Proof : Bolzano Weierstrass theorem

Question

As part of the complete proof the professor gave he proved this implication:

Let $ A \subset \mathbb{R}$ and every sequence $(a_n)_{n \in \mathbb{N}} $ in A has at least one accumulation point in A.

That implies, A is bounded and closed.

The proof he gave ( only the part where he proves, that A is bounded) :

A is bounded: Let's assume ( for a proof by contradiction ) that A is not bounded. Then:

$$ \exists (a_n)_{n \in \mathbb{N}} \in A : |a_n - a| \geq n \; > \forall n \in \mathbb{N} ( a \in A ) $$

We know, every sequence in A has an accumulation point in A, we'll call it x, then it follows:

$$ |x-a| \geq |a_n-a| - |a_n-x| \geq n - |x-a_n| $$

This means $|x-a_n| < 1$ for an infinite number of $n \in \mathbb{N}$ ( this follows from the definition of an accumulation point )

$$ \rightarrow |x-a| \geq n-1 $$

for an infinite amount of $ n \in \mathbb{N}$ , which is a contradiction.

q.e.d.

Right in the beginning, why does A not being bounded imply that there exists a sequence where the distance between a member of the sequence and the accumulation point is greater than all real numbers? Or am I reading this wrong?

Answer 1

The first displayed line in the proof makes no sense as written: $\forall n\in\Bbb N\,(a\in A)$ is a statement, not a number, so $n>\forall n\in\Bbb N\,(a\in A)$ is meaningless. The notation $(a_n)_{n\in\Bbb N}\in A$ is also incorrect: each of the points $a_n$ is an element of $A$, but the sequence itself certainly is not. This appears to be a very clumsy attempt to say the following:

There are a point $a\in A$ and a sequence $\langle a_n:n\in\Bbb N\rangle$ in $A$ such that $|a_n-a|\ge n$ whenever $n\in\Bbb N$.

If one insists on writing this with quantifiers, it’s

$$\exists a\in A\,\forall n\in\Bbb N\,\exists a_n\in A\,\big(|a_n-a|\ge n\big)\;.$$

Suppose that $A$ is bounded. Then by definition. there is an $n\in\Bbb N$ such that $|a|<n$ for each $a\in A$. Fix some $a\in A$; then for each $a'\in A$ we have

$$|a'-a|\le |a'|+|a|<2n\;.$$

Conversely, if there is an $a\in A$ such that $|a'-a|<n$ for some $n\in\Bbb N$, then

$$|a'|=|(a'-a)+a|\le |a'-a|+|a|<n+|a|\;,$$

so $|a'|<m$ for each $a'\in A$, where $m$ is the integer $n+\lceil|a|\rceil$. This shows that $A$ is bounded if and only if there are an $a\in A$ and an integer $n$ such that $|a'-a|<n$ for each $a'\in A$.

Now negate that: $A$ us unbounded if and only if for each $a\in A$ and $n\in\Bbb N$ there is an $a'\in A$ such that $|a'-a|\ge n$. In particular, if $A$ is unbounded we can pick any specific $a\in A$ and be assured that for each $n\in\Bbb N$ there is an $a_n\in A$ such that $|a_n-a|\ge n$. That gives us the sequence $\langle a_n:n\in\Bbb N\rangle$.

The rest of the argument also needs a little repair. We do know that the sequence has an accumulation point $x$, and we can indeed conclude that $|x-a|\ge n-|x-a_n|$ for each $n\in\Bbb N$. It is incorrect, however, to say that this means that $|x-a_n|<1$ for infinitely many $n\in\Bbb N$: this inequality has nothing to do with that fact. There are infinitely many $n\in\Bbb N$ such that $|x-a_n|<1$ because $x$ is an accumulation point of the sequence. This means that there are infinitely many $n\in\Bbb N$ such that

$$|x-a|\ge n-|x-a_n|>n-1\;,$$

which is impossible, since $|x-a|$ is finite.

Answer 2

This seems to be the proof that $A$ is bounded. That $A$ is closed is trivial under sequential compactness. A converging sequence in $A$ only has one accumulation point, its limit, and that has by assumption to be an element of $A$.

$A$ is bounded if $A\subset B(a,n)$ for some $a$ and $n\in\Bbb N$. The contraposition is that $A$ has elements outside every ball $B(a,n)$, select one and call it $a_n$. Then with the limit point $x$ of $(a_n)_n$ you get, as stated, that for infinitely many $n\in\Bbb N$ you get $|x-a_n|<1$. Then by triangle inequality $$ n\le|a-a_n|\le|a-x|+|x-a_n| <|a-x|+1 $$ and this can only be true for finitely many $n\in\Bbb N$, which is in contradiction to it being true for infinitely many.