# Cardinality and continuum question [closed]

How come the rationals are countable; but there is continuous space between the rationals? For example, between two fractions you can find another, such as (3/2)/10000000 between 1/10000000 and 2/10000000.

-
Why shouldn't this be the case? –  Chris Eagle Jul 12 at 18:37
What do you mean by "there is continuous space between the rationals"? –  Zev Chonoles Jul 12 at 18:37
I believe MathApprentice means "How can it be that the rationals are dense in an uncountable set, yet they are themselves only countable?". –  Austin Mohr Jul 12 at 19:21
@Austin: I think that the intended question may be even more basic: how can a densely ordered set be only countable? –  Brian M. Scott Jul 12 at 21:20

## closed as unclear what you're asking by Zev Chonoles, Andrey Rekalo, Amzoti, Chris Eagle, Asaf KaragilaJul 12 at 19:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking.If this question can be reworded to fit the rules in the help center, please edit the question.

There are different ways to measure the 'size' of a set. One way is by means of cardinality, namely by constructing a bijection with other sets. Since there is a bijection between the rationals and the natural numbers it follows, by definition, that the cardinality of the rationals is the same as that of the naturals. In other words, the rationals are countable.

The cardinality of a set has nothing to do with the shape or density properties of the set. Cardinality is totally blind to such issues. The fact that the rationals are nicely distributed in the real line, and that between any two rationals there is another rational, and even between any two real numbers there is a rational number (a property called 'density') have nothing to do with cardinality. These properties are topological/metric. There are notions of size of a set that take topology into account. For instance, a subset of the reals has $0$ measure if for every $\epsilon >0$ there is a cover of the subset by open intervals such that the sum of the intervals is $<\epsilon$. It is not hard to show that any countable set will have $0$ measure, so that the rationals are 'small' in that respect too.

You seem to be thinking strongly about the property of density. There is a way to see that in that respect that rationals are large. It's quite simply taking the cardinality of the closure of the rationals. The closure of the rationals is the reals, and thus the cardinality is that of the continuum.

So, being large or small all depends on how you measure. Different concepts of size measure different things. A set can be small with respect to one while large with respect to another (for instance, the Cantor set has cardinality $c$ but measure $0$), and there is nothing strange about that.

-
What does measure have to do with topology? Also, the specific notion of density here is order-theoretic, not metric. –  dfeuer Jul 12 at 19:35
@dfeuer Lebesgue measure is strongly related to the topology of $\mathbb R$. The density, in this case, is both order theoretic and metric. –  Ittay Weiss Jul 12 at 22:14
Also related to you answer: math.stackexchange.com/questions/40309/cardinality-density/… –  Asaf Karagila Jul 13 at 8:19
@Did: I think that Clive was informally making the point that even though $\Bbb Q$ is densely ordered, the order isn’t complete, and a lot of ‘stuff’ — the irrationals — that could be there is missing, while in $\Bbb R$ this is not the case. –  Brian M. Scott Jul 14 at 7:30