Let X be a connected subset of real numbers. If every elements of X is irrational then what is the cardinality of X? We know cardinality of irrational numbers is same as the cardinality of real numbers,which is denoted by 'a'.
But I don't know what will be the change of cardinality for imposing the connectedness condition..Is it same of different?
On
If you don't have a convenient general theorem to hand, you could proceed as follows:
Suppose $r$ is a rational number. Consider the function $f_r:X\to \{0,1\}$ for which $f_r(x)=0$ if $x\lt r$ and $f_r(x)=1$ if $x\gt r$. Then it should be easy to show that $f_r$ is continuous.
Since $X$ is connected, either $f_r^{-1}(0)$ or $f_r^{-1}(1)$ (or both) must be empty.
If this is true for all $r$ then $X$ cannot contain two points (which can be separated by a rational number), so must contain $0$ or $1$ points.
A subset of the real numbers is connected if and only if it is an interval. Since there exists a rational number between every two irrational numbers, a connected subset of the reals containing only irrational numbers can therefore at most have 1 element. Note that the empty set also satifies the condition.
So the cardinality is 0 or 1