# Non standard analysis

1. Sep 25, 2010

### jem05

hello everyone,

i have this (decreasing) nested sequence of open sets (intervals of infinitesimal radius) about 0.
i just want to know if the intersection of the sets described above is still bounded away from 0.
that is, is it true that the countable intersection of such balls with infinitesimal radius will not be just the singleton.
i failed to prove that, with saturation, and i researched, there was sth with *R having uncountable cofinality, (every countable set is bounded)
i just need help understanding this.
thank you.

2. Sep 25, 2010

### VeeEight

Really? Z, Q?

Each of your open balls are bounded, so wouldn't the intersection be bounded? I admit I know nothing of nsa, but I don't precisely see the question.

3. Sep 25, 2010

### Hurkyl

Staff Emeritus
Z is a bounded subset of *R -- any positive transfinite element would be an upper bound. Of course, *Z is unbounded.

If you know countable sets are bounded, then just invert the radius of your intervals.

4. Sep 25, 2010

### jem05

yeah sure, once i make my peace with ever countable set is bounded. but is that sth known about *R or because it seems vague to me.

5. Sep 25, 2010

### Hurkyl

Staff Emeritus
I'm not familiar with the acronym 'sth'.

6. Sep 26, 2010

### jem05

sorry, i meant sth :something

7. Sep 26, 2010

### Hurkyl

Staff Emeritus
I'm more familiar with the direct construction via ultra-powers -- after choosing well-behaved representatives, I'm pretty sure you can use a diagonal argument to construct an upper-bound on any increasing countable sequence.

I guess you understand things in terms of saturation? I forget precisely what that means, could you briefly describe it?

8. Sep 27, 2010

### jem05

ok thx i should try this out.
countable saturation says that a decreasing sequence of nonempty internal sets is always non empty. my problem is showing my sequence isnt only non empty, it must have more tha one element.

9. Sep 27, 2010

### Hurkyl

Staff Emeritus
Ah! This one's easy. Suppose that it does only have one element. Construct a new sequence where you've removed that element from all of your sets!

10. Sep 28, 2010

### jem05

ok, i used the ultrafilter way, and a sort of diagonalization, it's very un-intiutive though, not that pretty,...
anyway, thanks a lot, hurkyl
so as for the countable saturation method, let's say we have a nested decreasing sequence of intervals (ai,bi)
if intersection is {xo} take the intervals (ai, xo)
by the same reasoning, their intersection is nonempty, so the point of their intersection must also be in the intersection of the previous intervals (ai,bi).
am i correct?

if anyone is interseted in the other proof concerning the ultrafiletrs, let me know, i will post it.