Being a separable metrizable space, the real line with the usual topology has countable extent. A space has countable extent if all closed and discrete subsets of are countable. In general, the extent of a space is the least cardinality of a closed and discrete set in that space. In this brief note, we show that the real line has another countable extent property that is unique to the real line. We show that every uncountable subset of the real line has a two-sided limit point. As an application, it follows from this fact that the Sorgenfrey line is Lindelof and that all compact subspaces in the Sorgenfrey Line are countable (see the blog post A Note On The Sorgenfrey Line).
Given a space and given , the point is a limit point of a subset if every open set containing contains a point of distinct from . Equivalently, we can replace open set in this definition with members of a base (basic open sets). Thus, the limit here is from a topological perspective and not from a metric space perspective. A space has countable extent if all closed and discrete subsets are countable. This is equivalent to the statement that every uncountable subset has a limit point.
In the real line, we say that the point is a limit point of if every open interval containing contains a point of distinct from . We say is a two-sided limit point of if every open interval containing contains points of on both sides of (i.e. both intervals and contain points of ). Likewise, one-sided (left-sided, right-sided) limit point means every open interval containing contains points of on one side (left side, right side, respectively).
Every uncountable subset of has a two-sided limit point.
Proof. Let be uncountable. The set as a topological space is a Lindelof space. Thus has a limit point (in fact, has uncountably many limit points).
Suppose that has no two-sided limit points. Then the uncountably many limit points of must be either left-sided or right-sided limit points. We assume the case that has uncountably many right sided limit points. The proof for the left-sided case is similar. Let be the set of all right-sided limit points of .
Since points in are right-sided limits of , for each , there is a rational number such that . Matching up the rational numbers with uncountably many points in , there is a rational number such that the following set is uncountable.
The uncountable set has a limit point . Note that is a limit point of . For the rational number indicated above, it must be the case . Note that all points of are to the right of . If , then there would no points of in an open interval of , which goes against the fact that is a limit point of .
By a similar argument as in the preceding paragraph, there are no points of on the right side of . If there is some and then is an open interval containing the point that contains no points of , contradicting the fact that is a limit point of . Thus is only a left-sided limit point of .
Choose two points and from such that . We know that contains no points of . It follows that contains no points of . Since is a right-sided limit point of , would contain points of , a contradiction. Thus the set must have a two-sided limit point and the lemma is proved.