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 is a sequence in the closed segment then there exists a subsequence of that converges to a limit in .
Let be an arbitrary point in , then can contain and infinite number of members of and it can contain a finite number of members of .
Let: contains a finite number of members of
If is in , then has a finite number of members of , and any subsest of can only contain a finite number of members of . Hence, for any , all the points between and are also in .
Let and assume .
Take an arbitrarily small . is not in since is greater than . is in since is less than .
is a subset of . The larger set contains an infinite number of members of and the subset contains a finite number of members of . So, the complement of the subset contains an infinite number of members of .
Thus, every in , has a neighbourhood that contains an infinite number of members of .
Set . Take any member of in and make this the first member of a sequence.
Set . Take any member of in with a higher index than the previous member, and make this the next member of the sequence.
Set and so forth indefinitely.
This results in a subsequence of that converges to .
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.
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.