If $d(x,B)\leq t_1+t_2$ then $d(x,B^{t_1})\leq t_2$ when $B^{t_1}:=\{x:d(x,B)\leq t_1\}$

Question

Let $(X,d)$ be a metric space and $t_1, t_2 > 0$ be fixed and define $B^t:=\{x: d(x, B)\leq t\}$ for any $t > 0, B\subset X$. Suppose then that $B\subset X$ and $x\in X$ are fixed such that $d(x, B)\leq t_1 + t_2$ when $d(x, B) := \inf\{y\in B:d(x, y)\}$.

I am stuck trying to show that now $d(x, B^{t_1})\leq t_2$. If $x\in B^{t_1}$, there is nothing to show so we may suppose that $x\not\in B^{t_1}$. By assumption we then have that $x\in B^{t_1+t_2}\setminus B^{t_1}$ i.e. $t_1<d(x,B)\leq t_1+t_2$. At this point I would very much like to conclude the claim, but I can't figure out the exact sequence of inequalities I have to use.

Answer 1

This is not true. Consider $X=\{0,1\}$ with the discrete metric (i.e $d(0,1)=1$), $B=\{0\}$, $x=1$, and $t_1=t_2=\frac{1}{2}$. Then $B^{t_1} = B$, $d(x,B) = 1 = t_1+t_2$, but $d(x,B^{t_1}) = 1 > t_2$.