A topological space is Lindelof if every open cover has a countable subcollection that also is a cover of . A topological space is hereditarily Lindelof if every subspace of , with respect to the subspace topology, is a Lindelof space. In this post, we prove a theorem that gives two equivalent conditions for the hereditarily Lindelof property. We consider the following theorem.

**Theorem 1**

Let be a topological space. The following conditions are equivalent.

- The space is a hereditarily Lindelof space.
- Every open subspace of is Lindelof.
- For every uncountable subspace of , there exists a point such that every open subset of containing contains uncountably many points of .

This is an excellent exercise for the hereditarily Lindelof property and for transfinite induction (for one of the directions). The equivalence is the exercise 3.12.7(d) on page 224 of [1]. The equivalence of the 3 conditions of Theorem 1 is mentioned on page 182 (chapter d-8) of [2].

**Proof of Theorem 1**

The direction is immediate. The direction is straightforward.

We show . Suppose is a non-Lindelof subspace of . Let be an open cover of such that no countable subcollection of can cover . By a transfinite inductive process, choose a set of points and a collection of open sets such that for each , and . The inductive process is possible since no countable subcollection of can cover . Now let . Note that each can at most contain countably many points of , namely the points in .

For each , let be an open subset of such that . We can now conclude: for every point of , there exists an open set containing such that contains only countably many points of . This is the negation of condition 3.

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**Remarks**

Condition 3 indicates that every uncountable set has a certain special type of limit points. Let . We say is a limit point of the set if every open set containing contains a point of different from . Being a limit point of , we only know that each open set containing contain infinitely many points of (assuming a space). Thus the limit points indicated in condition 3 are a special type of limit points. According to the terminology of [1], if is a limit point of satisfying condition 3, then is said to be a condensation point of . According to Theorem 1, existence of condensation point in every uncountable set is a strong topological property (being equivalent to the hereditarily property). It is easy to see that of condition 3 holds, all but countably many points of any uncountable set is a condensation point of .

In some situations, we may not need the full strength of condition 3. In such situations, the following corollary may be sufficient.

**Corollary 2**

If the space is hereditarily Lindelof, then every uncountable subspace of contains one of its limit points.

As noted earlier, if every uncountable set contains one of its limits, then all but countably many points of any uncountable set are limit points. To contrast the hereditarily Lindelof property with the Lindelof property, consider the following theorem.

*Theorem 3*

If the space is Lindelof, then every uncountable subspace of has a limit point.

The condition “every uncountable subspace of has a limit point” has another name. When a space $latesatisfies this condition, it is said to have countable extent. The ideas in Corollary 2 and Theorem 3 are also discussed in this previous post.

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**Reference**

- Engelking, R.,
*General Topology, Revised and Completed edition*, Heldermann Verlag, Berlin, 1989. - Hart, K. P., Nagata J. I., Vaughan, J. E., editors,
*Encyclopedia of General Topology, First Edition*, Elsevier Science Publishers B. V, Amsterdam, 2003.

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