Set of irrational numbers denoted I or Qc is neither open or

Set of irrational numbers, denoted I or Q^c, is neither open or closed in R. x-axis is a closed subset of R^2.

Solution

(A) A set of irrational no. is neither closed nor open

Let us suppose a sequence of irrational no. /a

for every value of a ,/a is irrational

Let B=/a & assume B to be rational

So = B ×a ,Ba will be rational no. & equal to which is surely irrational, contradict the assumption .

Now let us take the case /a tends to zero,for given €>0,the archimedean property of the real numbers assures us that there is a natural no. Such that 1/N<€/ so that /a<€ fir every index a>N,since /a is a set of irrational no. converging(open to closed condition) to rational no.zero(fully closed)& not fully open ,hence a set of irrational no. Can not be open or closed.

(B) X axis is a closed subset of R^2(xy plane)

R^2 is a very large system consist of x axis alone ,y axis alone ,y axis alone & axis on between x & y axis.

X axis itself is sum of real & closed (finite) value system & xy plane comprises wholely x axis which implies that x axis is a close subset of R^2.