Interior of $\{(x,y) \mid y = 0\} \cup \{(x,y )\mid x>0 \ \text{and}\ y \neq 0\}$ I want to find the interior of the following subset of $\mathbb{R}^2$ $$\{(x, y) \mid y = 0\} \cup \{(x, y) \mid x>0 \ \text{and}\ y  \neq 0\}$$
I guess that the interior of the above set will be $\mathbb{R}_+ \times \mathbb{R}$. But I can not prove it.
But I have calculated the interior of each set seperately, i.e., $\text{Int } (\{(x, y) \mid y = 0\}) = \emptyset$ and $\text{Int } (\{(x, y) \mid x>0 \ \text{and}\ y  \neq 0\}) = \{(x,y) \mid x>0 \ \text{and}\ y  \neq 0\}$.
But I can not proceed further. Please help me.
 A: Let $A =\{(x, y) \mid y = 0\} \cup \{(x, y) \mid x>0 \ \text{and}\ y  \neq 0\}$
For any $(x,y) \in \mathbb R^2$ if $y =0$ then $(x,y) \in A$.  If $y\ne 0$ then $(x,y) \in A$ if and only if $x> 0$.
So $A = \{(x,y)| y=0; x \le 0\}\cup \{(x,y)| x > 0\}$.
It doesn't make much difference but it's easier to intuitively see that as the open half plane where $x > 0$ with the $x$ axis tacked on like a kite string.  And using intuition I can "see" that the interior is where the picture is solid and thick and that is the half plane where $x > 0$.
Okay... intuition.  That's not a proof.
And we come to your observation

I guess that the interior of the above set will be R+×R. But I can not prove it.

Well do it in steps:

*

*Prove that if $x \in \mathbb R^+\times \mathbb R$, that is if $(x,y)$ is such that $x > 0$ then it is an interior point of $A$.

That's pretty easy.  Let $r = x$ then for any $(u,v) \in B_r(x)$ we have $|x-u|=\sqrt{(x-u)^2 + (y-v)^2} = d((u,v),(x,y)) < r = x$ so $u > 0$.  If $v=0$ then $(u,v) \in A$.  And if $v \ne 0$ then $u > 0$ so $(u,v) \in A$. So $(x,y)$ is an interior point.


*Prove if $x \not \in \in \mathbb R^+\times \mathbb R$ the is if $(x,y)$ is such that $x \le 0$ then it is not an interior point of $A$.

Well to that in cases.
2a) if $x \le 0$ and $y \ne 0$ then $(x,y) \not \in A$ so $(x,y)$ can not be an interior point.
The only remaining option is


*$y =0$ and $x \le 0$.  But the for any $d > 0$ then the point $(x,\frac 12 d) \in B_d(x,0)$ [ because $d((x,\frac 12 d),(x,0)) = \frac 12 d > 0$] but $(x,\frac 12 d)\not \in A$ because $x \le 0$ but $y \ne 0$.

....

But I have calculated the interior of each set seperately,

Okay.  That's cool.  You are assuming we can assume there is some relation between $Int(M\cup N)$ and $Int(M) \cup Int(N)$.
But is there such a theory?
(There probably is)
