Basis for a topology on $\mathbb{R}$ given by the union of two families of sets

Question

Let

$$ K=\left\{\frac{1}{n}\quad |\quad n\in \mathbb{N}\smallsetminus \{0\}\right\}\subset \mathbb{R}$$

and consider the sets

$\mathcal{A}=\left\{(a,b) \quad | \quad a,b\in\mathbb{R}, \quad a<b\right\},\quad\quad \mathcal{B}=\left\{(a,b)\cap (\mathbb{R}\smallsetminus K) \quad | \quad a,b\in \mathbb{R}, \quad a<b\right\}$

(a) Prove that $\mathcal{A}\cup \mathcal{B}$ is a basis for a topology $\tau$ on $\mathbb{R}$.

(b) Prove that the euclidean topology $\tau_e$ on $\mathbb{R}$ is not finer than $\tau$.

(c) Determinate the interior, the closure and the boundary of $K$ in $(\mathbb{R}, \tau)$.

(d) Establish whether $(\mathbb{R},\tau)$ is a separable topological space.

(e) Show that $(\mathbb{R},\tau)$ is not $T_3$.

I've found that:

(a) $\mathbb{R}$ is union of elements of $\mathcal{A}$ because $\mathbb{R}=\bigcup\limits_{a<b} (a,b)$, so, $\mathbb{R}$ is union of elements of $\mathcal{A}\cup \mathcal{B}$.

Now, if I choose $U,V\in \mathcal{A}\cup\mathcal{B}$, I have three cases:

1. If $U=\emptyset$ or $V=\emptyset$, then $U\cap V=\emptyset$ that is union of elements of $\mathcal{A}$.
2. If $U=\mathbb{R}$ or $V=\mathbb{R}$, then $U\cap V=V$ (or $U\cap V=U)$, so, I have that $U\cap V\in \mathcal{A}\cup\mathcal{B}$ in both cases.
3. If $U\in \mathcal{A}$ and $V\in\mathcal{B}$, then

$$U\cap V=(a,b)\cap [(c,d)\cap (\mathbb{R}\smallsetminus K)]=(a,b)\cap(c,d)\cap (\mathbb{R}\smallsetminus K)=(\max\{a,c\}, \min\{b,d\})\cap (\mathbb{R}\smallsetminus K)$$ that is emptyset or it's an element of $\mathcal{B}$, so, even in this case we have $U\cap V\in \mathcal{A}\cup \mathcal{B}$.

So, $\mathcal{A}\cup\mathcal{B}$ is a basis for a topology on $\mathbb{R}$.

(b) Clearly, the euclidean topology is contained in $\tau$ because all the euclidean open sets are generated from the elements of the family $\mathcal{A}$. Moreover, I say that the two topology cannot coincide because in (e) it is requested to prove that $\tau$ is not $T_3$ and the euclidean topology is $T_3$ instead. It's not a great solution but I think it works. I would like to find an open set in $\tau$ that is not an euclidean open set, but I haven't succeeded yet.

(c) I've found that $K^\circ=\emptyset$ and $\overline{K}=\partial K=[0,+\infty)$. I'm not really sure of this answer.

(d) I think that it is because

$\mathcal{A}=\left\{(a,b) \quad | \quad a,b\in\mathbb{Q}, \quad a<b\right\}\cup \mathcal{B}=\left\{(a,b)\cap (\mathbb{R}\smallsetminus K) \quad | \quad a,b\in \mathbb{Q}, \quad a<b\right\}$ is a countable basis of $(\mathbb{R},\tau)$ and so $2$-countability implies separability.

(e) I built a counterexample choosing the point $x=\frac{2}{3}$ and the closed set $C=(-\infty,0]\cup K$, do they work?

Is it correct? Any suggestions for the unsure answers?

Answer 1

Some of this is correct, but there are some problems.

a.)

You correctly showed the union of the basis is $\mathbb R$, but you seem confused about what to do for the second part. Neither $\emptyset$ nor $\mathbb R$ are members of $\mathcal A\cup\mathcal B$, so cases 1 and 2 are irrelevant. If $U,V\in\mathcal A\cup \mathcal B$ then the three possibilities to consider are that both are in $\mathcal A$, both are in $\mathcal B$, or one is in each. You have only considered the third one.

You have asserted that $U\cap V\in \mathcal A\cup \mathcal B$, but that is neither true ($\emptyset \notin \mathcal A\cup \mathcal B$), nor what you have to prove.

What you must prove (may help to review the basic criteria B1 and B2 for when a set is a basis for a topology) is that if $x\in U\cap V$ for $U,V\in \mathcal A\cup \mathcal B$, then there is some $W\in \mathcal A\cup \mathcal B$, with $x\in W\subseteq U\cap V$.

As to how to prove this, your argument in 3 basically does the trick, once you generalize it for the cases $U,V\in \mathcal A$ and $U,V\in \mathcal V$. Since you have shown that either $U\cap V\in \mathcal A\cup \mathcal B$, or $U\cap V= \emptyset$, if we begin with $x\in U\cap V$ we must only consider the first of these cases, and then can take $W=U\cap V$.

b.)

Your argument is fine, but to be explicit, $\mathbb R\backslash K$ is open in $\tau$ (see comment on c) below) but is not Euclidean open.

c.)

$K^\circ$ is indeed empty (since no basis element is a subset of $K$), however $K$ is closed (since $\mathbb R\backslash K=\bigcup \mathcal B$ is open, as a union of basis members), so $\overline{K}=\partial K= K$.

d.)

This argument is correct, though you could also directly observe that $\mathbb Q$ intersects every basis element, and is therefore a countable dense subset, if you wanted to be explicit. The latter is slightly preferable because you avoid using the fact that second countability implies separability, which requires the (countable) axiom of choice.

e.)

This example does not work, since $x\in \left(\frac{7}{12},\frac{9}{12}\right)$, and $C\subseteq \left(-\infty,\frac{7}{12}\right)\cup \left(\frac{11}{12},\infty\right)$. More generally, for all the points $x\neq 0$, the set of closed Euclidean neighborhoods of $x$ form a neighborhood basis, so you must choose $x=0$ to get a counterexample.

Then you can take $C=K$, and note that if $U\ni 0$ and $V\supseteq K$ are open, then $V$ is Euclidean open, and $0$ is in the Euclidean closure of $V$, so every Euclidean neighborhood $(a,b)\ni 0$ intersects $V$. Since this intersection is open, and $K$ has no interior, we also have $((a,b)\backslash K)\cap V\neq \emptyset$, so every basic neighborhood of $0$ must intersect $V$, hence $U$ must intersect $V$.