Bolzano–Weierstrass theorem

**MathSucker** · April 29th 2011, 06:40 AM

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

**Plato** · April 29th 2011, 06:50 AM

Originally Posted by MathSucker

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

Some statements of this theorem vary. But here is one source.

**MathSucker** · April 29th 2011, 09:36 AM

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

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

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

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

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

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

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

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

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

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

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

This results in a subsequence of $(x_n)$ that converges to $x$ .

**HallsofIvy** · May 2nd 2011, 03:38 PM

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.

**Drexel28** · May 2nd 2011, 03:44 PM

There are very short proofs, but they may use machinery/terms above the level of your current course. What level is this at?

**MathSucker** · May 3rd 2011, 09:55 AM

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.

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