Consider $\mathbb{R}^2$ with the jungle river metric $d_j$.
$$d_j(x,y) = \begin{cases} |x_2-y_2|, & \text{if $x_1 = y_1$;} \\
|x_2| + |y_2| + |x_1-y_1|, & \text{if $x_1 \neq y_1 $} \end{cases}$$
Let
$$C = \{(x, 1) : x ∈ Q\}$$
What is the closure and the interior of C?
I think the closure is C as ${\overline{C}}$ = C ∪ {all limit points of C}. For any (x,1) in C we can find an "open ball" that only intersects C at (x,1) by taking B(x,ε) for ε<2. Does this mean C has no limit points and that ${\overline{C}}$=C?
For the interior I used the fact that if A ⊂ X then A$^◦$ = X \ ${\overline{(X \backslash A)}}$. I believe ${\overline{(\mathbb{R}^2 \backslash C)}}=\mathbb{R}^2$ which would make C$^◦$ = $∅$.
Best Answer
[Note: I realize this has been asked before this week, but as OP has given a specific thought process that is not duplicated elsewhere that I’m aware of, and that deserves addressing, I’m inclined to answer.]
While your conclusion is correct, your reasoning isn’t quite right for the closure of $C$.
The fact that you can take a ball around any point in $C$ and it only intersects $C$ in that one point doesn’t tell you that $C$ is closed, it tells you rather that $C$ is discrete. For example, the set $\{\frac{1}{n}\mid n\in \mathbb N\}\subset \mathbb R$ is discrete by this same argument, but is not closed, as it does not contain its limit point $0$.
Instead, you could observe that if $p\notin C$, then either $p=(x,y)$ for $y\neq 1$, whereby the open ball $B(p,|y-1|)$ does not intersect $C$, or $p=(x,1)$ for $x\notin \mathbb Q$, whereby the ball of radius $1$ does not intersect $C$. Therefore $\mathbb R^2\backslash C$ is open, hence $C$ is closed.
For the interior, what you have written is not wrong, but it somewhat begs the question of how you determined that $\overline{\mathbb R^2\backslash C}=\mathbb R^2$ in the first place. To conclude this, you could observe that $(x,1)=\lim_{n\to\infty} (x,1-\frac{1}{n})$.
Update.
On reflection, with a little care you could make your original argument for the closure work as well. The fact that every point $p\in C$ has some ball around it intersecting $C$ only at $p$, of the same radius independent of $p$, indicates a stronger condition than discreteness - what we might call $\varepsilon$-discreteness (here for $\varepsilon=2$). An $\varepsilon$-discrete metric space is complete (since every Cauchy sequence is eventually constant), hence any $\varepsilon$-discrete subset of a metric space must be closed.