Properties of the indiscrete rational extension of $\mathbb{R}$

Question

Let $X = \mathbb R$ equipped with the topology generated by open intervals of the form $(a,b)$ and set of the form $(a,b)\cap \mathbb Q.$ Then

1. $X$ is regular.

2. $X$ is normal

3. $X$ \ $\mathbb Q$ is dense in $ X$

4. $\mathbb Q$ is dense in $X$

My attempt is :

for (3) $(a-\epsilon,a+\epsilon) \cap \mathbb Q$ is nbd of a $\in \mathbb Q$ intersect with $X$ \ $\mathbb Q$ is empty, so $X$ \ $\mathbb Q$ is not dense in $X$

for (4) i think $\mathbb Q$ is dense in $X$

Please give me the counter example of (1) and (4).

Thank you.

Answer 1

Your solution to (3) appears correct to me (a tiny bit more later).

For (4) you just need to show that every nonempty open set contains a rational number. The nonempty open sets in this space are just unions of the sets you described (i.e., sets of the form $(a,b)$ or $(a,b) \cap \mathbb{Q}$ for $a<b$). So it really suffices to show that all of these sets contain a rational number: this shouldn't be too difficult.

For (1) and (2) I'll just leave a hint.

Hint: Note that $F = \mathbb{R} \setminus \mathbb{Q}$ is closed in this new space. (This is essentially what you proved in your solution to (3).) Can you separate $0$ and $F$ by disjoint open sets? (In a bit more detail, if $U$ is an open neighbourhood of $0$, find an irrational number $x$ such that every open neighbourhood of $x$ intersects $U$.)