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TIFR2010-Maths-A-5

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Let $f$ be an one to one function from the closed interval [0, 1] to the set of real numbers $R$, then.

  1. $f$ must be onto
  2. Range of $f$ must contain a rational number
  3. Range of $f$ must contain an irrational number
  4. Range of $f$ must contain both rational and irrational numbers
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Option C should be the correct option, i.e. Range of f must contain an irrational number.

Let R be the set of all real numbers,

P be the set of all rational numbers and,

Q be the set of all irrational numbers.

Clearly the sets P & Q are mutually exclusive & collectively exhaustive with respect to R.

That is, every real number is either a rational number or an irrational number.

The set of all rational numbers i.e P is countably infinite, but the set of all real numbers i.e. R is uncountable.

This implies that their set difference i.e. the set of all Irrational numbers ( set Q here) must be uncountable.

Why so??

If the irrational numbers were countably infinite, then the real numbers would have to be countably infinite as well, as the union of two countably infinite sets.

Now the the set of all the real numbers between 0 & 1 is uncountable(using Cantor's diagonalization theorem, I guess).

So our domain is uncountable.

Since f is a one to one function whose domain is uncountable, its range must also be uncountable.

But the set of rational numbers is countably infinite which is smaller than uncountable, so we will ran out of rational numbers if we try to map [0, 1] to the set of rational numbers.

Thus range of f must contain at least one irrational number.

We can also map [0, 1] to the irrational numbers only, but we can never map it to the rational number only using any one to one mapping, since the cardinality of [0, 1] is greater than the cardinality of P.

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