We discuss the Alexandroff double circle, which is a compact and non-metrizable space. A theorem about the hereditarily normality of a product space is also discussed.
Let and be the two concentric circles centered at the origin with radii 1 and 2, respectively. Specifically where . Let . Furthermore let be the natural homeomorphism. Figure 1 below shows the underlying set.
Figure 1 – Underlying Set
We define a topology on as follows:
- Points in are isolated.
- For each and for each positive integer , let be the open arc in whose center contains and has length (in the Euclidean topology on ). For each , an open neighborhood is of the form where
The following figure shows an open neighborhood at point in .
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 . We discuss the following points about the Alexandroff double circle.
- is a Hausdorff space.
- is not separable.
- is not hereditarily Lindelof.
- is compact.
- is sequentially compact.
- is not metrizable.
- is not perfectly normal.
- is completely normal (and thus hereditarily normal).
- is not hereditarily normal.
The proof that 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 consists of uncountably many singleton open subsets. For the same reason, is a non-Lindelof subspace, making the Alexandroff double circle not hereditarily Lindelof.
The property that is compact is closely tied to the compactness of the inner circle in the Euclidean topology. Note that the subspace topology of the Alexandroff double circle on is simply the Euclidean topology. Let be an open cover of consisting of open sets as defined above. Then there are finitely many basic open sets , , , from covering . These open sets cover the entire space except for the points , which can be covered by finitely many open sets in .
A space is sequentially compact if every sequence of points of has a subsequence that converges to a point in . 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 hinges on the sequentially compactness of and in the Euclidean topology. Let be a sequence of points in . If the set is a finite set, then is a constant sequence for some large enough integer . So assume that is an infinite set. Either is infinite or is infinite. If is infinite, then some subsequence of converges in in the Euclidean topology (hence in the Alexandroff double circle topology). If is infinite, then some subsequence of converges to in the Euclidean topology. Then this same subsequence converges to in the Alexandroff double circle topology.
Note that any compact metrizable space satisfies a long list of properties, which include separable, Lindelof, hereditarily Lindelof.
A space is perfectly normal if it is normal with the additional property that every closed set is a -set. For the Alexandroff double circle, the inner circle is not a -set, or equivalently the outer circle is not an -set. To see this, suppose that 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 . Let . At least one of the sets is uncountable. Let be one such. Consider , which is also uncountable and has a limit point in (in the Euclidean topology). Let be one such point (i.e. every Euclidean open set containing contains points of ). Then the point is a member of the closure of (Alexandroff double circle topology).
We first discuss the notion of separated sets. Let be a Hausdorff space. Let and . The sets and are said to be separated (are separated sets) if and . 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 is said to be completely normal if satisfies the property that for any two sets and that are separated, there are disjoint open sets and with and . 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 is completely normal, let and be separated sets. Thus we have and . Note that and are separated sets in the Euclidean space . Let and be disjoint Euclidean open subsets of with and .
For each , choose open (Alexandroff double circle open) with , and . Likewise, for each , choose open (Alexandroff double circle open) with , and . Then let and be defined by the following:
Because , the open sets and are disjoint. As a result, and are disjoint open sets in the Alexandroff double circle with and .
For the sake of completeness, we show that any completely normal space is hereditarily normal. Let be completely normal. Let . Let and be disjoint closed subsets of . Then in the space , and are separated. Note that and (where gives the closure in ). Then there are disjoint open subsets and of such that and . Now, and are disjoint open sets in such that and .
Thus we have established that the Alexandroff double circle is hereditarily normal.
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 ,
We produce a subspace that is not normal. To this end, let be a countable subset of such that . Let . Let . We show that is not normal.
Let and . These are two disjoint closed sets in . Let and be open in such that and . We show that .
For each integer , let . We claim that each is open in . To see this, pick . We know . There exist open and (open in ) such that . It is clear that . Thus each is open.
Furthermore, we have for each . Based in Result 7, is not a -set. So we have but . There exists but . Thus and is open.
Since , we have . Choose an open neighborhood of such that . since , there exists some such that . Hence . Since , . Thus .
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 -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 has a countable subset that is not closed and discrete and if has a closed set that is not a -set then has a subspace that is not normal.
The theorem can be restated as:
If is hereditarily normal, then either every countable subset of is closed and discrete or is perfectly normal.
The above theorem is due to Katetov and can be found in . It shows that the hereditarily normality of a cross product imposes quite strong restrictions on the factors. As a quick example, if both and are compact, for to be hereditarily normal, both and must be perfectly normal.
Another example. Let , the succesor of the first uncountable ordinal with the order topology. Note that is not perfectly normal since the point is not a point. Then for any compact space , is not hereditarily normal. Let , the successor of the first infinite ordinal with the order topology (essentially a convergent sequence with the limit point). The product 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 . The resulting space is the deleted Tychonoff plank and is not normal (see The Tychonoff Plank).
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