Let(X, d) be a metric space. the set {y ∈ X : d(x, y) ≤ r} is a closed ball centered at X and with radius r.
(a)Show that a closed ball is a closed set.
Denote the closed ball centered at $\displaystyle x$ of radius $\displaystyle r$ as $\displaystyle B_r(X,x)$. Now suppose that $\displaystyle \xi$ was a limit point of $\displaystyle B_r(X,x)$ but not an element of $\displaystyle B_r(X,x)$. Then every open ball around $\displaystyle \xi$ would contain another point of $\displaystyle B_r(X,x)$ besides $\displaystyle \xi$. Therefore $\displaystyle d(x,\xi)\le r+\varepsilon\quad\forall\varepsilon>0$. Assume that $\displaystyle d(x,\xi)>r\implies d(x,\xi)-r>0$, then choosing $\displaystyle \varepsilon=d(x,\xi)-r$ would derive a contradiction. Therefore $\displaystyle d(x,\xi)\le r$ and the conclusion follows.