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.

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


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

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