[proofplan] Unfold the definition of membership in the open ball $B(x_0,r)$ to obtain the strict inequality $d(x,x_0)<r$, which immediately gives $\rho>0$. To prove the inclusion, take an arbitrary point $y \in B(x,\rho)$ and use the triangle inequality to compare $d(y,x_0)$ with $d(y,x)+d(x,x_0)$. The definition of $\rho$ then forces $d(y,x_0)<r$, so $y$ lies in $B(x_0,r)$. [/proofplan]

[step:Use the open ball condition to prove the radius $\rho$ is positive]Since $x \in B(x_0,r)$, the definition of the open ball in the [metric space](/page/Metric%20Space) $(X,d)$ gives \begin{align*} d(x,x_0)<r. \end{align*} By subtracting the real number $d(x,x_0)$ from both sides, we obtain \begin{align*} 0<r-d(x,x_0)=\rho. \end{align*} Thus $\rho>0$.[/step]

[guided]The first point is to translate the geometric phrase "$x$ lies inside the ball centered at $x_0$ with radius $r$" into the metric inequality it means. By definition of an open ball in a metric space, \begin{align*} B(x_0,r)=\{z\in X:d(z,x_0)<r\}. \end{align*} Since $x\in B(x_0,r)$, we have \begin{align*} d(x,x_0)<r. \end{align*} The number $\rho$ was defined by \begin{align*} \rho:=r-d(x,x_0). \end{align*} Because $d(x,x_0)<r$, subtracting $d(x,x_0)$ from both sides gives \begin{align*} 0<r-d(x,x_0)=\rho. \end{align*} Therefore $\rho>0$. This is exactly the amount of distance between $x$ and the boundary radius $r$ measured from the center $x_0$.[/guided]

[step:Show every point within distance $\rho$ of $x$ remains inside $B(x_0,r)$] Let $y\in B(x,\rho)$ be arbitrary. By the definition of the open ball centered at $x$, we have \begin{align*} d(y,x)<\rho. \end{align*} Using the triangle inequality in the metric space $(X,d)$ for the points $y,x,x_0\in X$, we get \begin{align*} d(y,x_0)\le d(y,x)+d(x,x_0). \end{align*} Since $d(y,x)<\rho=r-d(x,x_0)$, adding $d(x,x_0)$ to both sides gives \begin{align*} d(y,x)+d(x,x_0)<r. \end{align*} Combining the two inequalities yields \begin{align*} d(y,x_0)<r. \end{align*} Hence $y\in B(x_0,r)$. Since $y\in B(x,\rho)$ was arbitrary, we conclude \begin{align*} B(x,\rho)\subset B(x_0,r). \end{align*} [/step]