Complete Ordered Fields and Complete Metric Spaces

Thread replies: 4
Thread images: 1

Anonymous
Complete Ordered Fields and Complete Metric Spaces 2017-09-06 07:08:09 Post No. 9152850
[Report] Image search: [Google]

File: aplus.jpg (7KB, 276x183px) Image search: [Google]

7KB, 276x183px

Complete Ordered Fields and Complete Metric Spaces Anonymous 2017-09-06 07:08:09 Post No. 9152850 [Report]

I am struggling to see whether or not all complete ordered fields are complete metric spaces as well.

Let there exist a set S, where we have multiplication, and addition such that it is a field. Let us give it a total ordering </= ($\leq$ if TeX works here). Then S is an ordered field.

Moreover, let S be a complete ordered field, ie: given any nonempty subset T of S, if the set of upper bounds of the set T is nonempty, then the set of upper bounds possesses a least element.

Claim: The same set S, when equipped with a metric d, is a metric space. Furthermore, (S,d) is a complete metric space (every cauchy sequence in it converges.)

I can't formally get a proof for this, nor can I think of any counter examples. This question is really interesting to me, and it's already taken away so much of my time from actual classwork. I'd really appreciate any help!

Anonymous 2017-09-06 07:47:26 Post No.9152941
[Report]

Anonymous 2017-09-06 07:47:26 Post No.9152941 [Report]

>>9152850
What have you tried to prove it?
I would think you could copy the proof (that all cauchy sequences are convergent) for R verbatim to any field with LUB property and it would work.

Anonymous 2017-09-06 08:38:58 Post No.9153038
[Report]

Anonymous 2017-09-06 08:38:58 Post No.9153038 [Report]

>>9152850
hey. you seem to know much about function spaces. can you help me with my qeustion ? :
Im not a mathematician.
I have an integral of a function, divided by a gamma function. It is a new function.

I interpret it like this: because I can integrate the function I can write the new function in a complete space without the loss of generality and write it as a linear combination of the basis functions. Then I can use the Parseval Identity to do scalar multiplication with functions in that space, right ?

I mean, can I just say that there is a such a space, in which this function can be written ?

Anonymous 2017-09-06 09:16:32 Post No.9153098
[Report]

Anonymous 2017-09-06 09:16:32 Post No.9153098 [Report]

>>9152850
Any complete ordered field is isomorphic to the real numbers, so yes.

I'm aware that Imgur.com will stop allowing adult images since 15th of May. I'm taking actions to backup as much data as possible. Read more on this topic here - https://archived.moe/talk/thread/1694/

I'm aware that Imgur.com will stop allowing adult images since 15th of May. I'm taking actions to backup as much data as possible.
Read more on this topic here - https://archived.moe/talk/thread/1694/