Interior closure and boundary of a set

Question

I am giving a solved example from a textbook about which I have some doubts:

The question is to find interior, closure and boundary of

$$B= \{(x,y)\in\mathbb R^2|y=0\} \subseteq \mathbb R^2$$

Solution:

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Let if there exist a $p(x_{0},y_{0})\in \mathrm{int} (B)$. Then there exists $r>0$ such that $B(p,r) \subset B$

Let $q=(x_0, \frac{r}{2})$. Then $d(p,q)= \frac{r}{2} < r$, so $q \in B(p,r)$. But $q$ does not belong to $B$ which is a contradiction. Thus $\mathrm{int}(B)=\emptyset$

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Let $C \in\overline{B}\setminus B$. Let $C = (\alpha, \beta)$. since $C \in B$, $|\beta|>0$. Then consider $B(C, \frac{|\beta|}{2})$.

$B(C,\frac{|\beta|}{2}) \cap B =\emptyset$, So C does not belong to $\overline{B}$, a contradiction. Thus $\overline{B}=B$ and boundary $B=B$.

I am studying maths as a hobby and do not have usual access to professors/fellow students to clear doubts. So I shall be grateful if following is clarified:

1. Does the initial statement of the question mean, though our given space is two dimensional $x, y$ plane...Set B is only the $x$ axis (since $y=0$)?

2. If so, why should we take $p(x_0, y_0)\in\mathrm{int}(B)$ instead of taking $p(x_0,0)\in \mathrm{int}(B)$?

3. Are we purposely taking $q=(x_0, \frac{r}{2})$ instead of taking $q=(x_0,0)$, so that it lies above the $x$ axis and therefore does not belong to $B$... to get the contradiction?

4. When we say $C\in\overline{B}\setminus B$, doesn't it mean $C$ belongs to $\overline{B}$ but $C$ does not belong to $B$ (and hence belongs to the boundary of $B$)? Then how do we say $C \in B$ in subsequent step?

5. Can the question be changed to:

   (a) $B= \{(x,y)\in \mathbb R^2:y=0\}$ in $\mathbb R$? If it can be, will in this case interior and closure of $B$ be $B$ itself?

Answer 1

You state that the solution says

Since $C \in B, |\beta|>0$

which is false, because if $c\in B$, then $\beta=0$. This is probably just a typo on your end, but still, the proof that $\overline B=B$ is false (even though the claim is correct).

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As far as your questions:

1. Yes.
2. You can equally take $(x_0, 0) \in \text{int}(B)$. In fact, if you take $(x_0, y_0)$, you know that since the point is an element of $B$, that $y_0=0$.
3. Yes.
4. See above. There's probably a typo in there somewhere. It should probably say "since $C\notin B, |\beta|>0$".
5. Certainly you cannot say the question like that, because $\{(x,y)\in\mathbb R^2|y=0\}$ is not a subset of $\mathbb R$, but it is a subset of $\mathbb R^2$.

5.1: You could say $B=\{x| x\in\mathbb R\}$, and in that case, the interiour of $B$ is $B$, and there is no boundary of $B$. But in that case, $B=\mathbb R$, so it isn't as interesting.