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Below we shall let be an arbitrary subset of such that there does not exist for any an interval .

**Lemma 1** For any interval , one can find some not containing any open intervals in bijective correspondence with .

*Proof*: is a continuous bijection. Moreover it is monotonically increasing, and its inverse is also continuous. Thus, by definition of continuity, if contains an open set , then would contain an open set too, which is not possible. One can via a map of the form map the interval bijectively and continuously to any other open interval.

**Lemma 2** All intervals in which are not singleton sets or empty sets, which include ones bounded below or above by infinity, are in bijective correspondence.

*Proof*: For any two such intervals and , one can easily find an injection from one of the other. Then, one can simply apply the Cantor-Schroder-Bernstein theorem (proven in [1]).

**Lemma 3** Let be an interval with endpoints with length . Let be a collection of intervals contained in which are mutually disjoint. Then, there does not exist a finite subcollection of , which we call , the sum of the lengths of which exceeds , or formally,*Proof*: Let be start and endpoints of for all . That the s are mutually disjoint implies that there exists a total order on any finite subcollection . Thus, we can order the s such that

From this, we have

which completes our proof.

**Proposition 1** The union of two disjoint countable sets is countable.

*Proof*: One can construct a bijection from to by multiplying non-negative numbers by and for negative numbers, multiplying by and subtracting . Then, one bijectively maps to by simply adding to all the elements. One can also add to all non-negative elements of and keep the remaining fixed to show that

Let be our two disjoint countable sets. We bijectively map them to and which are disjoint and such that has a bijective correspondence with countable set .

**Proposition 2** There exists an uncountable subset of that does not contain any open intervals.

*Proof*: is an example. If by contradiction, it were countable, then would be countable by Proposition 1, a contradiction.

**Proposition 3** If on some interval , there is a subset such that every open interval contained in contains infinitely many points of , then .

*Proof*: Suppose by contradiction. there exists some for which no sequence in converges to . This is only possible if there exists some neighborhood of , , such that . This, however, shows the existence of an open interval contained in which is disjoint with , which violates the hypothesis of the proposition.

**Proposition 4** Let be a subset of not containing any open intervals. Then, . This of course extends all to intervals too.

*Proof*: Take any . Take the balls centered at of radius for all . The hypothesis allows to pick an from each of those balls to construct a sequence converging to .

**Proposition 5** On any interval , subset is such that if and only if does not contain any open intervals.

*Proof*: If does contain an open interval, then obviously . This proves the only if direction. Now suppose that . Then if contains no open intervals, by Proposition 4, we would have , which we’ve assume to not be the case. Thus, would have to contain an open interval, which proves the if direction.

**Proposition 6** Let be any interval of . Let be a non-empty subset of such that its closure does not contain any open intervals. Then must contain an open interval, which means it is also uncountable.

*Proof*: It was proven in [2] that for a topological space on , any subset of is such that is partitioned by . Thus, the hypothesis that does not contain any open intervals means that , which tells us that does not contain any open intervals either.

Suppose by contradiction all points are such that all the neighborhoods of contain some element of , which would also mean that all the neighborhoods of contain infinitely many elements of . Suppose also that for every radius ball centered at some arbitrarily fixed , which we call , the closure of is not equal to . Then, by Proposition 5, would have to contain an open interval, which must be contained in the interior of , which is an open subset of . This concludes our proof.

**Theorem 1** Any open set in is necessarily a countable union of disjoint open intervals.

*Proof*: It is well known that the topology on is countably generated by open intervals where and . Because is countable, the aforementioned base of , which we shall call , being a subset of , must also be countable. By definition, every open set in can be expressed as the union of some subfamily of , which of course must also be countable. In fact, every open set in is a union of its connected components. Every connected component of an open set is an open set that associated with it the subfamily of sets in contained by it. By the Axiom of Choice, we map each connected open to an arbitrary open set in that it contains, and this map is an injective map from the set of connected components to our arbitrary open set in to , which means cardinality of the set of connected components is countable. This completes our proof.

**Lemma 4** A closed subset of an interval is uncountable if there is a way to, starting from only the interval , for all the intervals in the th iteration take two closed non-intersecting subintervals such that every time, the two resulting subintervals all contain any infinite number of elements of for all , with the process never terminating.

*Proof*: We associate every interval with a string of 0s and 1s. is assigned the empty string. Whenever we pick for an interval two disjoint closed intervals , where is the lower one, we assign the string formed by appending 0 to the string of and for , we append 1. Every infinite sequence from here corresponds directly to a chain of strictly decreasing subintervals of such that each of the here contain infinitely many elements of . This such chain of course also induces a decreasing chain of closed subsets of , which we shall denote with

By Cantor’s intersection theorem, (Theorem 1 of [2]), , which means that contains some element of . Thus, by Axiom of Choice, we can map each sequence to some element of found in its corresponding infinite intersection. As for two distinct sequences, there must be index at which they first differ, which implies that their corresponding sets defined via infinite intersection are disjoint. This shows that this function is injective. We have thus defined a bijection from a set known to be of the same cardinality of the real numbers to a subset of , which proves that is necessarily uncountable.

**References**