#mathanalysis

Fact: The set of real numbers, ℝ, is uncountable.

Proof: We will focus on the subset (0,1)⊆ℝ and show that this set cannot be countable. Since ℝ sits above an uncountable set it follows that ℝ is also uncountable. We have not yet proven the lemma a set containing an uncountable subset must be uncountable (dear reader, we would also start this by contradiction getting a 1-1 function from the "bigger" set to ℕ).

By contradiction, suppose ∃f:(0,1)↦ℕ which is 1-1. We can now "count" the elements of (0,1): a1=f -1(1), a2=f -1(2),….

We will go further and consider the decimal expansion of each an=0.an1an2an3.... That is, ank is the kth digit of the decimal expansion of an. We put these digits in the the following infinite table where the nth row holds the digits of an.

a11 a12 a13 a14 … a21 a22 a23 a24 … a31 a32 a33 a34 … a41 a42 a43 a44 … ⋮ ⋮ ⋮ ⋮ ⋱

We will now construct a "new" element b∈(0,1) by defining the decimal digits of b=0.b1b2b3b4... as follows. If ann=1, let bn=2. Otherwise, let bn=1. Doing this for every n∈ℕ defines every decimal digit of b and therefore defines b itself. Note also that bn≠ann ∀n.

Now we simply notice that b cannot be in the above "listing" of (0,1) because it cannot be in any of the rows. It cannot be in the jth row because the jth entry of that row is ajj≠bj. In this way we see that b∉{a1,a2,a3,...}=(0,1), which contradicts the construction of b∈(0,1).☐

Fact: There is an irrational number.

Proof: In fact, if p is any prime number, then √p is irrational.

To see that this is true, suppose by way of contradiction that √p=a/b where a,b∈ℕ and a and b are reduced so that gcd(a,b)=1.

Squaring both sides of our equation gives p=a2/b2 which can be rewritten as b2p=a2. This gives that p is a divisor of a2, and since p is prime, in fact we can conclude that p divides a. This means ∃c∈ℕ with cp=a.

Squaring both sides of this last equation gives c2p2=a2. Combining this with the previously known b2p=a2 gives b2p=c2p2. Dividing both sides of this by p gives b2=c2p, which now tells us that p must also divide b. This contradicts the fact that gcd(a,b)=1.☐

Proof: We will prove the first statement by extablishing set inclusion "both ways". The second statement follows from a slightly modified argument.

[⊆:] Let x∈(∩Aα)C. This means x∉∩Aα. By definition of ∩, there is an Aβ with x∉Aβ. This means x∈(Aβ)C⊆∪(Aα)C. So x∈∪(Aα)C.

[⊇:] Let y∈∪(Aα)C. This means there is an Aγ with y∈(Aγ)C. This means y∉Aγ. So y cannot be in ∩Aα which is contained in Aγ. That is to say y∈(∩Aα)C.☐

Proof: Let x < y be given real numbers. By the Archimedean property, there is an integer n such that n(y-x)>1. This means that the gap between ny and nx is wider than 1, meaning that we can find an integer in this gap.

Formally, if we consider the set A={m∈&Zopf;:m>nx}, then we have a nonempty set of integers which is bounded below. This set must have a least element, and we will call this element q. Note that q∈A implies that q>nx.

Also, q<ny. If this were not the case, we would have q-1≥ny-1>nx and then q-1∈A. But this would contradict that q is the least element of A.

We have q∈&Zopf; with ny>q>nx. This implies that y>q/n>x and q/n is an element of ℚ as desired.☐

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