Let's discuss Mathematics
- Thread starter ParadigmShifter
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ParadigmShifter
Random Nonsense Generator
zero of course
So do you consider that an impossibility, or just something no one has found?
IdiotsOpposite
Boom, headshot.
He's talking about the number 0. Part of 0 is as great in size as the whole of 0.
Of course, I think the Cantor Ternary Set works just as well, and it's not 0.
Actually I meant more along the lines of what Sauron said, like how {2, 3, ...} can get considered as a "part" of the "whole" {1, 2, 3, ...}, which are equal in terms of cardinality.
IdiotsOpposite
Boom, headshot.
Any infinite set should work for you then. I think the set [0,1] is equal to R in cardinality, and is a part of R.
One thing is, the cardinality of infinite sets is only somewhat analogous to the cardinality of finite sets. It's a rather arbitrary subset of the properties of finite set cardinality that applies to infinite sets, which were chosen because it is somewhat useful. The two types of cardinality do not measure the same property.
ParadigmShifter
Random Nonsense Generator
I think dutchfire did a proof in this thread that if a (EDIT: proper) subset of a set has the same cardinality, then both sets are infinite.
EDIT: Proper subset of course. A proper subset is a subset that is not equal to the original set.
EDIT: Proper subset of course. A proper subset is a subset that is not equal to the original set.
Problem is, we can name a part of the reals that has lower cardinality, such as the set {1}.
Perhaps a better example that fits the description is any space-filling curve. It would have infinite length over any intersecting enclosed area, so the part would be as long as the whole. But unlike the set of reals, a single point, isn't a curve or arc, so in some sense it isn't a part of the curve. Ergo every part of a space-filling curve is the same length as the whole (despite filling a smaller area).
ParadigmShifter
Random Nonsense Generator
Depends how you define "part". You seem to be using a topological style definition and arbitrarily exclude sets of measure 0 from a space filling curve but not from the set of reals.
I'm also starting to wonder whether we should start a new thread when the log10 of number of posts in this thread is greater than or equal to 3.
I'm also starting to wonder whether we should start a new thread when the log10 of number of posts in this thread is greater than or equal to 3.
IdiotsOpposite
Boom, headshot.
It's a question of how to interpret natural language, so some level or arbitrary decisions is unavoidable.
Things that are measure 0 aren't a part (in some sense), but if cardinality is the standard of measure, then a set of a single number has a measure of 1.
ParadigmShifter
Random Nonsense Generator
Is that true? I thought countable sets have measure 0. They definitely have dimension 0.
Not that familiar with measure theory, however.
Maybe I mean dimension rather than measure.
EDIT: Maybe that is Hausdorff dimension and not necessarily true for topological dimension. I have seen some definitions of dimension that have dim(Q) = 1, but that has Hausdorff dimension of 0. A vector space over Q can be 1 dimensional too.
EDIT2: OK, after some more research, it seems the Lebesgue measure of any countable set is 0 also
With respect to Rn, all 1-point sets are null, and therefore all countable sets are null. In particular, the set Q of rational numbers is a null set, despite being dense in R.
Null set = set of Lebesgue measure 0
http://en.wikipedia.org/wiki/Null_set
EDIT3: Obviously, any integral over a single point is going to be zero anyway, easily seen from the definition of an integral.
Not that familiar with measure theory, however.
Maybe I mean dimension rather than measure.
EDIT: Maybe that is Hausdorff dimension and not necessarily true for topological dimension. I have seen some definitions of dimension that have dim(Q) = 1, but that has Hausdorff dimension of 0. A vector space over Q can be 1 dimensional too.
EDIT2: OK, after some more research, it seems the Lebesgue measure of any countable set is 0 also
With respect to Rn, all 1-point sets are null, and therefore all countable sets are null. In particular, the set Q of rational numbers is a null set, despite being dense in R.
Null set = set of Lebesgue measure 0
http://en.wikipedia.org/wiki/Null_set
EDIT3: Obviously, any integral over a single point is going to be zero anyway, easily seen from the definition of an integral.
I just mean the quantity by which we judge something to be equal by. If that quantity is cardinality, then {1} is non zero. If that quantity is length, then a point is zero.
Yes, it's arbitrary that we compare numbers by cardinality, and curves by length, but we do most commonly.
Yes, it's arbitrary that we compare numbers by cardinality, and curves by length, but we do most commonly.
ParadigmShifter
Random Nonsense Generator
Well that's because the number of points on a continuous curve over an interval in R is infinite, so we need something else to use as a measure of its size.
IdiotsOpposite
Boom, headshot.
What, I can't claim that [0,2] is bigger than [0,1]?
ParadigmShifter
Random Nonsense Generator
Yes, using measure instead of dimension. They both have the same number of points though 
Just like the number of points in N, Z and Q are the same.
Just like the number of points in N, Z and Q are the same.Well there is a 1 to 1 correspondence between these two sets...
Speaking of which, how does one prove that for uncountable sets?
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