Re: Connectedness question
- From: Ryan Reich <ryanr@xxxxxxxxxxxx>
- Date: Thu, 09 Jun 2005 01:42:09 GMT
Michiel de Bondt wrote:
Ryan Reich wrote:
I think a course in topology is not required to transfer what is on stake here. Both in Q and R, there is a concept of distance: the distance between numbers a and b is |a - b|, the absolute difference of
some deep topology
both numbers. More generally, other sets can have a concept of
distance: the distance between a and b is denoted as d(a, b) usually. A
set is complete if the following holds: if a_1, a_2, a_3, ... is a
Cauchy sequence (google for definition), then it converges to a limit
within the set. Now 1/0!, 1/0!+1/1!, 1/0!+1/1!+1/2!, 1/0!+1/1!+1/2!,
... is a Cauchy sequence that converges to e = exp(1). But e is not a
rational number. So Q is not complete. R is complete, and in fact, R is
the smallest complete set that contains Q, and therefore called the completion of Q.
Please give more context when you quote; I didn't write the three words you cited and I have no idea what you purport to be responding to. In addition, although it IS true that both Q and R have metrizable topologies (and the hyperreals do not), that is not actually related to my explanation and, if you connect what I was trying to explain with what "none" was asking, is actually completely the opposite of what I said.
-- Ryan Reich ryanr@xxxxxxxxxxxx .
- Follow-Ups:
- Re: Connectedness question
- From: Michiel de Bondt
- Re: Connectedness question
- References:
- Connectedness question
- From: none
- Re: Connectedness question
- From: Marc Olschok
- Re: Connectedness question
- From: none
- Re: Connectedness question
- From: Ryan Reich
- Re: Connectedness question
- From: Michiel de Bondt
- Connectedness question
- Prev by Date: Re: Integral Inequality
- Next by Date: Re: Orlow cardinality question
- Previous by thread: Re: Connectedness question
- Next by thread: Re: Connectedness question
- Index(es):
Relevant Pages
|