Is $(\mathbb{R}^2, d)$ connected?

Question

$\mathbb{R}^2$ with the metric:

$$d:\mathbb{R}^2\times \mathbb{R}^2\to \mathbb{R}$$ defined by $d(x,y)=0,$ if $(x_1,y_1)=(x_2,y_2)$ and $d(x,y)=|x_1|+|x_2|+|y_1-y_2|$ if $(x_1,y_1)\neq(x_2,y_2),$ they ask me to investigate if $\mathbb{R}^2$ with this metric is connected. I think it is connected, for this I want to show that it is connected by paths, given two points $(x_1,y_1)\neq(x_2,y_2),$ I considered the function

$f:[0,1]\to\mathbb{R}^2,$ defined by $f(t)=(x_1+t(x_2-x_1),y_1+t(y_2-y_1))$. but I have not been able to prove that it is continuous.

Is this space connected? Can someone help me please? Thank you.

Answer 1

$(\mathbb{R}^2,d)$ is not connected, as every $(x,y) \in \mathbb{R}^2$ such that $x \neq 0$ is an isolated point. To see this, let $r = \frac{|x|}{2}$. I shall show that: $$ B_r(x,y) = \{x\} $$ Let $(x',y') \in \mathbb{R}^2$, and suppose $(x',y') \neq (x,y)$. Then: $$ d((x',y'),(x,y')) = |x| + |x'| + |y' - y| \geq |x| > r $$ so $(x',y') \notin B_r(x,y)$.