ReuvenLax

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 I haven't been blogging in a little while, and I want to start blogging about VSS related topics soon.  However, for now I'll throw out another math puzzle, though this one is much easier.

Remember that given a set U of real numbers, a real number x is said to be a limit point of U if every open interval around x contains a point of U other than x. 

Let V be a bounded, countabll-infinite set of real numbers, and let limit(V) be the set of limit points of V (note that limit(V) might not intersect V at all).  Is it possible for limit(V) to also be a countably-infinite set?

Published Monday, May 23, 2005 1:15 AM by ReuvenLax
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Comments

 

ReuvenLax said:

By countable, I mean a set that is in one-to-one correspondence with the natural numbers. Terminology is rarely consistent, so some calls this countable, some call it countable infinite, some say denumerable, etc.
May 23, 2005 12:16 PM
 

ReuvenLax said:

I accidentally deleted the first comment on this post. The complaint was that countable often is used to mean either countably infinite _or_ finite. In this case I'm strictly talking about countably infinite sets.
May 23, 2005 12:21 PM
 

Royhwa said:

I think...

Yes,it is possible to be.

If the number of members of V is N, the number of members of limit(V) would be 2*N-2.
May 23, 2005 8:41 PM
 

ReuvenLax said:

Nope. A finite set has no limit points, which is why I specified that both sets be infinite.
May 24, 2005 10:46 AM
 

psnarula said:

let V = {1/n}

Then V is both countably infinite and bounded.

The set of limit points of V is simply {0}. This set is finite and thus countable.
May 26, 2005 11:15 AM
 

ReuvenLax said:

Once, again by countable I mean countably-infinite. I'll edit the blog posting to make this more clear.
May 26, 2005 4:19 PM
 

psnarula said:

Oh sorry. Just modify my original comment as follows:

Let V = {1/n + 1/m} where n,m are natural numbers.

V is bounded by [0,2] and is can be shown to be countable by the same diagonalization argument Cantor used to show NxN is countable.

Then the set of accumulation points (limit points) of V is {1/n} since we can get arbitrarily close to each 1/n by letting m->infinity.
May 27, 2005 9:09 AM
 

ndiamond said:

Four really simple answers are the set of rationals in the interval 0 to 1. In these cases it happens that limit(V) = V.

(Yeah I didn't want to favour closed or open or shared or locked
oops
closed or open or left half-closed or right half-closed, so I just mentioned all four of these examples.)
May 29, 2005 8:37 PM
 

ReuvenLax said:

psnarula mentioned the set I had in mind. {1/n + 1/m : n,m :el: N} is probably the simplest set with this property.

The other suggestion, the set of rationals between 0 and 1 does _not_ work. If V is this set, limit(V) is the entire interval [0,1]; not a countable set.
May 31, 2005 4:52 AM
 

psnarula said:

now can we start talking about volume shadow copy? :-)
May 31, 2005 9:30 AM
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