I am looking for the most compact proof for the Bolzano–Weierstrass theorem,anybody have any links/ideas. Thanks.

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- Apr 29th 2011, 06:40 AMMathSuckerBolzano–Weierstrass theorem
I am looking for the most compact proof for the Bolzano–Weierstrass theorem,anybody have any links/ideas. Thanks.

- Apr 29th 2011, 06:50 AMPlato
Some statements of this theorem vary. But here is one source.

- Apr 29th 2011, 09:36 AMMathSucker
What do you think of this one? I want to be able to write it out in the shortest time possible. Is there anything I can trim out?

If $\displaystyle (x_n)$ is a sequence in the closed segment $\displaystyle [a,b]$ then there exists a subsequence of $\displaystyle (x_n)$ that converges to a limit in $\displaystyle [a,b]$.

Let $\displaystyle c$ be an arbitrary point in $\displaystyle [a,b]$, then $\displaystyle [a,c]$ can contain and infinite number of members of $\displaystyle (x_n)$ and it can contain a finite number of members of $\displaystyle (x_n)$.

Let: $\displaystyle D=\{c\in[a,b] \, : \,[a,c]$ contains a finite number of members of $\displaystyle (x_n)\}$

If $\displaystyle c$ is in $\displaystyle D$, then $\displaystyle [a,c]$ has a finite number of members of $\displaystyle (x_n)$, and any subsest of $\displaystyle [a,c]$ can only contain a finite number of members of $\displaystyle (x_n)$. Hence, for any $\displaystyle c \in D$, all the points between $\displaystyle a$ and $\displaystyle c$ are also in $\displaystyle D$.

Let $\displaystyle x=sup(D)$ and assume $\displaystyle x \in (a,b)$.

Take an arbitrarily small $\displaystyle \varepsilon$. $\displaystyle [a,x+ \varepsilon]$ is not in $\displaystyle D$ since $\displaystyle x+ \varepsilon$ is greater than $\displaystyle sup(D)$. $\displaystyle [a, x - \varepsilon]$ is in $\displaystyle D$ since $\displaystyle x - \varepsilon$ is less than $\displaystyle sup(D)$.

$\displaystyle [a, x - \varepsilon]$ is a subset of $\displaystyle [a, x + \varepsilon]$. The larger set contains an infinite number of members of $\displaystyle (x_n)$ and the subset contains a finite number of members of $\displaystyle (x_n)$. So, the complement of the subset contains an infinite number of members of $\displaystyle (x_n)$.

Thus, every $\displaystyle x$ in $\displaystyle [a,b]$, has a neighbourhood that contains an infinite number of members of $\displaystyle (x_n)$.

Set $\displaystyle \varepsilon=1$. Take any member of $\displaystyle (x_n)$ in $\displaystyle (x-1,x+1)$ and make this the first member of a sequence.

Set $\displaystyle \varepsilon=1/2$. Take any member of $\displaystyle (x_n)$ in $\displaystyle (x-1/2,x+1/2)$ with a higher index than the previous member, and make this the next member of the sequence.

Set $\displaystyle \varepsilon=1/4$ and so forth indefinitely.

This results in a subsequence of $\displaystyle (x_n)$ that converges to $\displaystyle x$. - May 2nd 2011, 03:38 PMHallsofIvy
The best proof based on what? It is, in fact, possible to take "Bolzano-Weierstrasse" as a defining axiom for the real numbers and then the other basic properties (monotone convergence, upper bound property, Heine-Borel, etc.) can be proved from that. But you can also prove "Bolzano-Weierstrasse" from any of them.

- May 2nd 2011, 03:44 PMDrexel28
There are very short proofs, but they may use machinery/terms above the level of your current course. What level is this at?

- May 3rd 2011, 09:55 AMMathSucker
Undergraduate level, but we're given zero worked solutions. The prof basically just writes out proofs on the board, hands us a list of questions, and then walks out. I was just looking for a proof for the "Bolzano-Weierstrasse" theorem that I could bang out on the exam in as little time possible. I found that one, and then wrote it out in my own words and was wondering if it made sense.