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Consider the following subsets of the plane $\mathbb{R}^2$:

$$X=\{(x,y)|y=0\}\cup \{(x,y)|x>0\text{ and}\; y=1/x\}$$

How to show that $A$ and $B$ are open in $X$ under subspace topology. Efforts:

Let's define $A=\{(x,y)|y=0\}$ and $B=\{(x,y)|x>0\text{ and}\; y=1/x\}$.

To show that $A$ is open I need to find an open set $N$ of $\mathbb{R}^2$ such that $X\cap N=A$. I am not able to proceed further.

I welcome any hints.

Thanks for reading.

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Note that both $A$ and $B$ are closed in $\mathbb R^2$ (why?), and so $M:=\mathbb R^2\backslash B$ and $N:=\mathbb R^2\backslash A$ are open in $\mathbb R^2$. Consequently, since $X=A\cup B$ and $A\cap B=\emptyset$, we have $A=X\cap M$ and $B=X\cap N$, showing that both $A$ and $B$ are also open in $X$.

Note that this implies that $A$ and $B$ are also closed in $X$, since $A=X\backslash B$ and $B=X\backslash A$.

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Hint: It might be easier to show that both $A,B$ are closed in $X$, and then since $X = A \cup B$, we immediately have that $A,B$ are both open in $X$.

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