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((\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\}


  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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