May 29th, 2021

On intervals of reals in relation to countable and uncountable sets

Originally published at 狗和留美者不得入内. You can comment here or there.

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

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

Proof: [公式] 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 [公式] , then [公式] would contain an open set too, which is not possible. One can via a map of the form [公式] map the interval [公式] bijectively and continuously to any other open interval. [公式]

Lemma 2 All intervals in [公式] 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 [公式] and [公式] , one can easily find an injection from one of the other. Then, one can simply apply the Cantor-Schroder-Bernstein theorem (proven in [1]). [公式]

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

[公式] From this, we have

[公式] which completes our proof. [公式]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Theorem 1 Any open set in [公式] is necessarily a countable union of disjoint open intervals.

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

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

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

[公式]By Cantor’s intersection theorem, (Theorem 1 of [2]), [公式] , which means that [公式] contains some element of [公式] . Thus, by Axiom of Choice, we can map each sequence to some element of [公式] found in its corresponding infinite intersection. As for two distinct sequences, there must be index [公式] 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 [公式] , which proves that [公式] is necessarily uncountable.

References

On the fundamental theorem of calculus

Originally published at 狗和留美者不得入内. You can comment here or there.

In [1], we defined the Riemann integral on intervals of [公式] . We shall now prove some theorems pertaining to it. Below, we will let [公式] denote the Riemann sum of [公式] associated with tagged partition [公式].

Definition 1 A piecewise function [公式] defined on a bounded interval [公式] is defined by partitioning [公式] into a finite number of sub-intervals [公式] and defining on each of the [公式] s a function [公式] such that for [公式] , [公式] . We say that [公式] is piecewise continuous if each of the [公式] s is continuous. One can, here, replace “continuous” with any other qualification of function, such as “smooth”.

Proposition 1 We have the following properties of Riemann integrable functions.

  1. If [公式] is Riemann integrable on two disjoint intervals [公式] , with the Riemann integrals equal to [公式] , then [公式] is Riemann integrable on [公式] , with[公式]
  2. Let [公式] be an interval containing interval [公式] . If [公式] is Riemann integrable on [公式] , it is Riemann integrable on both [公式] and [公式] , with [公式]
  3. If on any interval [公式] , both [公式] and [公式] are Riemann integrable, then [公式] is also Riemann integrable on [公式] , with [公式] . Moreover for any [公式] , [公式] is Riemann integrable on [公式] with [公式] . In other word, the Riemann integral on [公式] operator is a linear operator on the space of Riemann integrable functions on [公式] .

Proof: For (1), we take a tagged partition [公式] of [公式] satisfying [公式] and a tagged partition [公式] of [公式] satisfying [公式] . Formally, this says that any [公式] refinement of [公式] is such that [公式] and similar for [公式] . For [公式] we use the tagged partition [公式] . One easily verifies with the triangle inequality that any refinement [公式] of it satisfies [公式] .

For (2), take any tagged partition [公式] of [公式] that contains the endpoints of [公式] . The equivalence of (3) of Lemma 3 from [2] with Riemann integrability tells us that for any [公式] , there exists a refinement [公式] of [公式] such that for any [公式] which are refinements of [公式] ,

[公式] Both of [公式] can be split into tagged partitions of [公式] and [公式] , which we denote via [公式] and [公式] (where the second index represents whether or not the partition is defined on [公式] or [公式] , such that

[公式] If by contradiction [公式] were not Riemann integrable on one of [公式] , then there would by Lemma 3 of [2] exist an [公式] for which there does not exist a partition [公式] that is a refinement of [公式] such that the split of any two refinements of it into partitions on [公式] , which we denote as [公式] and [公式] , satisfies both

[公式] Setting [公式] and using the triangle inequality on the above would thus violate the existence of a refinement [公式] of [公式] which guarantees [公式] , which would imply that [公式] is not Riemann integrable on [公式] , a contradiction.

With similar logic, one proves Riemann integrability of [公式] on [公式] . The integral equality of [公式] then follows directly from [公式] .

The proof of (3) we leave as an exercise to the reader. [公式]

Proposition 2 Any continuous function on [公式] is Riemann integrable.

Proof: We construct a sequence of tagged partitions of [公式] . We let the [公式] th tagged partition be given by

[公式]

in which we have let [公式] be the subintervals of [公式] corresponding to the partition, where [公式] . As for the tags, we let [公式] , noting that by the Extreme Value Theorem, the infimum here exists. We notice that under the partial order of partition refinement, this sequence of tagged partitions is a chain. Moreover, the Riemann sums of these partitions form a bounded (by Extreme Value Theorem) monotonically non-decreasing sequence. Thus, by the Monotone Convergence Theorem, this sequence of Riemann sums converges to some [公式] . For any [公式] , we can find a partition from this sequence such that the distance between its Riemann sum and [公式] is less than [公式] . Since a refinement of a tagged partition where the tags correspond to infimum of [公式] on subintervals cannot decrease the Riemann sum, [公式] satisfies the definition of Riemann integral. We emphasize in this proof that the subintervals in the definition of Riemann sum are closed and are not disjoint at endpoints, which allows for two tags (of different subintervals) to have the same value. [公式]

Corollary 1 Any function [公式] continuous and bounded on an open or half-open interval in [公式] is Riemann integrable. Moreover, if we extend such a function to the closure of that interval via the limit and Riemann integrate this on the closure, the result is the same. Formally, [公式] . Because of this, we can disregard whether or not the endpoints are included in the interval itself and simply use [公式] to denote this Riemann integral.

Proof: We note that [公式] and [公式] are both closed intervals on which the Riemann integral exists and is zero. The Riemann integral exists on [公式] by Proposition 2. Applying (2) of Proposition 1 tells us that the Riemann integral exists and is equal to [公式] on each of [公式] . [公式]

Definition 2 Suppose that [公式] . Then, for any Riemann integrable [公式] on [公式] we define [公式]

Proposition 3 Any function [公式] piecewise continuous on any interval is Riemann integrable on that interval.

Proof: Corollary 1 tells that we can WLOG only consider closed intervals. Let the interval be [公式] , with [公式] , where the [公式] s are disjoint intervals on each of which [公式] is continuous. Proposition 2 tells us that [公式] is Riemann integrable on each of the [公式] s. Thus, we can apply (1) of Proposition 1 to derive that [公式] is Riemann integrable on [公式] .

One can also use the equivalence of (2) and (3) in Theorem 1 of [2], of which this is a very special case. [公式]

Proposition 4 On some interval [公式] , define Riemann integrable [公式] such that[公式] for all [公式] . Then,

[公式]

Proof: Trivial and left to the reader.

Theorem 1 (Mean value theorem for definite integrals) Let [公式] be a continuous function. Then, there exists [公式] such that [公式] Proof: The Extreme Value Theorem gives us infimum and supremum values of [公式] on [公式] , which we denote with [公式] and [公式] . This gives us

[公式] By the Intermediate Value Theorem, for every point in [公式] is such that there exists [公式] such that [公式] . Applying this to the value of [公式] yields the desired result. [公式]

Theorem 2 (Fundamental theorem of calculus, part I) Let [公式] be a continuous real-valued function defined on closed interval [公式] . Let [公式] be the function defined, for all [公式] , by

[公式] Then [公式] is uniformly continuous on [公式] and differentiable on the open interval [公式] , and

[公式] for all [公式] .

Proof: Take arbitrary [公式] . We can define for all [公式] such that [公式]

[公式] Applying Mean Value Theorem for Definite Integrals (Theorem 1) to this gives some [公式] such that

[公式]

As [公式] , [公式] . Because [公式] is continuous on [公式] , we thus have

[公式]

By the Extreme Value Theorem, there exists [公式] such that [公式] for all [公式] . Thus, by Proposition 4, for any [公式] , take [公式] and we have [公式] for all [公式] . This shows uniform continuity. [公式]

Definition 3 We say that [公式] is an antiderivative of [公式] on [公式] if on [公式] , [公式] .

Lemma 1 Any continuous antiderivative [公式] of the zero function on [公式] is a constant function on [公式].

Proof: Suppose by contradiction that some [公式] that is a continuous anti-derivative of the zero function on [公式] is not constant. Then there exists [公式] such that [公式] . The mean value theorem for differentiable functions tells us that there exists some [公式] such that [公式] , which means that [公式] would then not be a continuous function on [公式] with derivative equal to zero on [公式] . [公式]

Lemma 2 If a function [公式] defined on [公式] has antiderivatives [公式] on [公式] , then [公式] is a constant function on [公式] , where the values of [公式] at [公式] are simply defined via continuous extension.

Proof: The Fundamental Theorem of Calculus, Part I (Theorem 1) tells us that

[公式] is an antiderivative of [公式] on [公式] that is continuous on [公式] . Because the relation defined via constant difference is transitive, it suffices to simply show that [公式] is constant on [公式] .

That [公式] on [公式] tells us that [公式] is an antiderivative of the zero function on [公式] , from which we deduce via Lemma 1 that is a constant function. [公式]

Corollary 2 If [公式] is a real valued continuous function on [公式] and [公式] is a continuous antiderivative of [公式] on [公式] , then

[公式]

Proof: We define [公式] on [公式] . Fundamental Theorem of Calculus Part I tells us that it is a continuous antiderivative of [公式] on [公式] . We have that

[公式]

By Lemma 2, [公式] is a constant function on [公式] , we must have that [公式] , which completes our proof. [公式]

Theorem 2 (Fundamental theorem of calculus part II (Newton-Leibniz axiom)) If [公式] is a real valued function on [公式] and Riemann integrable on [公式] and [公式] is an antiderivative of [公式] on [公式] then

[公式]

Proof: Take an arbitrary partition [公式] . We have that

[公式] We can apply the mean value theorem for differentiability on the subintervals [公式] to assign tags such that [公式] which yields the Riemann sum

[公式]

By definition of Riemann integrable as given in (4) of Theorem 1 of [2], for all [公式] , there exists [公式] such that for any tagged partition [公式] of [公式] denoted by [公式] and [公式] for the tags such that its norm or mesh [公式] , we have

[公式]

Another way to put it, per Lemma 2, is that for all [公式] , there exists [公式] such that any tagged partitions [公式] with meshes both less than [公式] ,

[公式]

such that

[公式]

Thus, we can, setting [公式] , take the limit of [公式] to derive

[公式] which completes our proof. [公式]

References