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Prove that a set with an uncountable subset is itself uncountable.

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Let M be any infinite set and let set A be any countable set. Prove that .

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Prove that each of the following sets is countable:

(a) The set of all numbers with two distinct decimal representations (like 0.50000…and 0.49999…).

(b) The set of all rational points in the plane (that is, points with rational coordinates)

(c) The set of all rational intervals (That is, intervals with rational end points)

(d) The set of all polynomials with rational coefficients.

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A number is called algebraic if it is a root of a polynomial equation with rational coefficients. Prove that the set of all algebraic numbers is countable.

Prove the existence of uncountably many transcendental numbers, that is, numbers which are not algebraic. Hint: Use Theorems 2 and 5 of previous blog(s).

Prove that the set of all real functions (more generally, functions taking values in a set containing at least two elements) defined on a set M is of power greater than the power of M. In particular, prove that the power of the set of all real functions (continuous and discontinuous) defined in the interval is greater than c. Hint: Use the fact that the set of all characteristic functions (that is, functions taking values 0 and 1) on M is equivalent to the set of all subsets of M.

Give an indirect proof of the equivalence of the closed interval , the open interval and the half-open interval or . Hint: Use theorem 7 (the Cantor-Bernstein theorem).

Prove that the union of a finite set of a finite or countable number of sets each of power c is itself of power c.

Prove that the each of the following sets has the power of the continuum:

(a) The set of all infinite sequences of positive integers.

(b) The set of all ordered n-tuples of real numbers.

(c) The set of all infinite sequences of real numbers.

Develop a contradiction inherent in the notion of the “set of all sets which are not members of themselves”. Hint: Is this set a member of itself? Comment: Thus, we will be careful to sets which are “too big,” like the “set of all sets”.

We will go through the solutions of the above soon,

Cheers,

Nalin Pithwa

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