# Equivalency of some advanced calculus properties

1. Jun 25, 2007

### hsp

i really don't know how to prove that the following are equivalent:

2. Jun 26, 2007

### HallsofIvy

Staff Emeritus
Did you leave something out? Prove what are equivalent?

3. Jun 26, 2007

### Gib Z

Lol this looks like a scary ghost thread, with a missing passage from the OP and a double identical post from a moderator :D

4. Jun 26, 2007

### hsp

equivalence of properties

Gud afternun. I have this problem to prove that these following properties are equivalent:
Nested Interval Property
Bolzano-Wierstrass theorem
Monotonic sequence property
LUB property
Heine-Borel theorem
archimedean property and cauchy convergence
line connectedness
dedekind completeness

5. Jun 26, 2007

### Gib Z

Are you quite sure that they are equivalent..

6. Jun 26, 2007

### hsp

yes. I've seen the diagram/sketch of the proof on the book advanced calculus by Buck.

7. Jun 26, 2007

### HallsofIvy

Staff Emeritus
8. Jun 27, 2007

### mikecon0523

I have an answer for that.. LET US PRAY!

9. Jun 27, 2007

### ZioX

You interpreted it wrong, it seems. It is archimedean property and not archimedean convergence. Archimedean property is the fact that the reals contain no 'infinitesimals'. Disregarding the formal use of infinitesimal we can just say 'for any real number x there exists a natural number n such that n>x.

'The non-existence of nonzero infinitesimal real numbers is intuitively obvious. In axiomatic theory of real numbers, it is implied by the least upper bound property as follows. Denote by Z the set consiting of all positive infinitesimals, together with zero. This set is non-empty and is bounded above by 1 (or by any other positive non-infinitesimal, for that matter) and nonempty. Therefore, Z has a least upper bound c. Suppose that the real number c is positive. Is c itself an infinitesimal? If so, then 2c is also an infinitesimal (since n(2c) = (2n)c < 1), but that contradicts the fact that c is an upper bound of Z (since 2c > c when c is positive). Thus c is not infinitesimal, so neither is c/2 (by the same argument as for 2c, done the other way), but that contradicts the fact that among all upper bounds of Z, c is the least (since c/2 < c; but every x > c/2 can't be infinitesmal: nx > nc/2 > 1). Therefore, c is not positive, so c = 0 is the only infinitesimal.'

Last edited: Jun 27, 2007
10. Jun 28, 2007

### hsp

I just want to clear my statement. What I've said is "archimedean property and cauchy convergence property"

11. Jun 28, 2007

### HallsofIvy

Staff Emeritus
12. Jun 12, 2011

### itspixiejem

mikecon and hsp, I think i now both of you...=)) I also need answers for this topic. Please share! =))