Metric Space, closed sets in a closed ball

hebby · #1 $Old$ November 19th, 2009, 04:19 PM

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.

Focus · #2 $Old$ November 19th, 2009, 05:43 PM

Have you tried any of the questions you posted? If so post your working out. A tip to get you started, pick a point outside of the ball and find and open ball that contains it.

Drexel28 · #3 $Old$ November 19th, 2009, 06:30 PM

Quote:

Originally Posted by hebby $View Post$

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 $x$ of radius $r$ as $B_r(X,x)$ . Now suppose that $\xi$ was a limit point of $B_r(X,x)$ but not an element of $B_r(X,x)$ . Then every open ball around $\xi$ would contain another point of $B_r(X,x)$ besides $\xi$ . Therefore $d(x,\xi)\le r+\varepsilon\quad\forall\varepsilon>0$ . Assume that $d(x,\xi)>r\implies d(x,\xi)-r>0$ , then choosing $\varepsilon=d(x,\xi)-r$ would derive a contradiction. Therefore $d(x,\xi)\le r$ and the conclusion follows.