\alpha=X_\alpha for all \alpha \in S-F. The set of all such sets B is an equivalent base for the product topology.

Another equivalent base is defined using the projection maps. For each \alpha \in S, consider the map \pi_\alpha:\prod \limits_{\beta \in S} X_\beta \mapsto X_\alpha such that \pi_\alpha(f)=f_\alpha for each f in the product. In words, the function \pi_\alpha maps each point in the product space to its \alpha^{th} coordinate. The mapping \pi_\alpha is called the \alpha^{th} projection map. For each set U \subset X_\alpha, \pi_\alpha^{-1}(U) is the following set:

\pi_\alpha^{-1}(U)=\left\{f \in \prod \limits_{\beta \in S} X_\beta: \pi_\alpha(f)=f_\alpha \in U\right\}

Consider sets of the form \bigcap \limits_{\alpha \in F} \pi_\alpha^{-1}(U_\alpha) where F \subset S is finite and U_\alpha is open in X_\alpha for each \alpha \in F. The set of all such sets is another equivalent base for the product topology. If we only require that each U_\alpha \in \mathcal{B}_\alpha, a predetermined base for the coordinate space X_\alpha, we also obtain an equivalent base for the product topology.

Countable Product of First Countable Spaces
For i=1,2,3,\cdots, let X_i be a first countable space. We show that X=\prod \limits_{i=1}^{\infty}X_i is a first countable space.

Let \mathbb{N} be the set of positive integers. For each n \in \mathbb{N}, let [n]=\left\{1,2,\cdots,n\right\} and let \mathbb{N}^{[n]} be the set of all functions t:[n] \mapsto \mathbb{N}. For each i and each x \in X_i, let \mathcal{B}_x(i)=\left\{B_x(i,j): j \in \mathbb{N}\right\} be a countable local base at x.

Let f \in X=\prod \limits_{i=1}^{\infty}X_i. We wish to define a countable local base at f. For each n \in \mathbb{N}, define W_n to be:

W_n=\left\{\prod \limits_{i=1}^n B_{f(i)}(i,t(i)):t \in \mathbb{N}^{[n]}\right\}

Let W be the set of all subsets of the product space X of the following form:

\prod \limits_{j=1}^{\infty}V_j where there is some n \in \mathbb{N} such that \prod \limits_{j=1}^{n}V_j \in W_n and for all j>n, V_j=X_j.

Each W_n is countable and W is essentially the union of all the W_n. Thus W is countable. We claim that W is a local base at f. Let O \subset X be an open set containing f. We can assume that O=\prod \limits_{i=1}^\infty O_i where there is some n \in \mathbb{N} such that for each i \le n, O_i is open in X_i and for i>n, O_i=X_i.

For each i \le n, f(i) \in O_i. Choose some B_{f(i)}(i,t(i)) such that f(i) \in B_{f(i)}(i,t(i)) \subset O_i. Let V=\prod \limits_{i=1}^{\infty} V_i such that \prod \limits_{i=1}^{n} V_i=\prod \limits_{i=1}^n B_{f(i)}(i,t(i)) and V_i=X_i for all i>n. Then V \in W and f \in V \subset O. This completes the proof that X=\prod \limits_{i=1}^{\infty}X_i is a first countable space.

Uncountable Product
Let S be an uncountable index set. For \alpha \in S, let X_\alpha. We want to avoid the situation that all but countably many X_\alpha are one-point space. So we assume each coordinate space X_\alpha has at least two points, say, p_\alpha and q_\alpha with p_\alpha \ne q_\alpha. We show that X=\prod \limits_{\alpha \in S}X_\alpha is not first countable.

Let f \in \prod \limits_{\alpha \in S}X_\alpha. Let U_1,U_2, \cdots be open subsets of the product space such that for each i, f \in U_i. We show that there is some open set O such that f \in O and each U_i \nsubseteq O. For each i, there is a basic open set B_i=\bigcap \limits_{\alpha \in F_i} \pi_\alpha^{-1}(U_{\alpha,i}) such that f \in B_i \subset U_i.

Let F=F_1 \cup F_2 \cup \cdots. Since S is uncountable and F is countable, choose \gamma \in S-F. Since X_\gamma has at least two points p_\gamma and q_\gamma, choose one of them that is different from f_\gamma, say, p_\gamma. Choose two disjoint open subsets M_1 and M_2 of X_\gamma such that f_\gamma \in M_1 and p_\gamma \in M_2 of X_\gamma. Let O=\prod \limits_{\alpha \in S}O_\alpha such that O_\gamma=M_1 and O_\alpha=X_\alpha for all \alpha \ne \gamma. We have f \in O. For each i, there is g_i \in B_i \subset U_i such that g_i(\gamma)=p_\gamma. Thus each g_i \notin O. Thus there is no countable local base at f. Thus any product space with uncountably many factors, each of which has at least two points, is never first countable.

Reference

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

An example of a quotient space, II

This post presents another example of a quotient space of a first countable space. The resulting quotient space is not first countable. After we first define the space without referring to the concepts of quotient space, we show that this example is a quotient space. This example will be further discussed when we discuss sequential spaces.

For a previous discussion on quotient space in this blog, see An example of a quotient space, I. For more information on quotient spaces in general, see [2].

Let \mathbb{N} be the set of all positive integers. For each i \in \mathbb{N} and j \in \mathbb{N}, let V_{i,j}=\left\{(\frac{1}{i},\frac{1}{k}):k \ge j\right\}. Let V=\bigcup \limits_{i=1}^{\infty} V_{i,1}. Let H=\left\{(\frac{1}{i},0):i \in \mathbb{N}\right\}. For each i \in \mathbb{N}, let V_i=V_{i,1} \cup \left\{(\frac{1}{i},0)\right\}. Define the space Y as:

Y=\left\{(0,0)\right\} \cup H \cup V.

The set H is the horizontal part of the space and the set V is the vertical part of the space. The origin (0,0) is an additional point. In the topology for the space Y, each point in V is isolated. Each point (\frac{1}{i},0) \in H has an open neighborhood of the form

\left\{(\frac{1}{i},0)\right\} \cup V_{i,j} for some j \in \mathbb{N}

The open neighborhoods at (0,0) are obtained by removing finitely many V_i from Y and by removing finitely many isolated points in the V_i that remain. The open neighborhoods just defined form a base for a topology on the set Y, i.e. by taking unions of these open neighborhoods, we obtain all the open sets for this space.

We wish to discuss Y and its subspace Z=\left\{(0,0)\right\} \cup V. If we consider Z as a topological space in its own right (open sets in Z are of the form U \cap Z where U is open in Y), then in Z there are no infinite compact sets (all compact subsets of Z are finite). In particular, no sequence of points \left\{x_n \in V:n \in \mathbb{N}\right\} can converge to the point (0,0). However, with respect to Z, (0,0) is a limit point of V. This implies that the space Y is not first countable.

A space W is said to be a Frechet space if w \in \overline{A} where A \subset W, then there is some sequence \left\{w_n \in A:n \in \mathbb{N}\right\} that converges to the point w. Any first countable space is a Frechet space. If there is a local base \left\{U_n: n \in \mathbb{N}\right\} at w and if w \in \overline{A}, then whenever we pick w_n \in A \cap U_n, w_n would converge to w.

The space Y defined above is not a Frechet space. Note that (0,0) \in \overline{V} and no sequence of points in V can converge to (0,0).

Though the space Y defined above is not a first countable space, Y is a quotient space of a first countable space X (in fact, a subspace of the Euclidean plane). Let E=H \cup V where H and V are defined as in the definition of the space Y above. Let H_{-1} be the set H_{-1}=\left\{(\frac{1}{i},-1):i \in \mathbb{N}\right\}. Let X=E \cup H_{-1} \cup \left\{(0,-1)\right\} with the Euclidean topology inherited from the Euclidean plane.

Define a quotient space from X by collapsing each pair of points \left\{(\frac{1}{i},0),(\frac{1}{i},-1)\right\} into one point \left\{(\frac{1}{i},0)\right\}. This is done for each i \in \mathbb{N}. For convenience, the point \left\{(0,-1)\right\} is shifted to \left\{(0,0)\right\}. The resulting quotient space is the same as the space Y defined above.

The quotient space we just describe can also be described by the quotient map f:X \mapsto Y:

f((0,-1))=(0,0),

f((\frac{1}{i},-1))=(\frac{1}{i},0) for each (\frac{1}{i},-1) \in H_{-1},

f((\frac{1}{i},0))=(\frac{1}{i},0) for each (\frac{1}{i},0) \in H,

f(x)=x for each x \in V.

Then the following topology coincides with the topology on Y that we define earlier in the post:

\tau_f=\left\{U \subset Y: f^{-1}(U) \text{ is open in } X\right\}

Reference

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

An example of a quotient space, I

It is well known that in a compact Hausdorff space, if every point in the space is a G_\delta point, then the space is first countable (see The cardinality of compact first countable spaces, II). Outside of compact spaces, this is not true. We present an example of a space where every point is a G_\delta point but has no countable local base at any one point. This is an opportunity to introduce the notion of quotient space. Our example can be viewed as the image of a first countable space under a quotient map. We first define the example and then present a brief discussion of quotient spaces.

Let \mathbb{N} be the set of all positive integers. Let W=\left\{\frac{1}{n}: n \in \mathbb{N}\right\}. Let S_1=\mathbb{N} \times W and S_2=\mathbb{N} \times \left\{0\right\}. Consider X=S_1 \cup S_2 as a subspace of the Eucliean plane. In the Euclidean topology, each point is S_1 is isolated and an open neighborhood at each point (i,0) in S_2 is of the form:

    \left\{(i,0)\right\} \cup \left\{(i,\frac{1}{k}):k \ge n\right\} for some n \in \mathbb{N}

We use X to define a space Y by considering S_2 as one point (call this point p). Thus Y=S_1 \cup \left\{p\right\}. Points in S_1 remain isolated. An open neighborhood of p is of the form \left\{p\right\} \cup U^- where U^-=U-S_2 and U is a Euclidean open subset of X such that S_2 \subset U.

To facilitate the discussion, we describe the open sets at the point p in more details. Let \mathbb{N}^{\mathbb{N}} be the set of all functions f:\mathbb{N} \mapsto \mathbb{N}. For each i \in \mathbb{N} and j \in \mathbb{N}, let V_{i,j}=\left\{(i,\frac{1}{k}): k \ge j\right\}. An open neighborhood of p is of the following form:

    B_f=\left\{p\right\} \cup \bigcup \limits_{i=1}^{\infty} V_{i,f(i)}

To describe in words, each open neighborhood at p is obtained by removing finitely many points in each vertical segment V_{i,1}.

The resulting space Y is not first countable. We show that any countable set of open neighborhoods at p cannot be a local base at p. To this end, we consider B_{f_1},B_{f_2},B_{f_3},\cdots where each f_i \in \mathbb{N}^{\mathbb{N}}. We wish to find an open set O such that p \in O and each B_{f_i} is not a subset of O.

For each i \in \mathbb{N}, pick some g(i) \in \mathbb{N} such that g(i)>f_i(i). Consider O=B_g. Note that for each i \in \mathbb{N}, (i,\frac{1}{f_i(i)}) \in B_{f_i} and (i,\frac{1}{f_i(i)}) \notin O=B_g. This shows that no countable number of open neighborhoods can form a base at the point p.

However, every point in the space Y is a G_\delta point. For the point p, we have \left\{p\right\}=\bigcap \limits_{i=1}^\infty B_{h_i} where h_i \in \mathbb{N}^{\mathbb{N}} is a constant function mapping to i.

Quotient Spaces
We now present one definition of quotient spaces. Let (X,\tau) be a topological space. Let q:X \mapsto Y be a surjection, i.e. q(X)=Y. Consider the following collection of subsets of Y:

    \tau_q=\left\{O \subset Y: q^{-1}(O) \text{ is open in }X\right\}

The set \tau_q is a topology for the set Y. With this topology \tau_q on Y, the function q is continuous. In fact, \tau_q is the finest (largest or strongest) topology on Y that makes the function q continuous, i.e. if \tau_1 is another topology on Y making q a continuous function, then \tau_1 \subset \tau_q.

The topology \tau_q is called the quotient topology induced on Y by the mapping q. When Y is given such a quotient topology, it is called a quotient space of X (most of the times just called quotient space). The induced map q is called a quotient map. If Y has a quotient topology defined by a quotient map, then Y is said to be the quotient image of X. We plan to discuss quotient space and quotient topology in greater details in this blog. For further information, see [2].

In the example discussed in this post, the quotient map is to map each point in S_1 to itself and to map each point in S_2 to a single point p. In essence, we collapse the whole x-axis S_2 into one single point. The open sets for the point p are simply the Euclidean open sets containing S_2. This example shows that the quotient image of a first countable space needs not be first countable. Though the quotient image of a first countable space may not be first countable, it has the property that sequences suffice to define the topology (these are called sequential spaces, see [1]).

Reference

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

The cardinality of compact first countable spaces, III

In the real line every uncountable closed set has cardinality continuum, the cardinality of the real line (see Closed uncountable subsets of the real line). It follows that every uncountable compact subset of the real line has cardinality continuum. In this post we show that in any abstract topological space that is compact, Hausdorff and first countable, every uncountable compact space has cardinality continuum. In the previous post The cardinality of compact first countable spaces, I, we show that continuum is an upper bound of the cardinality of any compact, Hausdorff, first countable space. In this post, we show that continuum is a lower bound for any compact, Hausdorff, first countable space that is uncountable. This is accomplished by defining a subset that has the same candinalty as the Cantor set. Finding this “Cantor” subset is facilitated by the following lemmas. All spaces under consideration are Hausdorff.

Definitions
Let Y be a topological space. Let A \subset Y and p \in Y. The point p is said to be a complete accumulation point of A if for every open set O containing p, \lvert O \cap A \lvert=\lvert A \lvert. Since we are only interested in A being uncountable without requiring \lvert A \lvert to be any specific cardinal, we call p a complete accumulation point of A if for every open O containing p, O \cap A is uncountable. In particular, the point p a complete accumulation point of the space Y if for every open set O containing p, O \cap Y is uncountable.

Lemma 1
Let Y be a Lindelof space such that Y is uncountable. Then Y has a complete accumulation point.

Proof. Suppose Y is Lindelof and suppose that every point of Y is not a complete accumulation point. Then for each y \in Y, there is an open set O_y containing y such that O_y \cap Y is countable. Then the set \mathcal{O} of all O_y is an open cover of Y. Since Y is Lindelof, \mathcal{O} has a countable subcover. This means that Y is countable. \blacksquare

Lemma 2
Let Y be a Lindelof space such that Y is uncountable and first countable. Then Y has at least two complete accumulation points.

Proof. Suppose Y is Lindelof, uncountable and first countable. By Lemma 1, there is p \in Y that is a complete accumulation point of Y. Let \left\{U_n:n=1,2,3,\cdots\right\} be a local base at p. Observe that there is some n such that Y-U_n is uncountable. Otherwise, the whole space Y would be countable. By Lemma 1, Y-U_n would have a complete accumulation point q. \blacksquare

Lemma 3
Let Y be a compact space such that Y is uncountable and first countable. Then there are disjoint compact subsets K_0 and K_1 such that both K_0 and K_1 are uncountable.

Proof. Suppose Y is a compact space such that Y is uncountable and first countable. By Lemma 2, there are two complete accumulation points p_0 and p_1 in Y. Choose disjoint open sets U_0 and U_1 containing p_0 and p_1, respectively, such that \overline{U_0} \cap \overline{U_1}=\phi. The desired compact sets are K_0=\overline{U_0} and K_1=\overline{U_1}. \blacksquare

Theorem 4
Let X be a compact space such that X is uncountable and first countable. Then there is a compact subset C of X such that \lvert C \lvert=2^\omega.

Proof. The induction process is similar to the process in constructing the middle third Cantor set in The Cantor set, I. We start with the compact space X and find two disjoint compact sets B_0 and B_1 that are uncountable using Lemma 3. For B_0, we apply Lemma 3 to obtain uncountable disjoint compact subsets B_{00} and B_{01}. Likewise obtain uncountable disjoint compact subsets B_{10} and B_{11} of B_1.

In each compact set in each subsequent step, we find two disjoint compact subsets that are uncountable. Suppose that at step n, we have obtained compact sets B_g for each n-length sequence of zeros and ones. Then we apply Lemma 3 to obtain two uncountable disjoint compact sets B_{g0} and B_{g1} of B_g.

Let A_0=X. For each integer n>0, let A_n be the union of all B_g over all n-length sequences g of zeros and ones. Let C^*=\bigcap \limits_{n=0}^{\infty}A_n. We choose a subset of C^* of cardinality continuum.

For each f \in 2^\omega, let f_n be the n-length sequence of zeros and ones that agrees with the first n coordinates of f. Then, for each f \in 2^\omega, choose a point x_f in \bigcap \limits_{n=1}^{\infty}B_{f_n}. Let C=\left\{x_f: f \in 2^\omega \right\}. Note that x_f \ne x_g for f,g \in 2^\omega with f \ne g. Thus C is a subset of the compact space X such that \lvert C \lvert=2^\omega.

The previous posts on compact first countable spaces
The cardinality of compact first countable spaces, I
The cardinality of compact first countable spaces, II