0
$\begingroup$

Imagine infinite intersection of $\bigcap_{n=1}^\infty\left(0,\,\frac1n\right)$ where $n$ is positive natural number.

Assume infinite intersection of $\left(0,\,\frac1n\right)$ is empty set. In other words, given any positive real number $p$ less than 1, there exists some $N$ such that $\frac1N$ is less than $p$. But then we have actually made one to one mapping from real numbers to rational numbers. But such a mapping is impossible for the difference of cardinality between real and rational numbers. So the infinite intersection of $\left(0,\,\frac1n\right)$ is not empty set. What's wrong with it?

$\endgroup$
0

2 Answers 2

Reset to default
1
$\begingroup$

You won't get a one-to-one mapping form $\mathbb R_+$ to $\mathbb Q$ in this way.

Every positive real is larger than some $\frac1n$, but nothing in what you say implies that two different positive reals are larger than different $\frac1n$s.

(In fact, you don't get a concrete mapping at all unless you either appeal to the Axiom of Choice to pick out for each of the $p$s one of the many $\frac1n$s that are smaller than it, or decide on a concrete way to choose between them -- say, always pick the smallest $n$ such that $\frac1n<p$. But in the latter case it is easy to see that, for example, the real numbers $\frac12+\frac17\pi$ and $\frac12+\frac18\pi$ both map to $\frac12$, so your correspondence is not one-to-one).

$\endgroup$
3
  • $\begingroup$ Ok, now I understand the argument above is false. But what about this. For any 1/n, there exists some real number p less than 1/n. Then, there exists infinite real number in the infinite intersection of (0,1/n). What's wrong with it then? $\endgroup$
    – PHilo
    Commented Oct 9, 2016 at 6:58
  • 1
    $\begingroup$ @PHilo: Just because you have found a real number smaller then $\frac1n$ doesn't mean that it will also be smaller than $\frac1m$ for a different $m$, so you have no reason to think you have actually found anything in the intersection. $\endgroup$
    – hmakholm left over Monica
    Commented Oct 9, 2016 at 7:02
  • $\begingroup$ Because for every p < 1/n there is another 1/m < p so p is not in the intersection. $\endgroup$
    – fleablood
    Commented Oct 9, 2016 at 7:44
0
$\begingroup$

Assume infinite intersection of (0,1/n) is empty set. In other words, given any positive real number p less than 1, there exists some N such that 1/N is less than p. But then we have actually made one to one mapping from real numbers to rational numbers. But such a mapping is impossible for the difference of cardinality between real and rational numbers. So the infinite intersection of (0,1/n) is not empty set. What's wrong with it?

The claim that the mapping is one-to-one is unjustified and false. Otherwise (assuming AC) the argument seems correct.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .