Alexandroff Double Circle

We discuss the Alexandroff double circle, which is a compact and non-metrizable space. A theorem about the hereditarily normality of a product space Y_1 \times Y_2 is also discussed.

Let C_1 and C_2 be the two concentric circles centered at the origin with radii 1 and 2, respectively. Specifically C_i=\left\{(x,y) \in \mathbb{R}^2: x^2 + y^2 =i \right\} where i=1,2. Let X=C_1 \cup C_2. Furthermore let f:C_1 \rightarrow C_2 be the natural homeomorphism. Figure 1 below shows the underlying set.

Figure 1 – Underlying Set


We define a topology on X as follows:

  • Points in C_2 are isolated.
  • For each x \in C_1 and for each positive integer j, let O(x,j) be the open arc in C_1 whose center contains x and has length \frac{1}{j} (in the Euclidean topology on C_1). For each x \in C_1, an open neighborhood is of the form B(x,j) where
      \text{ }

      B(x,j)=O(x,j) \cup (f(O(x,j))-\left\{f(x) \right\}).

    The following figure shows an open neighborhood at point in C_1.

Figure 2 – Open Neighborhood


A List of Results

It can be verified that the open neighborhoods defined above form a base for a topology on X. We discuss the following points about the Alexandroff double circle.

  1. X is a Hausdorff space.
  2. X is not separable.
  3. X is not hereditarily Lindelof.
  4. X is compact.
  5. X is sequentially compact.
  6. X is not metrizable.
  7. X is not perfectly normal.
  8. X is completely normal (and thus hereditarily normal).
  9. X \times X is not hereditarily normal.

The proof that X \times X is not hereditarily normal can be generalized. We discuss this theorem after presenting the proof of Result 9.

Results 1, 2, 3

It is clear that the Alexandroff double circle is a Hausdorff space. It is not separable since the outer circle C_2 consists of uncountably many singleton open subsets. For the same reason, C_2 is a non-Lindelof subspace, making the Alexandroff double circle not hereditarily Lindelof. \blacksquare


Result 4

The property that X is compact is closely tied to the compactness of the inner circle C_1 in the Euclidean topology. Note that the subspace topology of the Alexandroff double circle on C_1 is simply the Euclidean topology. Let \mathcal{U} be an open cover of X consisting of open sets as defined above. Then there are finitely many basic open sets B(x_1,j_1), B(x_2,j_2), \cdots, B(x_n,j_n) from \mathcal{U} covering C_1. These open sets cover the entire space except for the points f(x_1), f(x_2), \cdots,f(x_n), which can be covered by finitely many open sets in \mathcal{U}. \blacksquare


Result 5

A space W is sequentially compact if every sequence of points of W has a subsequence that converges to a point in W. The notion of sequentially compactness and compactness coincide for the class of metric spaces. However, in general these two notions are distinct.

The sequentially compactness of the Alexandroff double circle X hinges on the sequentially compactness of C_1 and C_2 in the Euclidean topology. Let \left\{x_n \right\} be a sequence of points in X. If the set \left\{x_n: n=1,2,3,\cdots \right\} is a finite set, then \left\{x_n: n>m \right\} is a constant sequence for some large enough integer m. So assume that A=\left\{x_n: n=1,2,3,\cdots \right\} is an infinite set. Either A \cap C_1 is infinite or A \cap C_2 is infinite. If A \cap C_1 is infinite, then some subsequence of \left\{x_n \right\} converges in C_1 in the Euclidean topology (hence in the Alexandroff double circle topology). If A \cap C_2 is infinite, then some subsequence of \left\{x_n \right\} converges to x \in C_2 in the Euclidean topology. Then this same subsequence converges to f^{-1}(x) in the Alexandroff double circle topology. \blacksquare


Result 6

Note that any compact metrizable space satisfies a long list of properties, which include separable, Lindelof, hereditarily Lindelof. \blacksquare


Result 7

A space is perfectly normal if it is normal with the additional property that every closed set is a G_\delta-set. For the Alexandroff double circle, the inner circle C_1 is not a G_\delta-set, or equivalently the outer circle C_2 is not an F_\sigma-set. To see this, suppose that C_2 is the union of countably many sets, we show that the closure of at least one of the sets goes across to the inner circle C_1. Let C_2=\bigcup \limits_{i=1}^\infty T_n. At least one of the sets is uncountable. Let T_j be one such. Consider f^{-1}(T_j), which is also uncountable and has a limit point in C_1 (in the Euclidean topology). Let t be one such point (i.e. every Euclidean open set containing t contains points of f^{-1}(T_j)). Then the point t is a member of the closure of T_j (Alexandroff double circle topology). \blacksquare


Result 8

We first discuss the notion of separated sets. Let T be a Hausdorff space. Let E \subset T and F \subset T. The sets E and F are said to be separated (are separated sets) if E \cap \overline{F}=\varnothing and F \cap \overline{E}=\varnothing. In other words, two sets are separated if each one does not meet the closure of the other set. In particular, any two disjoint closed sets are separated. The space T is said to be completely normal if T satisfies the property that for any two sets E and F that are separated, there are disjoint open sets U and V with E \subset U and F \subset V. Thus completely normality implies normality.

It is a well know fact that if a space is completely normal, it is hereditarily normal (actually the two notions are equivalent). Note that any metric space is completely normal. In particular, any Euclidean space is completely normal.

To show that the Alexandroff double circle X is completely normal, let E \subset X and F \subset X be separated sets. Thus we have E \cap \overline{F}=\varnothing and F \cap \overline{E}=\varnothing. Note that E \cap C_1 and F \cap C_1 are separated sets in the Euclidean space C_1. Let G_1 and G_2 be disjoint Euclidean open subsets of C_1 with E \cap C_1 \subset G_1 and F \cap C_1 \subset G_2.

For each x \in E \cap C_1, choose open U_x (Alexandroff double circle open) with x \in U_x, U_x \cap C_1 \subset G_1 and U_x \cap \overline{F}=\varnothing. Likewise, for each y \in F \cap C_1, choose open V_y (Alexandroff double circle open) with y \in V_y, V_y \cap C_1 \subset G_2 and V_y \cap \overline{E}=\varnothing. Then let U and V be defined by the following:

    U=\biggl(\bigcup \limits_{x \in E \cap C_1} U_x \biggr) \cup \biggl(E \cap C_2 \biggr)

    \text{ }

    V= \biggl(\bigcup \limits_{y \in F \cap C_1} V_y \biggr) \cup \biggl(F \cap C_2 \biggr)

Because G_1 \cap G_2 =\varnothing, the open sets U_x and V_y are disjoint. As a result, U and V are disjoint open sets in the Alexandroff double circle with E \subset U and F \subset V.

For the sake of completeness, we show that any completely normal space is hereditarily normal. Let T be completely normal. Let Y \subset T. Let H \subset Y and K \subset Y be disjoint closed subsets of Y. Then in the space T, H and K are separated. Note that H \cap cl_T(K)=\varnothing and K \cap cl_T(H)=\varnothing (where cl_T gives the closure in T). Then there are disjoint open subsets O_1 and O_2 of T such that H \subset O_1 and K \subset O_2. Now, O_1 \cap Y and O_2 \cap Y are disjoint open sets in Y such that H \subset O_1 \cap Y and K \subset O_2 \cap Y.

Thus we have established that the Alexandroff double circle is hereditarily normal. \blacksquare

For the proof that a space is completely normal if and only if it is hereditarily normal, see Theorem 2.1.7 in page 69 of [1],

Result 9

We produce a subspace Y \subset X \times X that is not normal. To this end, let D=\left\{d_n:n=1,2,3,\cdots \right\} be a countable subset of X such that \overline{D}-D\ne \varnothing. Let y \in \overline{D}-D. Let Y=X \times X-C_1 \times \left\{y \right\}. We show that Y is not normal.

Let H=C_1 \times (X-\left\{y \right\}) and K=C_2 \times \left\{y \right\}. These are two disjoint closed sets in Y. Let U and V be open in Y such that H \subset U and K \subset V. We show that U \cap V \ne \varnothing.

For each integer j, let U_j=\left\{x \in X: (x,d_j) \in U \right\}. We claim that each U_j is open in X. To see this, pick x \in U_j. We know (x,d_j) \in U. There exist open A and B (open in X) such that (x,d_j) \in A \times B \subset U. It is clear that x \in A \subset U_j. Thus each U_j is open.

Furthermore, we have C_1 \subset U_j for each j. Based in Result 7, C_1 is not a G_\delta-set. So we have C_1 \subset \bigcap \limits_{j=1}^\infty U_j but C_1 \ne \bigcap \limits_{j=1}^\infty U_j. There exists t \in \bigcap \limits_{j=1}^\infty U_j but t \notin C_1. Thus t \in C_2 and \left\{t \right\} is open.

Since (t,y) \in K, we have (t,y) \in V. Choose an open neighborhood B(y,k) of y such that \left\{t \right\} \times B(y,k) \subset V. since y \in \overline{D}, there exists some d_j such that (t,d_j) \in \left\{t \right\} \times B(y,k). Hence (t,d_j) \in V. Since t \in U_j, (t,d_j) \in U. Thus U \cap V \ne \varnothing. \blacksquare


Generalizing the Proof of Result 9

The proof of Result 9 requires that one of the factors has a countable set that is not discrete and the other factor has a closed set that is not a G_\delta-set. Once these two requirements are in place, we can walk through the same proof and show that the cross product is not hereditarily normal. Thus, the statement that is proved in Result 9 is the following.

If Y_1 has a countable subset that is not closed and discrete and if Y_2 has a closed set that is not a G_\delta-set then Y_1 \times Y_2 has a subspace that is not normal.

The theorem can be restated as:

If Y_1 \times Y_2 is hereditarily normal, then either every countable subset of Y_1 is closed and discrete or Y_2 is perfectly normal.

The above theorem is due to Katetov and can be found in [2]. It shows that the hereditarily normality of a cross product imposes quite strong restrictions on the factors. As a quick example, if both Y_1 and Y_2 are compact, for Y_1 \times Y_2 to be hereditarily normal, both Y_1 and Y_2 must be perfectly normal.

Another example. Let W=\omega_1+1, the succesor of the first uncountable ordinal with the order topology. Note that W is not perfectly normal since the point \omega_1 is not a G_\delta point. Then for any compact space Y, W \times Y is not hereditarily normal. Let C=\omega+1, the successor of the first infinite ordinal with the order topology (essentially a convergent sequence with the limit point). The product W \times C is the Tychonoff plank and based on the discussion here is not hereditarily normal. Usually the Tychonoff plank is shown to be not hereditarily normal by removing the cornor point (\omega_1,\omega). The resulting space is the deleted Tychonoff plank and is not normal (see The Tychonoff Plank).



  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Przymusinski, T. C., Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
  3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.


\copyright \ \ 2012


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