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Number Theory Level 3

Which of the following statements are wrong?

(a) Each point on the real line corresponds to a unique real number.
(b) There are more real numbers between any two consecutive integers than the whole set of integers.
(c) We can't say (b) above since we can't compare two infinities.
(d) Every bounded interval is a finite set.
(e) Some bounded intervals are finite sets.
(f) There are finite irrational numbers between any two given irrational numbers.

a, d, f b, d, f c, f Some other combination

1 solution

Omkar Kamat
Dec 29, 2014

This question is easy if you know the basics of real analysis.

A) This follows from the fact that each real number is the limsup of a sequence of rational numbers and the limsup of a sequence is unique by definition. (limsup= limit of the supremum of the sequence)

B) If we try to list the reals in between two integers, say 1 and 2, we can always find 1 that is not in the list. This is a standard result in real analysis which is called Cantor's Diagonalisation.

C) We can compare 2 infinities as shown in the last point. So this option is wrong.

D) An open interval (a,b) a<b can contain infinitely many points. However, they are bounded by b+1 above and a-1 below. So this is wrong. Consider all rationals of the form 1/n where n is an integer. This sequence is clearly bounded. However, there are infinitely many rationals here.

E) This can be true for certain sets but not ALL sets. There are many examples you could create to show that this is true.

F) Consider 2 rational numbers A and B. Assume A<B. The rationals are dense in the real numbers, i.e. between any 2 real numbers, we can find a rational. So we can find a rational C such that A<C<B. Now consider (A+C)/2 .This is irrational and is between A and B. We can repeat this ad infitum to get infinitely many irrationals.

We can repeat this argument again and again to get infinitely many irrationals between the two original irrational numbers.

So the wrong options are C,D and F. So 3 of them are wrong.

Your argument for F is false; consider $A = -\sqrt{2}, B = \sqrt{2}$ . But F is indeed false; the easiest way is to show that there are uncountably many reals between $A,B$ , but only countably many rationals (since the set of rationals is countable), so there are still uncountably many left, and those that are left are irrationals.

Ivan Koswara - 6 years, 5 months ago

Yes. Thank you. Using the cardinalities of the rationals and the irrationals is the most intuitive thing to do. I was just trying to give some intuition as to why the results might be true as not everyone would have done real analysis.

Omkar Kamat - 6 years, 5 months ago

Bounded interval is always of the form $[a,b]$ ; that is, $\{x | a \le x \le b\}$ . You might have meant bounded sets for D and E. But still, D is still false and E is still true (consider $[a,a]$ ).

Ivan Koswara - 6 years, 5 months ago

I think I was talking about bounded sets not closed sets as bounded sets include clopen and open sets as well. So my answers are more general than what is required.

Omkar Kamat - 6 years, 5 months ago

An interval is always in the form of $(a,b)$ , which may or may not include the endpoints. (Which means "bounded interval" is superfluous as all intervals except $(-\infty,\infty)$ are bounded, and also means I misinterpreted it earlier.) So yes, you're looking for "bounded sets", not "bounded intervals".

Ivan Koswara - 6 years, 5 months ago

I think b is also wrong because between any 2 integers there are infinite rational nos and the whole set of integers is also infinite. So we can't actually say that there are more real numbers between any two consecutive integers than the whole set of Integers.

Sarthak Pal - 6 years, 5 months ago

No b is right. Look up on cardinalities and cantors diagonalisation argument to find out why. It's one of the things you learn in real analysis.

Omkar Kamat - 6 years, 5 months ago

DANG IT! I missed D and that was clearly incorrect! nice question

michael bye - 5 years, 6 months ago

Sleekit ...... i missed that last spot..

Matthew Huckabey - 2 years, 5 months ago

It does make sense, as the result is irrelevant, to answer some other combination.

Matthew Huckabey - 2 years, 5 months ago

How do you know this is real?

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