1. (Or 0; some textbooks consider the empty set to be connected, and some do not.)
Let X be a subset of the irrational numbers with cardinality > 1. We prove that X is not connected. Let x and y be distinct elements of X; assume without loss of generality that x < y. Choose any rational number q such that x < y < q. (For example, choose an integer n sufficiently large that 0 < 1/n < y-x. Then some multiple of 1/n lies between x and y.) Now consider the sets A = X ∩ (-∞, q) and B = X ∩ (q, ∞). Since q is rational, q ∉ X, hence X = A ∪ B. But A and B are disjoint open subsets of X, so this proves that X is not connected.
Let X be a subset of the irrational numbers with cardinality > 1. We prove that X is not connected. Let x and y be distinct elements of X; assume without loss of generality that x < y. Choose any rational number q such that x < y < q. (For example, choose an integer n sufficiently large that 0 < 1/n < y-x. Then some multiple of 1/n lies between x and y.) Now consider the sets A = X ∩ (-∞, q) and B = X ∩ (q, ∞). Since q is rational, q ∉ X, hence X = A ∪ B. But A and B are disjoint open subsets of X, so this proves that X is not connected.
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Since Q is dense in R then the cardinality of X must be 1.
To see why, since X is connected then it must be an interval (a,b) of R. All we have to show is there is a rational number p/q in (a,b). To find one just select and integer q large enough so that the interval (aq, bq) contains an integer p (we can do this since lim q(b-a) gets arbitrarily large as q does). Then since p is in (aq, bq), p/q is in (a,b). Contradicting that all numbers in X are irrational.
To see why, since X is connected then it must be an interval (a,b) of R. All we have to show is there is a rational number p/q in (a,b). To find one just select and integer q large enough so that the interval (aq, bq) contains an integer p (we can do this since lim q(b-a) gets arbitrarily large as q does). Then since p is in (aq, bq), p/q is in (a,b). Contradicting that all numbers in X are irrational.