I want this short proof of the Bolzano-Weierstrass Theorem checked please

[Eclair_de_XII]

TL;DR Summary

Any bounded, infinite subset of the real numbers must contain at least one limit point.

Let $X$ be a bounded subset of $\mathbb{R}$ with infinite cardinality. We consider a countably-infinite subset of $X$. We write this set as a sequence to be denoted $\{a_n\}_{n\in\mathbb{N}}$.

Now define $A$ to be the set of points in the sequence with the property that for each $a_n\in A$, there is an open set that owns $a_n$ with the property that the intersection of the open set w.r.t. to the sequence is empty. If this condition is fulfilled, then choose the open set $U_n$ s.t. there is a point $a_k\in\{a_n\}_{n\in\mathbb{N}}$ s.t. $a_k=\sup U_n$.

Suppose $A^C=\emptyset$ and assume $\inf\{a_n\}_{n\in\mathbb{N}}\in A$. Suppose $\sup\{a_n\}_{n\in\mathbb{N}}\in A$. By definition, there is a point $a_k$ s.t. $a_k>\sup\{a_n\}_{n\in\mathbb{N}}$, which is a contradiction. Hence, $\sup\{a_n\}_{n\in\mathbb{N}}\in A^C$, contrary to supposition.

 

[Infrared]

Summary:: Any bounded, infinite subset of the real numbers must contain at least one limit point.

Let $X$ be a bounded subset of $\mathbb{R}$ with infinite cardinality. We consider a countably-infinite subset of the subset contained in $X$ whose complement w.r.t. $X$ has finite cardinality.

This can't happen if $X$ is uncountable.

Now define $A$ to be the set of points in the sequence with the property that for each $a_n\in A$, there is an open set $U_n>0$ owning $a_n$. Moreover, the intersection $U_n\cap\{a_i\}_{i\in\mathbb{N}}$ is empty and there is a point $a_k\in\{a_n\}_{n\in\mathbb{N}}$ s.t. $a_k=\sup U_n$.

I don't understand this definition of $A$ at all. That intersection can't be empty since it contains $a_n$, if I'm reading correctly. What's the relevance of its supremum being an element of the sequence?

Suppose $A^C=\emptyset$ and assume $\inf\{a_n\}_{n\in\mathbb{N}}\in A$. Suppose $\sup\{a_n\}_{n\in\mathbb{N}}\in A$. By definition, there is a point $a_k$ s.t. $a_k>\sup\{a_n\}_{n\in\mathbb{N}}$, which is a contradiction. Hence, $\sup\{a_n\}_{n\in\mathbb{N}}\in A^C$, contrary to supposition.

I'm having a hard time following, but it looks like you're trying to argue that either the supremum or infimum of a bounded sequence is a limit point of that sequence, which is not true. For (counter-)example, consider $\{-1,1,1/2,1/3,1/4,1/5,...\}.$ The sup and inf are 1 and -1, respectively, but 0 is the only limit point.

 

[Eclair_de_XII]

That intersection can't be empty since it contains $a_n$, if I'm reading correctly.

You're right. Maybe I should redefine $U_n$ to be equal to the difference of the open set whose right end-point is equal to another point in the sequence and the singleton $\{a_n\}$.

I'm having a hard time following, but it looks like you're trying to argue that either the supremum or infimum of a bounded sequence is a limit point of that sequence, which is not true.

My argument is as follows:

Assume that the infimum of the sequence is a point in the sequence. Otherwise, it is a limit point. If $\{a_n\}_{n\in\mathbb{N}}$ contains no limit points, then each point $a_n$ must have an open ball that contains no other element of the sequence other than itself. If this is possible, then we can choose an open ball whose right end-point is equal to another point in the sequence, and so on. If the supremum is an element of the sequence, then this is a contradiction. If it is not an element of the sequence, then it is a limit point.

 

[Eclair_de_XII]

You know, I realize that the argument I based my proof on is flawed for the reason that there would be no point larger than the supremum of a given sequence, regardless of whether or not the supremum is a limit point. I'll come up with a new argument/proof soon.

 

[Eclair_de_XII]

Let $X\subset\mathbb{R}$ be bounded and infinite. Choose a countably infinite subset of $X$ and call it $\{a_n\}_{n\in\mathbb{N}}$, or $A$ for short.

Suppose no point in $A$ is a limit point. Consider the open interval $(\inf A,\sup A)$ which contains all but two points in $A$. Denote $d_1:=\sup A-\inf A$.

Now consider the set $A_1:=A\setminus\{\sup A,\inf A\}$. We note that $\sup A_1<\sup A$ and $\inf A_1>\inf A$. Hence, $d_2:=\sup A_1-\inf A_1$ must be less than $d_1$. Continuing in this fashion, we obtain a sequence of decreasing points $\{d_m\}_{m\in\mathbb{N}}$ that converge to zero.

Let $\epsilon>0$. Then there exists $N\in\mathbb{N}$ s.t. if $m\geq N$ then $\leq d_{m-1}<\epsilon$.

We define the set $A_m$ to be the set obtained from nixing the infimum and supremum from $A_{m-1}$, and moreover, the smallest open interval containing $A_m$ is $(\inf A_{m-1},\sup A_{m-1})$.

The set $A_m$ must contain infinitely many points of $A$ since its complement w.r.t. $A$ has finite cardinality. Construct a subsequence from $A_m$ and denote it $\{a_{n_i}\}_{i\in\mathbb{N}}$. For each $k,l$, $|a_{n_l}-a_{n_k}|<d_{m}<d_{m-1}$.

Hence, this subsequence is Cauchy and thus convergent. As a result, there is a point $x_0\in X\setminus A$ s.t. there is $N_2\in\mathbb{N}$ s.t. if $k\geq N_2$, $x_0\in B(a_{n_k},\epsilon)$.

 

[fresh_42]

Before I get into details (means I am lazy), did you check your ideas with the two proofs on Wikipedia?
https://en.wikipedia.org/wiki/Bolzano–Weierstrass_theorem#Proof

 

[Eclair_de_XII]

I have not. But it does sound somewhat like the proof that implements those nested intervals, with the flaw that the distance of the nested intervals in my proof might not converge to zero. For example, if we have the union of sequences $\{\frac{1}{2n}\}_{n\in\mathbb{N}}\cup\{1+\frac{1}{n}\}_{n\in\mathbb{N}}$, the algorithm does not even work.

I might have to make the assumption that neither the supremum or infimum of any given $A_n$ is a limit point. And if the sequence of $\inf A_n,\sup A_n$ do not converge onto each other, then they'd necessarily converge to the left and right end-points, respectively, of some interval $(a,b)$.

Note that $b=\inf\{\sup A_n\}$ and $a=\sup\{\inf A_n\}$. Then necessarily, $\inf A_n$ must converge to $a$ and $\sup A_n$ must converge to $b$. This proves the existence of two limit points; the proof above is just a special case of this when $b=a$.

I'm pretty sure this whole proof could be formalized better, and shortened, to boot. For example, I could say that the sequence of $\{\inf A_n\}$ is bounded from above and is increasing, so it must converge to its supremum, etc. And I might not even need to consider the sequence $\{\sup A_n\}$.

 

[Eclair_de_XII]

Let $X\subset\mathbb{R}$ be bounded and infinite. Choose a countably infinite subset of $X$ and call it $\{a_n\}_{n\in\mathbb{N}}$, or $A$ for short.

Suppose the supremum of $A$ is not a limit point. Consider the sequence $A_1:=A\setminus\{\sup A\}$ and assume that the supremum of $A_1$ is not a limit point. We note that $\sup A_1<\sup A$. Continuing in this fashion, we obtain a sequence of decreasing points $\{\sup A_m\}_{m\in\mathbb{N}}$ that is bounded from below. Hence, this sequence must converge to its infimum $a\in X$.

I realize now that this is just a rephrasing of the first proof in the Wikipedia page you linked. Of course, in hindsight, I realize that it might be possible for the second proof I have in this topic to work provided it's reworked into a proof by contradiction.

 

[Eclair_de_XII]

Let $X\subset\mathbb{R}$ be bounded and infinite. Choose a countably-infinite subset of $X$ and call it $\{a_n\}_{n\in\mathbb{N}}=:A$. Moreover, the open interval $(\inf A,\sup A)$ must own infinitely many points of $A$, since the latter is a subset of the former.

Suppose that $X'=\emptyset$. Then it must follow that $A'=\emptyset$ also, and so, neither the infimum nor supremum of $A$ are limit points; they must belong to $A$, as a result. Define the set $A_1:=A\setminus\{\inf A,\sup A\}$. It must follow that $\inf A_1$ and $\sup A_1$ must be less than $\inf A$ and $\sup A$ respectively, and so, $A_1$ can be contained within a smaller open interval than $A$. Moreover, $\inf A_1\in X$, so it is not a limit point. We can nix the infimum and supremum of $A_1$ to form $A_2$.

Continuing in this fashion, we define a sequence of intervals $\{A_n\}_{n\in\mathbb{N}}$ and a sequence of strictly decreasing points of the form $d_n:=\sup A_n-\inf A_n$. The sequence $\{d_n\}_{n\in\mathbb{N}}$ is bounded from below by zero, and so must converge to its infimum.

Suppose $\inf \{d_n\}>0$. Then there is an open interval whose right endpoint is a limit point; specifically, it is converged onto by the sequence $\{\sup A_n\}_{n\in\mathbb{N}}$, contrary to the fact that $X$ does not have any limit points.

Let $\epsilon>0$. Then there is $N>0$ s.t. if $n\geq N$, $d_n=\sup A_n-\inf A_n<\epsilon$. We note that $A_n$ has infinite cardinality since its complement wrt $A$ has cardinality equal to $2n$. No two elements in any subsequence contained entirely within $A_n$ can differ by more than $d_n$ and by proxy, $\epsilon$. It follows that any such subsequence is Cauchy and thus, converges to some value $a\in A_n\subset A\subset X$, contrary to supposition.