Is the set of irrational numbers compact?

This set is not closed, therefore non-compact. Every irrational point has rationals in every neighborhood, and thus cannot be exterior.

What are the compact subsets of Q?

HINT: A subset of Q is compact in Q if and only if it is compact as a subset of R, since compactness is an inherent property of the set and its relative topology, not of how it sits inside some other space.

Is the set of rational numbers compact?

Answer is No . A subset K of real numbers R is compact if it is closed and bounded . But the set of rational numbers Q is neither closed nor bounded that’s why it is not compact.

What is the set of all irrational numbers?

Thus the set of all irrational numbers is uncountable.

Is the set of rational numbers compact as a subset of the real numbers?

Answer is No . A subset K of real numbers R is compact if it closed and bounded . But the set of rational numbers Q is neither closed nor bounded that’s why it is not compact.

Is the set of irrational numbers in [ 0, 1 ] uncountable?

An uncountable set is a set, which has infinitely many members. The basic idea of proving that is to show that by averaging between every two different numbers there exists a number in between.

How to find the limit of an irrational number?

Apply the above with X = K and A = (a, b) ∩ Q. Choose a sequence of rational points in A converging to an irrational point of [a, b] ⊂ R. Then by the lemma, the limit, which is an irrational number, lies in ¯ K = K ( K is a compact subset of a Hausdorff space, hence is closed).

Which is an irrational number in the Hausdorff space?

Then by the lemma, the limit, which is an irrational number, lies in ¯ K = K ( K is a compact subset of a Hausdorff space, hence is closed). But we have assumed K ⊂ Q, a contradiction. Claim. For all ϵ > 0, the set [ − ϵ, ϵ] ∩ Q is non-compact.