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I'm having a bit of a struggle with the following proof.

Statement: Prove that $N((0,0);1)$ is an open set in $\mathbb{R}\times\mathbb{R}$ with metric $d((x_1,x_2),(y_1,y_2))=|x_1-y_1|+|x_2-y_2|$.

Attempt: Let $x\in N((0,0);1)$. We need $N(x;\epsilon)=\{(a_1,a_2):|x_1-a_1|+|x_2-a_2|<\epsilon \} \subseteq N((0,0);1)=\{(y_1,y_2):|y_1|+|y_2|<1 \}$. For this to occur, we must have $|x_1-a_1|+|x_2-a_2|<\epsilon\implies |a_1|+|a_2|<1$. Notice $|x_1-a_1|+|x_2-a_2|< |x_1|+|a_1|+|x_2|+|a_2|<\epsilon$. Pick $\epsilon=|x_1|+|x_2|+1$. This is greater than 0, for $0<|x_1|+|x_2|<1$. Then, $N(x;\epsilon)=\{(a_1,a_2):|a_1|+|a_2|<1 \}$, so $N((0,0);1)$ is open.

I feel as though it's not valid for me to have chosen $\epsilon$ as I did; could someone point me in the right direction? Thanks, exam in a few hours, so any and all help is appreciated.

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$|a_1|+|a_2| \leq |x_1-a_1|+|x_1-a_1|+|x_1|+|x_2| <\epsilon+|x_1|+|x_2|<1$ if $0<\epsilon <1-(|x_1|+|x_2|)$.

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