Question

Show that the set of rational numbers with the subspace topology of R is disconnected

Answer 1

Let a$\in$ R be an irrational number.

Then a$\notin$ Q (rationals)

Let us suppose the sets :

P = (- $\infty$ , a ) $\cap$ Q

T = ( a, $\infty$ ) $\cap$ Q

Let x $\in P$

Let B($\in$ ,x) be an open ball in of x in Q

Then, for all x in P, there exists $\in$ from R such that B($\in$ ,x) lies in P if $\in = a-x$

Similarly, for all x in T , there exists $\in$ from R such that B ($\in$ , x ) lies in T if $\in = x-a$

So , there open neighborhoods of P and T in Q, hence P and T are open sets in Q.

Now,

P $\cup$ T = Q , P$\cap$ T = $\varnothing$ where P and T are non-empty open sets.

So, P and T are a separation in Q.

Hence, the result.