# Describing open neighborhoods

Let $d$ be a metric on $\mathbb{R}^2$ defined as $$d((x_1,y_1),(x_2,y_2))=\begin{cases} |y_1-y_2| \mbox{ if } x_1=x_2 \\ 1+|y_1-y_2| \mbox{ if } x_1 \neq x_2 \end{cases}$$.

Let $N((x,y),\epsilon)$ be an open neighborhood in $(\mathbb{R}^2, d)$. If $0< \epsilon \leq 1$ then we have two cases. If $x_1 = x_2$, then we get a line segment between $0<y\leq 1$. If $x_1\neq x_2$, then we get a box with $y \in (1,2]$ and $x_1$, $x_2$ on the x-axis obviously not equal.

I'm not sure if the above is correct. Nor am I sure of $\epsilon > 1$.

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You need to consider all points that lie in $B = B((x,y), \epsilon)$.
If $\epsilon \leq 1$, then if $(x',y') \in B$, you must have $x=x'$, as the distance is $\geq1$ whenever $x \neq x'$. Hence $B = \{x\}\times (y-\epsilon, y+\epsilon)$ in this case.
If $1 < \epsilon$, then if $(x',y') \in B$, there are two cases to consider, (1) $x'=x$, in which case $\{x\}\times (y-\epsilon, y+\epsilon) \subset B$, and (2) $x'\neq x$, in which case $\{x'\}\times (y-(\epsilon-1), y+(\epsilon-1)) \subset B$. Putting these together gives $B = \{x\}\times (y-\epsilon, y+\epsilon) \cup \mathbb{R} \times (y-(\epsilon-1), y+(\epsilon-1))$.
To describe the geometric shape, for $\epsilon = 1$, an $\epsilon$-ball around a point $(x, y)$ is a vertical line segment of length 2 centered around the point, together with a horizontal line going through the point and infinitely long. For smaller $\epsilon$, the horizontal line disappears, and the vertical line becomes shorter. For larger $\epsilon$, the vertical line segment grows longer, and the horisontal line becomes a horizontal band, $(\epsilon - 1)$ wide. copper.hat comes with a slightly different picture since he assumes the ball to be open, while I assume it to be closed. –  Arthur Oct 25 '12 at 7:33