I have searched in math exchange, but the examples I have found does not help me to be sure that that my solution of the following exercise is correct. I am quite sure about my solution, but not 100% sure.

Let $\rho(s,t)$ the discrete metric on $\mathbb{R}^2$, let $x=(x_1,x_2), y=(y_1,y_2)$ and let define the following distance $$ d(x,y)=\rho(x_1,y_1)+|x_2-y_2|\,. $$ Goal: find the closure of $A:=\{x\in \mathbb{R^2}:0<x_1<1, 0<x_2<1\}$.

In my opinion, $\bar A:=\{x\in \mathbb{R^2}:0<x_1<1, 0\le x_2\le 1\}$. I get this solution by thinking that the closure of a set contains all its points plus its limit points. Thus I searched the limit points. We can see that every ball centered to any point of coordinate $(z,0)$ and $(z,1),0<z<1$ contains at least one point of $A$, so we can include them in the closure. On the other hand, every ball centered at any point with coordinate $x_1=0$ or $x_1=1$ does not contain any point of $A$ and therefore they don't belong to $\bar A$. In addition, those points do not even seem to be adherence point of $A$.

Finally, to test that $\bar A$ is actually closed , it is enough to check if its complement is open. Actually, this seems to be case for me.

Is my reasoning correct?


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