## 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 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

### 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 . We shall now prove some theorems pertaining to it. Below, we will let denote the Riemann sum of associated with tagged partition .

Definition 1 A piecewise function defined on a bounded interval is defined by partitioning into a finite number of sub-intervals and defining on each of the s a function such that for , . We say that is piecewise continuous if each of the 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 is Riemann integrable on two disjoint intervals , with the Riemann integrals equal to , then is Riemann integrable on , with
2. Let be an interval containing interval . If is Riemann integrable on , it is Riemann integrable on both and , with
3. If on any interval , both and are Riemann integrable, then is also Riemann integrable on , with . Moreover for any , is Riemann integrable on with . In other word, the Riemann integral on operator is a linear operator on the space of Riemann integrable functions on .

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

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

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

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

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

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

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

Proposition 2 Any continuous function on is Riemann integrable.

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

in which we have let be the subintervals of corresponding to the partition, where . As for the tags, we let , 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 . For any , we can find a partition from this sequence such that the distance between its Riemann sum and is less than . Since a refinement of a tagged partition where the tags correspond to infimum of on subintervals cannot decrease the Riemann sum, 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.

Corollary 1 Any function continuous and bounded on an open or half-open interval in 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, . Because of this, we can disregard whether or not the endpoints are included in the interval itself and simply use to denote this Riemann integral.

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

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

Proposition 3 Any function 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 , with , where the s are disjoint intervals on each of which is continuous. Proposition 2 tells us that is Riemann integrable on each of the s. Thus, we can apply (1) of Proposition 1 to derive that is Riemann integrable on .

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

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

Proof: Trivial and left to the reader.

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

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

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

Then is uniformly continuous on and differentiable on the open interval , and

for all .

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

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

As , . Because is continuous on , we thus have

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

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

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

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

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

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

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

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

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

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

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

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

Proof: Take an arbitrary partition . We have that

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

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

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

such that

Thus, we can, setting , take the limit of to derive

which completes our proof.

References