Consider $A = \{x \in \mathbb{Q} \mid x < \pi \}$. Formally prove that $\sup A = \pi$.

Question

I wasn't sure how to go about solving this, but I had a couple of ideas:

1.) Utilizing the fact that the rationales are a subset of the reals to say that $\sup (A) - \varepsilon < x < \pi$. However, I run into a brick wall. For this to happen, $\mathbb{Q}$ must have a least upper bound, which it does not by nature.

2.) Write $x = \frac{m}{n}$ and try to prove to the contrary that $x \geq \pi$. This should lead to a contradiction that should say that there will be a number greater than $x$ that is less than $\pi$, but I am not sure how to go about this strategy fully.

Answer 1

With a bit of research you may be able to prove that $3<\pi<4$. Hence $4$ is an upper bound for $A$ and $2$ is not. We conclude that a least upper bound exists. So let $a=\sup A$.

If $a<\pi$, then $\frac1{\pi-a}>0$. By the Archimedean property, there exists $n\in\Bbb N$ with $n>\frac1{\pi-a}$. Show that the set $\{\,k\in\Bbb N\mid \frac kn>a\,\}$ is non-empty, hence it has a minimal element $m$. Conclude that $a<\frac mn <\pi$, contradicting the upper-bound property of $a$.

If $a>\pi$, proceed similarly to find $\pi<\frac mn<a$, thus exhibiting a smaller upper bound than $a$.

Conclude that $a=\pi$.

Answer 2

We want to prove that $sup(A) = \pi$, so let's start:

We need two things.

1) $\pi$ is upper bound of set A. It is quite obvious, because $A=\{x\in Q : x<\pi \} $, so $\forall_{x \in A} : x \leq \pi $

And it ends our proof, that $\pi$ is an upper bound of A.

2) There is no "better" upper bound. Formally speaking: $\forall_{\epsilon>0} \exists_{x\in A} : \pi - \epsilon < a $

There, we have to use very important property of rational numbers : density. If we have $p,q \in Q$ such that $p<q$, there exists $ r \in Q$ such that $p<r<q$.

We also need one more thing. That between every irrational numbers, there lies another irrational, and rational. ( By that we can conclude, that between any 2 numbers, there lies rational and irrational between them)

Having all those things, let's proceed.

We want to find $x \in Q$ such that $\pi - \epsilon < x < \pi$

Due to our observation above, it s obvious that we can find rational x with these properties.

Answer 3

It might not seem like it but this exercise is meant to be easy. Even trivial. Just do it by definitions.

To prove $\pi = \sup A$ we must prove two things i) $\pi$ is an upper bound of $A$. and ii) if $b < \pi$ then $b$ is not an upper bound of $A$.

Pf of i) For all $a \in A$ then $a < \pi$ so $\pi$ is an upper bound of $A$.

Pf of ii) between any two real numbers $x,y$ so that $x < y$ the is a rational number $q$ so that $x < q < y$. This is because $\mathbb Q$ is dense in $\mathbb R$.

So if $b < \pi$ there is a $q$ so that $b < q < \pi$. So $q < \pi$ so $q \in A$. So $b$ is not an upper bound of $A$.

So $\pi$ is the least upper bound of $A$ and $\pi = \sup A$.

.....

that's all the is to it.

Okay, sometime you must prove that between any to real numbers there is a rational. But presumably you have already proven that.

Lemma 1: If $M > 0$ there is a natural number $n$ so that $n > M$.

If not $\mathbb N$ is bounded above so $\sup \mathbb N$ exist. Thus $\sup \mathbb N-1$ is not an upper bound of $\mathbb N$ so there is a natural number, $m$, so that $\sup \mathbb N - 1 < m$. But $m \le \sup \mathbb N$. So $\sup\mathbb N-1 < m \le \mathbb N$. So $\sup \mathbb N < m + 1 \in \mathbb N$ which is a contradiction.

Cor: If $\epsilon > 0$ the is a $n$ so that $\frac 1n < \epsilon$.

Pf: Let $n > \frac 1\epsilon > 0$. Then $0 < \frac 1n < \epsilon$.

Lemma 2: If $x < y$ then there exists a rational $q$ so that $x < q < y$.

Pf: Let $n$ be a natural number so that $0 \frac 1n < y-x$. Let $A = \{\frac mn| m\in \mathbb Z; \frac mn \le x\}$. $A$ is bounded above by $x$. $A$ is not empty because... if $x \ge 0$ then $-\frac 1n \in A$. If $x < 0$ then there is an $m > n|x|$ and $-\frac mn < x$. So $\sup A - \frac 1{2n}$ is not an upper bound of $A$ so there is an $m$ so that $\sup A - \frac 1{2n} < \frac mn \le \sup A \le x < x + \frac 1n < y$. So $\frac {m+1}n\not \in A$ and $x < \frac {m+1}n \le x + \frac 1n < y$.